How can I split date column into date and time in R?
I have this DATE column with Date (20150601) + Hour (02) and so on till the next year. I need to split this one column in two corresponding columns (Date and Hour) in R. As I understand I need stringr package.
What I have
DATE
- 2015060102
- 2015060104
- 2015060106
- 2015060109
- 2015060110
rstudio
asked Nov 21 '18 at 15:47
Yerzhan
11
add a comment |
I have this DATE column with Date (20150601) + Hour (02) and so on till the next year. I need to split this one column in two corresponding columns (Date and Hour) in R. As I understand I need stringr package.
What I have
DATE
- 2015060102
- 2015060104
- 2015060106
- 2015060109
- 2015060110
rstudio
asked Nov 21 '18 at 15:47
Yerzhan
11
add a comment |
I have this DATE column with Date (20150601) + Hour (02) and so on till the next year. I need to split this one column in two corresponding columns (Date and Hour) in R. As I understand I need stringr package.
What I have
DATE
- 2015060102
- 2015060104
- 2015060106
- 2015060109
- 2015060110
rstudio
asked Nov 21 '18 at 15:47
Yerzhan
11
I have this DATE column with Date (20150601) + Hour (02) and so on till the next year. I need to split this one column in two corresponding columns (Date and Hour) in R. As I understand I need stringr package.
What I have
DATE
- 2015060102
- 2015060104
- 2015060106
- 2015060109
- 2015060110
rstudio
rstudio
asked Nov 21 '18 at 15:47
Yerzhan
11
asked Nov 21 '18 at 15:47
Yerzhan
11
asked Nov 21 '18 at 15:47
Yerzhan
11
asked Nov 21 '18 at 15:47
Yerzhan
11
asked Nov 21 '18 at 15:47
Yerzhan
11
11
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
If you are just after a column, each of them a string, you can use substr for this task:
input_vals <- "2015060102"
date_col <- substr(input_vals, 1, 8)
hour_col <- substr(input_vals, 9, 10)
answered Nov 21 '18 at 16:07
rocksteady
7110
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you are just after a column, each of them a string, you can use substr for this task:
input_vals <- "2015060102"
date_col <- substr(input_vals, 1, 8)
hour_col <- substr(input_vals, 9, 10)
answered Nov 21 '18 at 16:07
rocksteady
7110
add a comment |
If you are just after a column, each of them a string, you can use substr for this task:
input_vals <- "2015060102"
date_col <- substr(input_vals, 1, 8)
hour_col <- substr(input_vals, 9, 10)
answered Nov 21 '18 at 16:07
rocksteady
7110
add a comment |
If you are just after a column, each of them a string, you can use substr for this task:
input_vals <- "2015060102"
date_col <- substr(input_vals, 1, 8)
hour_col <- substr(input_vals, 9, 10)
answered Nov 21 '18 at 16:07
rocksteady
7110
If you are just after a column, each of them a string, you can use substr for this task:
input_vals <- "2015060102"
date_col <- substr(input_vals, 1, 8)
hour_col <- substr(input_vals, 9, 10)
answered Nov 21 '18 at 16:07
rocksteady
7110
answered Nov 21 '18 at 16:07
rocksteady
7110
answered Nov 21 '18 at 16:07
rocksteady
7110
answered Nov 21 '18 at 16:07
rocksteady
7110
7110
add a comment |
add a comment |
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