Replace all NA values for variable with one row equal to 0
Slightly difficult to phrase, as far as I saw none of the similar questions answered my problem.
I have a data.frame such as:
df1 <- data.frame(id = rep(c("a", "b"), each = 4),
val = c(NA, NA, NA, NA, 1, 2, 2, 3))
df1
id val
1 a NA
2 a NA
3 a NA
4 a NA
5 b 1
6 b 2
7 b 2
8 b 3
and I want to get rid of all the NA values (easy enough using e.g. filter() ) but make sure that if this removes all of one id value (in this case it removes every instance of "a") that one extra row is inserted of (e.g.) a = 0
so that:
id val
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
obviously easy enough to do this in a roundabout way but I was wondering if there's a tidy/elegant way to do this. I thought tidyr::complete() might help but not entirely sure how to apply it to a case like this
I don't care about the order of the rows
Cheers!
r na complete
add a comment |
Slightly difficult to phrase, as far as I saw none of the similar questions answered my problem.
I have a data.frame such as:
df1 <- data.frame(id = rep(c("a", "b"), each = 4),
val = c(NA, NA, NA, NA, 1, 2, 2, 3))
df1
id val
1 a NA
2 a NA
3 a NA
4 a NA
5 b 1
6 b 2
7 b 2
8 b 3
and I want to get rid of all the NA values (easy enough using e.g. filter() ) but make sure that if this removes all of one id value (in this case it removes every instance of "a") that one extra row is inserted of (e.g.) a = 0
so that:
id val
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
obviously easy enough to do this in a roundabout way but I was wondering if there's a tidy/elegant way to do this. I thought tidyr::complete() might help but not entirely sure how to apply it to a case like this
I don't care about the order of the rows
Cheers!
r na complete
So you want to add rows with 0 only if all the values for particularid
is 0?
– Ronak Shah
3 hours ago
only if they're all NA for a particular id
– Robert Hickman
3 hours ago
1
@RobertHickman There seems to be some confusion about your desired output. Could you update your question with the expected output based on thisdf1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
? Thanks to @VivekKalyanarangan for the data.
– markus
2 hours ago
add a comment |
Slightly difficult to phrase, as far as I saw none of the similar questions answered my problem.
I have a data.frame such as:
df1 <- data.frame(id = rep(c("a", "b"), each = 4),
val = c(NA, NA, NA, NA, 1, 2, 2, 3))
df1
id val
1 a NA
2 a NA
3 a NA
4 a NA
5 b 1
6 b 2
7 b 2
8 b 3
and I want to get rid of all the NA values (easy enough using e.g. filter() ) but make sure that if this removes all of one id value (in this case it removes every instance of "a") that one extra row is inserted of (e.g.) a = 0
so that:
id val
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
obviously easy enough to do this in a roundabout way but I was wondering if there's a tidy/elegant way to do this. I thought tidyr::complete() might help but not entirely sure how to apply it to a case like this
I don't care about the order of the rows
Cheers!
r na complete
Slightly difficult to phrase, as far as I saw none of the similar questions answered my problem.
I have a data.frame such as:
df1 <- data.frame(id = rep(c("a", "b"), each = 4),
val = c(NA, NA, NA, NA, 1, 2, 2, 3))
df1
id val
1 a NA
2 a NA
3 a NA
4 a NA
5 b 1
6 b 2
7 b 2
8 b 3
and I want to get rid of all the NA values (easy enough using e.g. filter() ) but make sure that if this removes all of one id value (in this case it removes every instance of "a") that one extra row is inserted of (e.g.) a = 0
so that:
id val
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
obviously easy enough to do this in a roundabout way but I was wondering if there's a tidy/elegant way to do this. I thought tidyr::complete() might help but not entirely sure how to apply it to a case like this
I don't care about the order of the rows
Cheers!
r na complete
r na complete
asked 3 hours ago
Robert Hickman
15019
15019
So you want to add rows with 0 only if all the values for particularid
is 0?
– Ronak Shah
3 hours ago
only if they're all NA for a particular id
– Robert Hickman
3 hours ago
1
@RobertHickman There seems to be some confusion about your desired output. Could you update your question with the expected output based on thisdf1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
? Thanks to @VivekKalyanarangan for the data.
– markus
2 hours ago
add a comment |
So you want to add rows with 0 only if all the values for particularid
is 0?
– Ronak Shah
3 hours ago
only if they're all NA for a particular id
– Robert Hickman
3 hours ago
1
@RobertHickman There seems to be some confusion about your desired output. Could you update your question with the expected output based on thisdf1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
? Thanks to @VivekKalyanarangan for the data.
– markus
2 hours ago
So you want to add rows with 0 only if all the values for particular
id
is 0?– Ronak Shah
3 hours ago
So you want to add rows with 0 only if all the values for particular
id
is 0?– Ronak Shah
3 hours ago
only if they're all NA for a particular id
– Robert Hickman
3 hours ago
only if they're all NA for a particular id
– Robert Hickman
3 hours ago
1
1
@RobertHickman There seems to be some confusion about your desired output. Could you update your question with the expected output based on this
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
? Thanks to @VivekKalyanarangan for the data.– markus
2 hours ago
@RobertHickman There seems to be some confusion about your desired output. Could you update your question with the expected output based on this
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
? Thanks to @VivekKalyanarangan for the data.– markus
2 hours ago
add a comment |
8 Answers
8
active
oldest
votes
Another idea using dplyr
,
library(dplyr)
df1 %>%
group_by(id) %>%
mutate(val = ifelse(row_number() == 1 & all(is.na(val)), 0, val)) %>%
na.omit()
which gives,
# A tibble: 5 x 2
# Groups: id [2]
id val
<fct> <dbl>
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
1
(+1) Seems like the most robust answer here. Would be marginally more concise usingreplace(val, all(is.na(val)) * 1, 0)
instead of theifelse(...)
.
– Mikko Marttila
1 hour ago
@MikkoMarttila Good suggestion. I usually try and avoidifelse
in general
– Sotos
1 hour ago
add a comment |
We may do
df1 %>% group_by(id) %>% do(if(all(is.na(.$val))) replace(.[1, ], 2, 0) else na.omit(.))
# A tibble: 5 x 2
# Groups: id [2]
# id val
# <fct> <dbl>
# 1 a 0
# 2 b 1
# 3 b 2
# 4 b 2
# 5 b 3
After grouping by id
, if everything in val
is NA
, then we leave only the first row with the second element replaced by 0, otherwise the same data is returned after applying na.omit
.
In a more readable format that would be
df1 %>% group_by(id) %>%
do(if(all(is.na(.$val))) data.frame(id = .$id[1], val = 0) else na.omit(.))
(Here I presume that you indeed want to get rid of all NA
values; otherwise there is no need for na.omit
.)
1
@markus, right, I had assumed that that's the goal. Thanks!
– Julius Vainora
3 hours ago
It looks like op wants to retain the first row and replace theval
column of that row with0
where allval
isNA
for a group. Check my ans pls. Agree with @markus, it does seem tricky
– Vivek Kalyanarangan
2 hours ago
1
@VivekKalyanarangan, that's what I initially thought, but "and I want to get rid of all the NA values" suggests otherwise.
– Julius Vainora
2 hours ago
add a comment |
df1[is.na(df1)] <- 0
df1[!(duplicated(df1$id) & df1$val == 0), ]
id val
1 a 0
5 b 1
6 b 2
7 b 2
8 b 3
5
Would this work forid
s that containNA
s and non-NA
s? Try withdf1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
– markus
3 hours ago
I think this is the best so far (I'll leave it open for another hour or so to see) would maybe change to df %>% replace(is.na(.), 0) %>% .[!(duplicated(.$id) & .$val == 0), ]
– Robert Hickman
2 hours ago
add a comment |
Base R option is to find groups with all NA
s and transform
them by changing their val
to 0 and select only unique
rows so that there is only one row per group. We rbind
this dataframe with the groups which are !all_NA
.
all_NA <- with(df1, ave(is.na(val), id, FUN = all))
rbind(unique(transform(df1[all_NA, ], val = 0)), df1[!all_NA, ])
# id val
#1 a 0
#5 b 1
#6 b 2
#7 b 2
#8 b 3
dplyr
option looks ugly but one way is to make two groups of dataframes one with groups of all NA
values and other with groups of all non-NA values. For groups with all NA
values we add row with it's id
and val
as 0 and bind this to the other group.
library(dplyr)
bind_rows(df1 %>%
group_by(id) %>%
filter(all(!is.na(val))),
df1 %>%
group_by(id) %>%
filter(all(is.na(val))) %>%
ungroup() %>%
summarise(id = unique(id),
val = 0)) %>%
arrange(id)
# id val
# <fct> <dbl>
#1 a 0
#2 b 1
#3 b 2
#4 b 2
#5 b 3
add a comment |
Changed the df
to make example more exhaustive -
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4),
val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
library(dplyr)
df1 %>%
group_by(id) %>%
mutate(case=sum(is.na(val))==n(), row_num=row_number() ) %>%
mutate(val=ifelse(is.na(val)&case,0,val)) %>%
filter( !(case&row_num!=1) ) %>%
select(id, val)
Output
id val
<fct> <dbl>
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
6 c NA
7 c 2
8 c NA
9 c 3
add a comment |
Here is an option too:
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate_all(funs(replace(.,is.na(.),0))) %>%
slice(4:nrow(.))
This gives:
id val
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
Alternative:
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate_all(funs(replace(.,is.na(.),0))) %>%
unique()
3
where did 4 come from?
– Sotos
3 hours ago
The solution produces four 0s. We're only interested in having 1?
– NelsonGon
3 hours ago
What if one group has 4 and another 3?
– Sotos
3 hours ago
Sorry I only answered based on the question. Maybe then we could twist things up, not sure though!
– NelsonGon
2 hours ago
Consider this example -df1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
I think here OP wants to removeNA
values forA
group only, not the rest
– Vivek Kalyanarangan
2 hours ago
|
show 2 more comments
Here is a base R solution.
res <- lapply(split(df1, df1$id), function(DF){
if(anyNA(DF$val)) {
i <- is.na(DF$val)
DF$val[i] <- 0
DF <- rbind(DF[i & !duplicated(DF[i, ]), ], DF[!i, ])
}
DF
})
res <- do.call(rbind, res)
row.names(res) <- NULL
res
# id val
#1 a 0
#2 b 1
#3 b 2
#4 b 2
#5 b 3
Edit.
A dplyr
solution could be the following.
It was tested with the original dataset posted by the OP, with the dataset in Vivek Kalyanarangan's answer and with the dataset in markus' comment, renamed df2
and df3
, respectively.
library(dplyr)
na2zero <- function(DF){
DF %>%
group_by(id) %>%
mutate(val = ifelse(is.na(val), 0, val),
crit = val == 0 & duplicated(val)) %>%
filter(!crit) %>%
select(-crit)
}
na2zero(df1)
na2zero(df2)
na2zero(df3)
Rui, try withdf1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
. Unfortunately your solution doesn't return a data frame with only three rows.
– markus
3 hours ago
@markus No, it doesn't. TheNA
is replaced by a0
and the other value ofval
is notNA
so both must be in the output. At least that's how I'm understanding the OP's problem.
– Rui Barradas
2 hours ago
Fair enough. People are reading the question differently.
– markus
1 hour ago
add a comment |
Another base approach, one that doesn't maintain the order of the rows and takes advantage of factors remembering lost values:
df1 <- na.omit(df1)
df1 <- rbind(
df1,
data.frame(
id = levels(df1$id)[!levels(df1$id) %in% df1$id],
val = 0)
)
I do personally prefer the dplyr approach given by Sotos, as I don't like rbind
-ing data.frames back together so it's a matter of taste, but this isn't unbearably complicated by my eye. It's easy enough to adapt to a character id
column with a unique(df1$id)
variable.
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
Another idea using dplyr
,
library(dplyr)
df1 %>%
group_by(id) %>%
mutate(val = ifelse(row_number() == 1 & all(is.na(val)), 0, val)) %>%
na.omit()
which gives,
# A tibble: 5 x 2
# Groups: id [2]
id val
<fct> <dbl>
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
1
(+1) Seems like the most robust answer here. Would be marginally more concise usingreplace(val, all(is.na(val)) * 1, 0)
instead of theifelse(...)
.
– Mikko Marttila
1 hour ago
@MikkoMarttila Good suggestion. I usually try and avoidifelse
in general
– Sotos
1 hour ago
add a comment |
Another idea using dplyr
,
library(dplyr)
df1 %>%
group_by(id) %>%
mutate(val = ifelse(row_number() == 1 & all(is.na(val)), 0, val)) %>%
na.omit()
which gives,
# A tibble: 5 x 2
# Groups: id [2]
id val
<fct> <dbl>
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
1
(+1) Seems like the most robust answer here. Would be marginally more concise usingreplace(val, all(is.na(val)) * 1, 0)
instead of theifelse(...)
.
– Mikko Marttila
1 hour ago
@MikkoMarttila Good suggestion. I usually try and avoidifelse
in general
– Sotos
1 hour ago
add a comment |
Another idea using dplyr
,
library(dplyr)
df1 %>%
group_by(id) %>%
mutate(val = ifelse(row_number() == 1 & all(is.na(val)), 0, val)) %>%
na.omit()
which gives,
# A tibble: 5 x 2
# Groups: id [2]
id val
<fct> <dbl>
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
Another idea using dplyr
,
library(dplyr)
df1 %>%
group_by(id) %>%
mutate(val = ifelse(row_number() == 1 & all(is.na(val)), 0, val)) %>%
na.omit()
which gives,
# A tibble: 5 x 2
# Groups: id [2]
id val
<fct> <dbl>
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
answered 2 hours ago
Sotos
28.2k51640
28.2k51640
1
(+1) Seems like the most robust answer here. Would be marginally more concise usingreplace(val, all(is.na(val)) * 1, 0)
instead of theifelse(...)
.
– Mikko Marttila
1 hour ago
@MikkoMarttila Good suggestion. I usually try and avoidifelse
in general
– Sotos
1 hour ago
add a comment |
1
(+1) Seems like the most robust answer here. Would be marginally more concise usingreplace(val, all(is.na(val)) * 1, 0)
instead of theifelse(...)
.
– Mikko Marttila
1 hour ago
@MikkoMarttila Good suggestion. I usually try and avoidifelse
in general
– Sotos
1 hour ago
1
1
(+1) Seems like the most robust answer here. Would be marginally more concise using
replace(val, all(is.na(val)) * 1, 0)
instead of the ifelse(...)
.– Mikko Marttila
1 hour ago
(+1) Seems like the most robust answer here. Would be marginally more concise using
replace(val, all(is.na(val)) * 1, 0)
instead of the ifelse(...)
.– Mikko Marttila
1 hour ago
@MikkoMarttila Good suggestion. I usually try and avoid
ifelse
in general– Sotos
1 hour ago
@MikkoMarttila Good suggestion. I usually try and avoid
ifelse
in general– Sotos
1 hour ago
add a comment |
We may do
df1 %>% group_by(id) %>% do(if(all(is.na(.$val))) replace(.[1, ], 2, 0) else na.omit(.))
# A tibble: 5 x 2
# Groups: id [2]
# id val
# <fct> <dbl>
# 1 a 0
# 2 b 1
# 3 b 2
# 4 b 2
# 5 b 3
After grouping by id
, if everything in val
is NA
, then we leave only the first row with the second element replaced by 0, otherwise the same data is returned after applying na.omit
.
In a more readable format that would be
df1 %>% group_by(id) %>%
do(if(all(is.na(.$val))) data.frame(id = .$id[1], val = 0) else na.omit(.))
(Here I presume that you indeed want to get rid of all NA
values; otherwise there is no need for na.omit
.)
1
@markus, right, I had assumed that that's the goal. Thanks!
– Julius Vainora
3 hours ago
It looks like op wants to retain the first row and replace theval
column of that row with0
where allval
isNA
for a group. Check my ans pls. Agree with @markus, it does seem tricky
– Vivek Kalyanarangan
2 hours ago
1
@VivekKalyanarangan, that's what I initially thought, but "and I want to get rid of all the NA values" suggests otherwise.
– Julius Vainora
2 hours ago
add a comment |
We may do
df1 %>% group_by(id) %>% do(if(all(is.na(.$val))) replace(.[1, ], 2, 0) else na.omit(.))
# A tibble: 5 x 2
# Groups: id [2]
# id val
# <fct> <dbl>
# 1 a 0
# 2 b 1
# 3 b 2
# 4 b 2
# 5 b 3
After grouping by id
, if everything in val
is NA
, then we leave only the first row with the second element replaced by 0, otherwise the same data is returned after applying na.omit
.
In a more readable format that would be
df1 %>% group_by(id) %>%
do(if(all(is.na(.$val))) data.frame(id = .$id[1], val = 0) else na.omit(.))
(Here I presume that you indeed want to get rid of all NA
values; otherwise there is no need for na.omit
.)
1
@markus, right, I had assumed that that's the goal. Thanks!
– Julius Vainora
3 hours ago
It looks like op wants to retain the first row and replace theval
column of that row with0
where allval
isNA
for a group. Check my ans pls. Agree with @markus, it does seem tricky
– Vivek Kalyanarangan
2 hours ago
1
@VivekKalyanarangan, that's what I initially thought, but "and I want to get rid of all the NA values" suggests otherwise.
– Julius Vainora
2 hours ago
add a comment |
We may do
df1 %>% group_by(id) %>% do(if(all(is.na(.$val))) replace(.[1, ], 2, 0) else na.omit(.))
# A tibble: 5 x 2
# Groups: id [2]
# id val
# <fct> <dbl>
# 1 a 0
# 2 b 1
# 3 b 2
# 4 b 2
# 5 b 3
After grouping by id
, if everything in val
is NA
, then we leave only the first row with the second element replaced by 0, otherwise the same data is returned after applying na.omit
.
In a more readable format that would be
df1 %>% group_by(id) %>%
do(if(all(is.na(.$val))) data.frame(id = .$id[1], val = 0) else na.omit(.))
(Here I presume that you indeed want to get rid of all NA
values; otherwise there is no need for na.omit
.)
We may do
df1 %>% group_by(id) %>% do(if(all(is.na(.$val))) replace(.[1, ], 2, 0) else na.omit(.))
# A tibble: 5 x 2
# Groups: id [2]
# id val
# <fct> <dbl>
# 1 a 0
# 2 b 1
# 3 b 2
# 4 b 2
# 5 b 3
After grouping by id
, if everything in val
is NA
, then we leave only the first row with the second element replaced by 0, otherwise the same data is returned after applying na.omit
.
In a more readable format that would be
df1 %>% group_by(id) %>%
do(if(all(is.na(.$val))) data.frame(id = .$id[1], val = 0) else na.omit(.))
(Here I presume that you indeed want to get rid of all NA
values; otherwise there is no need for na.omit
.)
edited 2 hours ago
answered 3 hours ago
Julius Vainora
32.6k75979
32.6k75979
1
@markus, right, I had assumed that that's the goal. Thanks!
– Julius Vainora
3 hours ago
It looks like op wants to retain the first row and replace theval
column of that row with0
where allval
isNA
for a group. Check my ans pls. Agree with @markus, it does seem tricky
– Vivek Kalyanarangan
2 hours ago
1
@VivekKalyanarangan, that's what I initially thought, but "and I want to get rid of all the NA values" suggests otherwise.
– Julius Vainora
2 hours ago
add a comment |
1
@markus, right, I had assumed that that's the goal. Thanks!
– Julius Vainora
3 hours ago
It looks like op wants to retain the first row and replace theval
column of that row with0
where allval
isNA
for a group. Check my ans pls. Agree with @markus, it does seem tricky
– Vivek Kalyanarangan
2 hours ago
1
@VivekKalyanarangan, that's what I initially thought, but "and I want to get rid of all the NA values" suggests otherwise.
– Julius Vainora
2 hours ago
1
1
@markus, right, I had assumed that that's the goal. Thanks!
– Julius Vainora
3 hours ago
@markus, right, I had assumed that that's the goal. Thanks!
– Julius Vainora
3 hours ago
It looks like op wants to retain the first row and replace the
val
column of that row with 0
where all val
is NA
for a group. Check my ans pls. Agree with @markus, it does seem tricky– Vivek Kalyanarangan
2 hours ago
It looks like op wants to retain the first row and replace the
val
column of that row with 0
where all val
is NA
for a group. Check my ans pls. Agree with @markus, it does seem tricky– Vivek Kalyanarangan
2 hours ago
1
1
@VivekKalyanarangan, that's what I initially thought, but "and I want to get rid of all the NA values" suggests otherwise.
– Julius Vainora
2 hours ago
@VivekKalyanarangan, that's what I initially thought, but "and I want to get rid of all the NA values" suggests otherwise.
– Julius Vainora
2 hours ago
add a comment |
df1[is.na(df1)] <- 0
df1[!(duplicated(df1$id) & df1$val == 0), ]
id val
1 a 0
5 b 1
6 b 2
7 b 2
8 b 3
5
Would this work forid
s that containNA
s and non-NA
s? Try withdf1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
– markus
3 hours ago
I think this is the best so far (I'll leave it open for another hour or so to see) would maybe change to df %>% replace(is.na(.), 0) %>% .[!(duplicated(.$id) & .$val == 0), ]
– Robert Hickman
2 hours ago
add a comment |
df1[is.na(df1)] <- 0
df1[!(duplicated(df1$id) & df1$val == 0), ]
id val
1 a 0
5 b 1
6 b 2
7 b 2
8 b 3
5
Would this work forid
s that containNA
s and non-NA
s? Try withdf1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
– markus
3 hours ago
I think this is the best so far (I'll leave it open for another hour or so to see) would maybe change to df %>% replace(is.na(.), 0) %>% .[!(duplicated(.$id) & .$val == 0), ]
– Robert Hickman
2 hours ago
add a comment |
df1[is.na(df1)] <- 0
df1[!(duplicated(df1$id) & df1$val == 0), ]
id val
1 a 0
5 b 1
6 b 2
7 b 2
8 b 3
df1[is.na(df1)] <- 0
df1[!(duplicated(df1$id) & df1$val == 0), ]
id val
1 a 0
5 b 1
6 b 2
7 b 2
8 b 3
answered 3 hours ago
Adamm
832517
832517
5
Would this work forid
s that containNA
s and non-NA
s? Try withdf1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
– markus
3 hours ago
I think this is the best so far (I'll leave it open for another hour or so to see) would maybe change to df %>% replace(is.na(.), 0) %>% .[!(duplicated(.$id) & .$val == 0), ]
– Robert Hickman
2 hours ago
add a comment |
5
Would this work forid
s that containNA
s and non-NA
s? Try withdf1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
– markus
3 hours ago
I think this is the best so far (I'll leave it open for another hour or so to see) would maybe change to df %>% replace(is.na(.), 0) %>% .[!(duplicated(.$id) & .$val == 0), ]
– Robert Hickman
2 hours ago
5
5
Would this work for
id
s that contain NA
s and non-NA
s? Try with df1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
– markus
3 hours ago
Would this work for
id
s that contain NA
s and non-NA
s? Try with df1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
– markus
3 hours ago
I think this is the best so far (I'll leave it open for another hour or so to see) would maybe change to df %>% replace(is.na(.), 0) %>% .[!(duplicated(.$id) & .$val == 0), ]
– Robert Hickman
2 hours ago
I think this is the best so far (I'll leave it open for another hour or so to see) would maybe change to df %>% replace(is.na(.), 0) %>% .[!(duplicated(.$id) & .$val == 0), ]
– Robert Hickman
2 hours ago
add a comment |
Base R option is to find groups with all NA
s and transform
them by changing their val
to 0 and select only unique
rows so that there is only one row per group. We rbind
this dataframe with the groups which are !all_NA
.
all_NA <- with(df1, ave(is.na(val), id, FUN = all))
rbind(unique(transform(df1[all_NA, ], val = 0)), df1[!all_NA, ])
# id val
#1 a 0
#5 b 1
#6 b 2
#7 b 2
#8 b 3
dplyr
option looks ugly but one way is to make two groups of dataframes one with groups of all NA
values and other with groups of all non-NA values. For groups with all NA
values we add row with it's id
and val
as 0 and bind this to the other group.
library(dplyr)
bind_rows(df1 %>%
group_by(id) %>%
filter(all(!is.na(val))),
df1 %>%
group_by(id) %>%
filter(all(is.na(val))) %>%
ungroup() %>%
summarise(id = unique(id),
val = 0)) %>%
arrange(id)
# id val
# <fct> <dbl>
#1 a 0
#2 b 1
#3 b 2
#4 b 2
#5 b 3
add a comment |
Base R option is to find groups with all NA
s and transform
them by changing their val
to 0 and select only unique
rows so that there is only one row per group. We rbind
this dataframe with the groups which are !all_NA
.
all_NA <- with(df1, ave(is.na(val), id, FUN = all))
rbind(unique(transform(df1[all_NA, ], val = 0)), df1[!all_NA, ])
# id val
#1 a 0
#5 b 1
#6 b 2
#7 b 2
#8 b 3
dplyr
option looks ugly but one way is to make two groups of dataframes one with groups of all NA
values and other with groups of all non-NA values. For groups with all NA
values we add row with it's id
and val
as 0 and bind this to the other group.
library(dplyr)
bind_rows(df1 %>%
group_by(id) %>%
filter(all(!is.na(val))),
df1 %>%
group_by(id) %>%
filter(all(is.na(val))) %>%
ungroup() %>%
summarise(id = unique(id),
val = 0)) %>%
arrange(id)
# id val
# <fct> <dbl>
#1 a 0
#2 b 1
#3 b 2
#4 b 2
#5 b 3
add a comment |
Base R option is to find groups with all NA
s and transform
them by changing their val
to 0 and select only unique
rows so that there is only one row per group. We rbind
this dataframe with the groups which are !all_NA
.
all_NA <- with(df1, ave(is.na(val), id, FUN = all))
rbind(unique(transform(df1[all_NA, ], val = 0)), df1[!all_NA, ])
# id val
#1 a 0
#5 b 1
#6 b 2
#7 b 2
#8 b 3
dplyr
option looks ugly but one way is to make two groups of dataframes one with groups of all NA
values and other with groups of all non-NA values. For groups with all NA
values we add row with it's id
and val
as 0 and bind this to the other group.
library(dplyr)
bind_rows(df1 %>%
group_by(id) %>%
filter(all(!is.na(val))),
df1 %>%
group_by(id) %>%
filter(all(is.na(val))) %>%
ungroup() %>%
summarise(id = unique(id),
val = 0)) %>%
arrange(id)
# id val
# <fct> <dbl>
#1 a 0
#2 b 1
#3 b 2
#4 b 2
#5 b 3
Base R option is to find groups with all NA
s and transform
them by changing their val
to 0 and select only unique
rows so that there is only one row per group. We rbind
this dataframe with the groups which are !all_NA
.
all_NA <- with(df1, ave(is.na(val), id, FUN = all))
rbind(unique(transform(df1[all_NA, ], val = 0)), df1[!all_NA, ])
# id val
#1 a 0
#5 b 1
#6 b 2
#7 b 2
#8 b 3
dplyr
option looks ugly but one way is to make two groups of dataframes one with groups of all NA
values and other with groups of all non-NA values. For groups with all NA
values we add row with it's id
and val
as 0 and bind this to the other group.
library(dplyr)
bind_rows(df1 %>%
group_by(id) %>%
filter(all(!is.na(val))),
df1 %>%
group_by(id) %>%
filter(all(is.na(val))) %>%
ungroup() %>%
summarise(id = unique(id),
val = 0)) %>%
arrange(id)
# id val
# <fct> <dbl>
#1 a 0
#2 b 1
#3 b 2
#4 b 2
#5 b 3
edited 3 hours ago
answered 3 hours ago
Ronak Shah
32.6k103753
32.6k103753
add a comment |
add a comment |
Changed the df
to make example more exhaustive -
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4),
val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
library(dplyr)
df1 %>%
group_by(id) %>%
mutate(case=sum(is.na(val))==n(), row_num=row_number() ) %>%
mutate(val=ifelse(is.na(val)&case,0,val)) %>%
filter( !(case&row_num!=1) ) %>%
select(id, val)
Output
id val
<fct> <dbl>
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
6 c NA
7 c 2
8 c NA
9 c 3
add a comment |
Changed the df
to make example more exhaustive -
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4),
val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
library(dplyr)
df1 %>%
group_by(id) %>%
mutate(case=sum(is.na(val))==n(), row_num=row_number() ) %>%
mutate(val=ifelse(is.na(val)&case,0,val)) %>%
filter( !(case&row_num!=1) ) %>%
select(id, val)
Output
id val
<fct> <dbl>
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
6 c NA
7 c 2
8 c NA
9 c 3
add a comment |
Changed the df
to make example more exhaustive -
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4),
val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
library(dplyr)
df1 %>%
group_by(id) %>%
mutate(case=sum(is.na(val))==n(), row_num=row_number() ) %>%
mutate(val=ifelse(is.na(val)&case,0,val)) %>%
filter( !(case&row_num!=1) ) %>%
select(id, val)
Output
id val
<fct> <dbl>
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
6 c NA
7 c 2
8 c NA
9 c 3
Changed the df
to make example more exhaustive -
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4),
val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
library(dplyr)
df1 %>%
group_by(id) %>%
mutate(case=sum(is.na(val))==n(), row_num=row_number() ) %>%
mutate(val=ifelse(is.na(val)&case,0,val)) %>%
filter( !(case&row_num!=1) ) %>%
select(id, val)
Output
id val
<fct> <dbl>
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
6 c NA
7 c 2
8 c NA
9 c 3
answered 3 hours ago
Vivek Kalyanarangan
4,8911827
4,8911827
add a comment |
add a comment |
Here is an option too:
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate_all(funs(replace(.,is.na(.),0))) %>%
slice(4:nrow(.))
This gives:
id val
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
Alternative:
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate_all(funs(replace(.,is.na(.),0))) %>%
unique()
3
where did 4 come from?
– Sotos
3 hours ago
The solution produces four 0s. We're only interested in having 1?
– NelsonGon
3 hours ago
What if one group has 4 and another 3?
– Sotos
3 hours ago
Sorry I only answered based on the question. Maybe then we could twist things up, not sure though!
– NelsonGon
2 hours ago
Consider this example -df1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
I think here OP wants to removeNA
values forA
group only, not the rest
– Vivek Kalyanarangan
2 hours ago
|
show 2 more comments
Here is an option too:
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate_all(funs(replace(.,is.na(.),0))) %>%
slice(4:nrow(.))
This gives:
id val
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
Alternative:
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate_all(funs(replace(.,is.na(.),0))) %>%
unique()
3
where did 4 come from?
– Sotos
3 hours ago
The solution produces four 0s. We're only interested in having 1?
– NelsonGon
3 hours ago
What if one group has 4 and another 3?
– Sotos
3 hours ago
Sorry I only answered based on the question. Maybe then we could twist things up, not sure though!
– NelsonGon
2 hours ago
Consider this example -df1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
I think here OP wants to removeNA
values forA
group only, not the rest
– Vivek Kalyanarangan
2 hours ago
|
show 2 more comments
Here is an option too:
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate_all(funs(replace(.,is.na(.),0))) %>%
slice(4:nrow(.))
This gives:
id val
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
Alternative:
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate_all(funs(replace(.,is.na(.),0))) %>%
unique()
Here is an option too:
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate_all(funs(replace(.,is.na(.),0))) %>%
slice(4:nrow(.))
This gives:
id val
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
Alternative:
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate_all(funs(replace(.,is.na(.),0))) %>%
unique()
edited 2 hours ago
answered 3 hours ago
NelsonGon
815217
815217
3
where did 4 come from?
– Sotos
3 hours ago
The solution produces four 0s. We're only interested in having 1?
– NelsonGon
3 hours ago
What if one group has 4 and another 3?
– Sotos
3 hours ago
Sorry I only answered based on the question. Maybe then we could twist things up, not sure though!
– NelsonGon
2 hours ago
Consider this example -df1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
I think here OP wants to removeNA
values forA
group only, not the rest
– Vivek Kalyanarangan
2 hours ago
|
show 2 more comments
3
where did 4 come from?
– Sotos
3 hours ago
The solution produces four 0s. We're only interested in having 1?
– NelsonGon
3 hours ago
What if one group has 4 and another 3?
– Sotos
3 hours ago
Sorry I only answered based on the question. Maybe then we could twist things up, not sure though!
– NelsonGon
2 hours ago
Consider this example -df1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
I think here OP wants to removeNA
values forA
group only, not the rest
– Vivek Kalyanarangan
2 hours ago
3
3
where did 4 come from?
– Sotos
3 hours ago
where did 4 come from?
– Sotos
3 hours ago
The solution produces four 0s. We're only interested in having 1?
– NelsonGon
3 hours ago
The solution produces four 0s. We're only interested in having 1?
– NelsonGon
3 hours ago
What if one group has 4 and another 3?
– Sotos
3 hours ago
What if one group has 4 and another 3?
– Sotos
3 hours ago
Sorry I only answered based on the question. Maybe then we could twist things up, not sure though!
– NelsonGon
2 hours ago
Sorry I only answered based on the question. Maybe then we could twist things up, not sure though!
– NelsonGon
2 hours ago
Consider this example -
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
I think here OP wants to remove NA
values for A
group only, not the rest– Vivek Kalyanarangan
2 hours ago
Consider this example -
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
I think here OP wants to remove NA
values for A
group only, not the rest– Vivek Kalyanarangan
2 hours ago
|
show 2 more comments
Here is a base R solution.
res <- lapply(split(df1, df1$id), function(DF){
if(anyNA(DF$val)) {
i <- is.na(DF$val)
DF$val[i] <- 0
DF <- rbind(DF[i & !duplicated(DF[i, ]), ], DF[!i, ])
}
DF
})
res <- do.call(rbind, res)
row.names(res) <- NULL
res
# id val
#1 a 0
#2 b 1
#3 b 2
#4 b 2
#5 b 3
Edit.
A dplyr
solution could be the following.
It was tested with the original dataset posted by the OP, with the dataset in Vivek Kalyanarangan's answer and with the dataset in markus' comment, renamed df2
and df3
, respectively.
library(dplyr)
na2zero <- function(DF){
DF %>%
group_by(id) %>%
mutate(val = ifelse(is.na(val), 0, val),
crit = val == 0 & duplicated(val)) %>%
filter(!crit) %>%
select(-crit)
}
na2zero(df1)
na2zero(df2)
na2zero(df3)
Rui, try withdf1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
. Unfortunately your solution doesn't return a data frame with only three rows.
– markus
3 hours ago
@markus No, it doesn't. TheNA
is replaced by a0
and the other value ofval
is notNA
so both must be in the output. At least that's how I'm understanding the OP's problem.
– Rui Barradas
2 hours ago
Fair enough. People are reading the question differently.
– markus
1 hour ago
add a comment |
Here is a base R solution.
res <- lapply(split(df1, df1$id), function(DF){
if(anyNA(DF$val)) {
i <- is.na(DF$val)
DF$val[i] <- 0
DF <- rbind(DF[i & !duplicated(DF[i, ]), ], DF[!i, ])
}
DF
})
res <- do.call(rbind, res)
row.names(res) <- NULL
res
# id val
#1 a 0
#2 b 1
#3 b 2
#4 b 2
#5 b 3
Edit.
A dplyr
solution could be the following.
It was tested with the original dataset posted by the OP, with the dataset in Vivek Kalyanarangan's answer and with the dataset in markus' comment, renamed df2
and df3
, respectively.
library(dplyr)
na2zero <- function(DF){
DF %>%
group_by(id) %>%
mutate(val = ifelse(is.na(val), 0, val),
crit = val == 0 & duplicated(val)) %>%
filter(!crit) %>%
select(-crit)
}
na2zero(df1)
na2zero(df2)
na2zero(df3)
Rui, try withdf1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
. Unfortunately your solution doesn't return a data frame with only three rows.
– markus
3 hours ago
@markus No, it doesn't. TheNA
is replaced by a0
and the other value ofval
is notNA
so both must be in the output. At least that's how I'm understanding the OP's problem.
– Rui Barradas
2 hours ago
Fair enough. People are reading the question differently.
– markus
1 hour ago
add a comment |
Here is a base R solution.
res <- lapply(split(df1, df1$id), function(DF){
if(anyNA(DF$val)) {
i <- is.na(DF$val)
DF$val[i] <- 0
DF <- rbind(DF[i & !duplicated(DF[i, ]), ], DF[!i, ])
}
DF
})
res <- do.call(rbind, res)
row.names(res) <- NULL
res
# id val
#1 a 0
#2 b 1
#3 b 2
#4 b 2
#5 b 3
Edit.
A dplyr
solution could be the following.
It was tested with the original dataset posted by the OP, with the dataset in Vivek Kalyanarangan's answer and with the dataset in markus' comment, renamed df2
and df3
, respectively.
library(dplyr)
na2zero <- function(DF){
DF %>%
group_by(id) %>%
mutate(val = ifelse(is.na(val), 0, val),
crit = val == 0 & duplicated(val)) %>%
filter(!crit) %>%
select(-crit)
}
na2zero(df1)
na2zero(df2)
na2zero(df3)
Here is a base R solution.
res <- lapply(split(df1, df1$id), function(DF){
if(anyNA(DF$val)) {
i <- is.na(DF$val)
DF$val[i] <- 0
DF <- rbind(DF[i & !duplicated(DF[i, ]), ], DF[!i, ])
}
DF
})
res <- do.call(rbind, res)
row.names(res) <- NULL
res
# id val
#1 a 0
#2 b 1
#3 b 2
#4 b 2
#5 b 3
Edit.
A dplyr
solution could be the following.
It was tested with the original dataset posted by the OP, with the dataset in Vivek Kalyanarangan's answer and with the dataset in markus' comment, renamed df2
and df3
, respectively.
library(dplyr)
na2zero <- function(DF){
DF %>%
group_by(id) %>%
mutate(val = ifelse(is.na(val), 0, val),
crit = val == 0 & duplicated(val)) %>%
filter(!crit) %>%
select(-crit)
}
na2zero(df1)
na2zero(df2)
na2zero(df3)
edited 2 hours ago
answered 3 hours ago
Rui Barradas
16.1k41730
16.1k41730
Rui, try withdf1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
. Unfortunately your solution doesn't return a data frame with only three rows.
– markus
3 hours ago
@markus No, it doesn't. TheNA
is replaced by a0
and the other value ofval
is notNA
so both must be in the output. At least that's how I'm understanding the OP's problem.
– Rui Barradas
2 hours ago
Fair enough. People are reading the question differently.
– markus
1 hour ago
add a comment |
Rui, try withdf1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
. Unfortunately your solution doesn't return a data frame with only three rows.
– markus
3 hours ago
@markus No, it doesn't. TheNA
is replaced by a0
and the other value ofval
is notNA
so both must be in the output. At least that's how I'm understanding the OP's problem.
– Rui Barradas
2 hours ago
Fair enough. People are reading the question differently.
– markus
1 hour ago
Rui, try with
df1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
. Unfortunately your solution doesn't return a data frame with only three rows.– markus
3 hours ago
Rui, try with
df1 <- data.frame(id = rep(c("a", "b"), each = 2), val = c(NA, 1, 2, 3))
. Unfortunately your solution doesn't return a data frame with only three rows.– markus
3 hours ago
@markus No, it doesn't. The
NA
is replaced by a 0
and the other value of val
is not NA
so both must be in the output. At least that's how I'm understanding the OP's problem.– Rui Barradas
2 hours ago
@markus No, it doesn't. The
NA
is replaced by a 0
and the other value of val
is not NA
so both must be in the output. At least that's how I'm understanding the OP's problem.– Rui Barradas
2 hours ago
Fair enough. People are reading the question differently.
– markus
1 hour ago
Fair enough. People are reading the question differently.
– markus
1 hour ago
add a comment |
Another base approach, one that doesn't maintain the order of the rows and takes advantage of factors remembering lost values:
df1 <- na.omit(df1)
df1 <- rbind(
df1,
data.frame(
id = levels(df1$id)[!levels(df1$id) %in% df1$id],
val = 0)
)
I do personally prefer the dplyr approach given by Sotos, as I don't like rbind
-ing data.frames back together so it's a matter of taste, but this isn't unbearably complicated by my eye. It's easy enough to adapt to a character id
column with a unique(df1$id)
variable.
add a comment |
Another base approach, one that doesn't maintain the order of the rows and takes advantage of factors remembering lost values:
df1 <- na.omit(df1)
df1 <- rbind(
df1,
data.frame(
id = levels(df1$id)[!levels(df1$id) %in% df1$id],
val = 0)
)
I do personally prefer the dplyr approach given by Sotos, as I don't like rbind
-ing data.frames back together so it's a matter of taste, but this isn't unbearably complicated by my eye. It's easy enough to adapt to a character id
column with a unique(df1$id)
variable.
add a comment |
Another base approach, one that doesn't maintain the order of the rows and takes advantage of factors remembering lost values:
df1 <- na.omit(df1)
df1 <- rbind(
df1,
data.frame(
id = levels(df1$id)[!levels(df1$id) %in% df1$id],
val = 0)
)
I do personally prefer the dplyr approach given by Sotos, as I don't like rbind
-ing data.frames back together so it's a matter of taste, but this isn't unbearably complicated by my eye. It's easy enough to adapt to a character id
column with a unique(df1$id)
variable.
Another base approach, one that doesn't maintain the order of the rows and takes advantage of factors remembering lost values:
df1 <- na.omit(df1)
df1 <- rbind(
df1,
data.frame(
id = levels(df1$id)[!levels(df1$id) %in% df1$id],
val = 0)
)
I do personally prefer the dplyr approach given by Sotos, as I don't like rbind
-ing data.frames back together so it's a matter of taste, but this isn't unbearably complicated by my eye. It's easy enough to adapt to a character id
column with a unique(df1$id)
variable.
answered 16 mins ago
CriminallyVulgar
1
1
add a comment |
add a comment |
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So you want to add rows with 0 only if all the values for particular
id
is 0?– Ronak Shah
3 hours ago
only if they're all NA for a particular id
– Robert Hickman
3 hours ago
1
@RobertHickman There seems to be some confusion about your desired output. Could you update your question with the expected output based on this
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
? Thanks to @VivekKalyanarangan for the data.– markus
2 hours ago