Compare two lists and replace values in list one with values in list 2
2
I have two lists.
a = [1,2,3,4,0,4,5,6,3,6,0,5,6,8,0,3]
b = [1,2, None,4,5,4,5,6,3,6,7,5,6,8,4, None]
I want a resulting list like this.
new_list = [1,2,3,4,5,4,5,6,3,6,7,5,6,8,4,3]
List b is only replacing the 0's in list a and not touching the other elements for example the None is not being replaced.
How can I get about doing this?
Thanks in advance. Please do let me know if you need any other information.
I have tried the following and it does not work.
_ = dict(zip(a,b))
for k,v in _.items():
if k == 0:
a = a.replace(k,v)
python
asked Nov 23 '18 at 0:15
MattMatt
546
add a comment |
2
I have two lists.
a = [1,2,3,4,0,4,5,6,3,6,0,5,6,8,0,3]
b = [1,2, None,4,5,4,5,6,3,6,7,5,6,8,4, None]
I want a resulting list like this.
new_list = [1,2,3,4,5,4,5,6,3,6,7,5,6,8,4,3]
List b is only replacing the 0's in list a and not touching the other elements for example the None is not being replaced.
How can I get about doing this?
Thanks in advance. Please do let me know if you need any other information.
I have tried the following and it does not work.
_ = dict(zip(a,b))
for k,v in _.items():
if k == 0:
a = a.replace(k,v)
python
asked Nov 23 '18 at 0:15
MattMatt
546
2
What have you tried so far?
– Ian Quah
Nov 23 '18 at 0:17
- = dict(zip(a,b)) for k,v in _.items(): if k == 0: a = a.replace(k,v)
– Matt
Nov 23 '18 at 0:21
Fora = [0, 1, 0]andb = [None, 2, 3], what do you expect the result to be?
– lifebalance
Nov 23 '18 at 0:58
add a comment |
2
2
2
I have two lists.
a = [1,2,3,4,0,4,5,6,3,6,0,5,6,8,0,3]
b = [1,2, None,4,5,4,5,6,3,6,7,5,6,8,4, None]
I want a resulting list like this.
new_list = [1,2,3,4,5,4,5,6,3,6,7,5,6,8,4,3]
List b is only replacing the 0's in list a and not touching the other elements for example the None is not being replaced.
How can I get about doing this?
Thanks in advance. Please do let me know if you need any other information.
I have tried the following and it does not work.
_ = dict(zip(a,b))
for k,v in _.items():
if k == 0:
a = a.replace(k,v)
python
asked Nov 23 '18 at 0:15
MattMatt
546
I have two lists.
a = [1,2,3,4,0,4,5,6,3,6,0,5,6,8,0,3]
b = [1,2, None,4,5,4,5,6,3,6,7,5,6,8,4, None]
I want a resulting list like this.
new_list = [1,2,3,4,5,4,5,6,3,6,7,5,6,8,4,3]
List b is only replacing the 0's in list a and not touching the other elements for example the None is not being replaced.
How can I get about doing this?
Thanks in advance. Please do let me know if you need any other information.
I have tried the following and it does not work.
_ = dict(zip(a,b))
for k,v in _.items():
if k == 0:
a = a.replace(k,v)
python
python
asked Nov 23 '18 at 0:15
MattMatt
546
asked Nov 23 '18 at 0:15
MattMatt
546
edited Nov 23 '18 at 0:24
Matt
asked Nov 23 '18 at 0:15
MattMatt
546
asked Nov 23 '18 at 0:15
MattMatt
546
asked Nov 23 '18 at 0:15
MattMatt
546
546
2
What have you tried so far?
– Ian Quah
Nov 23 '18 at 0:17
- = dict(zip(a,b)) for k,v in _.items(): if k == 0: a = a.replace(k,v)
– Matt
Nov 23 '18 at 0:21
Fora = [0, 1, 0]andb = [None, 2, 3], what do you expect the result to be?
– lifebalance
Nov 23 '18 at 0:58
add a comment |
2
What have you tried so far?
– Ian Quah
Nov 23 '18 at 0:17
- = dict(zip(a,b)) for k,v in _.items(): if k == 0: a = a.replace(k,v)
– Matt
Nov 23 '18 at 0:21
Fora = [0, 1, 0]andb = [None, 2, 3], what do you expect the result to be?
– lifebalance
Nov 23 '18 at 0:58
2
2
What have you tried so far?
– Ian Quah
Nov 23 '18 at 0:17
What have you tried so far?
– Ian Quah
Nov 23 '18 at 0:17
- = dict(zip(a,b)) for k,v in _.items(): if k == 0: a = a.replace(k,v)
– Matt
Nov 23 '18 at 0:21
- = dict(zip(a,b)) for k,v in _.items(): if k == 0: a = a.replace(k,v)
– Matt
Nov 23 '18 at 0:21
For
a = [0, 1, 0] and b = [None, 2, 3], what do you expect the result to be?– lifebalance
Nov 23 '18 at 0:58
For
a = [0, 1, 0] and b = [None, 2, 3], what do you expect the result to be?– lifebalance
Nov 23 '18 at 0:58
add a comment |
4 Answers
4
active
oldest
votes
3
You can use zip and a simple list comprehension to generate a new list by picking elements of a if they're not 0 or elements of b if the corresponding a element is zero:
a = [1, 2, 3, 4, 0, 4, 5, 6, 3, 6, 0, 5, 6, 8, 0, 3]
b = [1, 2, None, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, None]
result = [y if x == 0 else x for x, y in zip(a, b)]
print(result)
# [1, 2, 3, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, 3]
answered Nov 23 '18 at 0:23
sliderslider
8,23811129
Fora = [0, 1, 0]andb = [None, 2, 3], result will be[None, 1, 3]...is that what the OP wants?
– lifebalance
Nov 23 '18 at 1:04
I think so. If you don't wantNone, you can change the condition toy if x == 0 and y is not None ....
– slider
Nov 23 '18 at 1:16
add a comment |
0
try with this:
a = [1, 0, 3, 4, 0, 6, 7, 0, 9]
b = [1, 2, None, 4, 5, 6, None, 8, 9]
k = 0 # Counter
c = [0] * len(a) # Creating the 'c' list
for n in a: # Reading 'a' list
if n != 0:
c[k] = a[k] # Copying 'a' list objects, when 'n' != 0, in 'c' list
else:
c[k] = b[k] # Copying 'b' list objects, when 'n' == 0, in 'c' list
k = k + 1
add a comment |
0
Or enumerate + a loop + indexing:
l=[i for i,v in enumerate(a) if v==0]
for i in l:
a[i]=b[i]
And now:
print(a)
Is:
[1, 2, 3, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, 3]
answered Nov 23 '18 at 0:47
U9-ForwardU9-Forward
14.5k21338
You are effectively iterating over the length of the list twice to achieve the end result. That's inefficient.
– lifebalance
Nov 23 '18 at 1:32
add a comment |
0
out =
for ea, eb in zip(a, b):
res = ea
# if element in a is 0 and corresponding element in b is not None
if ea == 0 and eb:
res = eb
out.append(res)
assert out == [1,2,3,4,5,4,5,6,3,6,7,5,6,8,4,3]
And for a = [0, 1, 0] and b = [None, 2, 3] this will generate
out == [0, 1, 3]
answered Nov 23 '18 at 0:30
lifebalancelifebalance
91921444
Wheres your explanation?
– U9-Forward
Nov 23 '18 at 0:48
The code is self-explanatory, yet added a comment at the risk of stating the obvious
– lifebalance
Nov 23 '18 at 1:30
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
You can use zip and a simple list comprehension to generate a new list by picking elements of a if they're not 0 or elements of b if the corresponding a element is zero:
a = [1, 2, 3, 4, 0, 4, 5, 6, 3, 6, 0, 5, 6, 8, 0, 3]
b = [1, 2, None, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, None]
result = [y if x == 0 else x for x, y in zip(a, b)]
print(result)
# [1, 2, 3, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, 3]
answered Nov 23 '18 at 0:23
sliderslider
8,23811129
Fora = [0, 1, 0]andb = [None, 2, 3], result will be[None, 1, 3]...is that what the OP wants?
– lifebalance
Nov 23 '18 at 1:04
I think so. If you don't wantNone, you can change the condition toy if x == 0 and y is not None ....
– slider
Nov 23 '18 at 1:16
add a comment |
3
You can use zip and a simple list comprehension to generate a new list by picking elements of a if they're not 0 or elements of b if the corresponding a element is zero:
a = [1, 2, 3, 4, 0, 4, 5, 6, 3, 6, 0, 5, 6, 8, 0, 3]
b = [1, 2, None, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, None]
result = [y if x == 0 else x for x, y in zip(a, b)]
print(result)
# [1, 2, 3, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, 3]
answered Nov 23 '18 at 0:23
sliderslider
8,23811129
Fora = [0, 1, 0]andb = [None, 2, 3], result will be[None, 1, 3]...is that what the OP wants?
– lifebalance
Nov 23 '18 at 1:04
I think so. If you don't wantNone, you can change the condition toy if x == 0 and y is not None ....
– slider
Nov 23 '18 at 1:16
add a comment |
3
3
3
You can use zip and a simple list comprehension to generate a new list by picking elements of a if they're not 0 or elements of b if the corresponding a element is zero:
a = [1, 2, 3, 4, 0, 4, 5, 6, 3, 6, 0, 5, 6, 8, 0, 3]
b = [1, 2, None, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, None]
result = [y if x == 0 else x for x, y in zip(a, b)]
print(result)
# [1, 2, 3, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, 3]
answered Nov 23 '18 at 0:23
sliderslider
8,23811129
You can use zip and a simple list comprehension to generate a new list by picking elements of a if they're not 0 or elements of b if the corresponding a element is zero:
a = [1, 2, 3, 4, 0, 4, 5, 6, 3, 6, 0, 5, 6, 8, 0, 3]
b = [1, 2, None, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, None]
result = [y if x == 0 else x for x, y in zip(a, b)]
print(result)
# [1, 2, 3, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, 3]
answered Nov 23 '18 at 0:23
sliderslider
8,23811129
answered Nov 23 '18 at 0:23
sliderslider
8,23811129
answered Nov 23 '18 at 0:23
sliderslider
8,23811129
answered Nov 23 '18 at 0:23
sliderslider
8,23811129
8,23811129
Fora = [0, 1, 0]andb = [None, 2, 3], result will be[None, 1, 3]...is that what the OP wants?
– lifebalance
Nov 23 '18 at 1:04
I think so. If you don't wantNone, you can change the condition toy if x == 0 and y is not None ....
– slider
Nov 23 '18 at 1:16
add a comment |
Fora = [0, 1, 0]andb = [None, 2, 3], result will be[None, 1, 3]...is that what the OP wants?
– lifebalance
Nov 23 '18 at 1:04
I think so. If you don't wantNone, you can change the condition toy if x == 0 and y is not None ....
– slider
Nov 23 '18 at 1:16
For
a = [0, 1, 0] and b = [None, 2, 3], result will be [None, 1, 3]...is that what the OP wants?– lifebalance
Nov 23 '18 at 1:04
For
a = [0, 1, 0] and b = [None, 2, 3], result will be [None, 1, 3]...is that what the OP wants?– lifebalance
Nov 23 '18 at 1:04
I think so. If you don't want
None, you can change the condition to y if x == 0 and y is not None ....– slider
Nov 23 '18 at 1:16
I think so. If you don't want
None, you can change the condition to y if x == 0 and y is not None ....– slider
Nov 23 '18 at 1:16
add a comment |
0
try with this:
a = [1, 0, 3, 4, 0, 6, 7, 0, 9]
b = [1, 2, None, 4, 5, 6, None, 8, 9]
k = 0 # Counter
c = [0] * len(a) # Creating the 'c' list
for n in a: # Reading 'a' list
if n != 0:
c[k] = a[k] # Copying 'a' list objects, when 'n' != 0, in 'c' list
else:
c[k] = b[k] # Copying 'b' list objects, when 'n' == 0, in 'c' list
k = k + 1
add a comment |
0
try with this:
a = [1, 0, 3, 4, 0, 6, 7, 0, 9]
b = [1, 2, None, 4, 5, 6, None, 8, 9]
k = 0 # Counter
c = [0] * len(a) # Creating the 'c' list
for n in a: # Reading 'a' list
if n != 0:
c[k] = a[k] # Copying 'a' list objects, when 'n' != 0, in 'c' list
else:
c[k] = b[k] # Copying 'b' list objects, when 'n' == 0, in 'c' list
k = k + 1
add a comment |
0
0
0
try with this:
a = [1, 0, 3, 4, 0, 6, 7, 0, 9]
b = [1, 2, None, 4, 5, 6, None, 8, 9]
k = 0 # Counter
c = [0] * len(a) # Creating the 'c' list
for n in a: # Reading 'a' list
if n != 0:
c[k] = a[k] # Copying 'a' list objects, when 'n' != 0, in 'c' list
else:
c[k] = b[k] # Copying 'b' list objects, when 'n' == 0, in 'c' list
k = k + 1
try with this:
a = [1, 0, 3, 4, 0, 6, 7, 0, 9]
b = [1, 2, None, 4, 5, 6, None, 8, 9]
k = 0 # Counter
c = [0] * len(a) # Creating the 'c' list
for n in a: # Reading 'a' list
if n != 0:
c[k] = a[k] # Copying 'a' list objects, when 'n' != 0, in 'c' list
else:
c[k] = b[k] # Copying 'b' list objects, when 'n' == 0, in 'c' list
k = k + 1
answered Nov 23 '18 at 0:41
user10693394
add a comment |
add a comment |
0
Or enumerate + a loop + indexing:
l=[i for i,v in enumerate(a) if v==0]
for i in l:
a[i]=b[i]
And now:
print(a)
Is:
[1, 2, 3, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, 3]
answered Nov 23 '18 at 0:47
U9-ForwardU9-Forward
14.5k21338
You are effectively iterating over the length of the list twice to achieve the end result. That's inefficient.
– lifebalance
Nov 23 '18 at 1:32
add a comment |
0
Or enumerate + a loop + indexing:
l=[i for i,v in enumerate(a) if v==0]
for i in l:
a[i]=b[i]
And now:
print(a)
Is:
[1, 2, 3, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, 3]
answered Nov 23 '18 at 0:47
U9-ForwardU9-Forward
14.5k21338
You are effectively iterating over the length of the list twice to achieve the end result. That's inefficient.
– lifebalance
Nov 23 '18 at 1:32
add a comment |
0
0
0
Or enumerate + a loop + indexing:
l=[i for i,v in enumerate(a) if v==0]
for i in l:
a[i]=b[i]
And now:
print(a)
Is:
[1, 2, 3, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, 3]
answered Nov 23 '18 at 0:47
U9-ForwardU9-Forward
14.5k21338
Or enumerate + a loop + indexing:
l=[i for i,v in enumerate(a) if v==0]
for i in l:
a[i]=b[i]
And now:
print(a)
Is:
[1, 2, 3, 4, 5, 4, 5, 6, 3, 6, 7, 5, 6, 8, 4, 3]
answered Nov 23 '18 at 0:47
U9-ForwardU9-Forward
14.5k21338
answered Nov 23 '18 at 0:47
U9-ForwardU9-Forward
14.5k21338
answered Nov 23 '18 at 0:47
U9-ForwardU9-Forward
14.5k21338
answered Nov 23 '18 at 0:47
U9-ForwardU9-Forward
14.5k21338
14.5k21338
You are effectively iterating over the length of the list twice to achieve the end result. That's inefficient.
– lifebalance
Nov 23 '18 at 1:32
add a comment |
You are effectively iterating over the length of the list twice to achieve the end result. That's inefficient.
– lifebalance
Nov 23 '18 at 1:32
You are effectively iterating over the length of the list twice to achieve the end result. That's inefficient.
– lifebalance
Nov 23 '18 at 1:32
You are effectively iterating over the length of the list twice to achieve the end result. That's inefficient.
– lifebalance
Nov 23 '18 at 1:32
add a comment |
0
out =
for ea, eb in zip(a, b):
res = ea
# if element in a is 0 and corresponding element in b is not None
if ea == 0 and eb:
res = eb
out.append(res)
assert out == [1,2,3,4,5,4,5,6,3,6,7,5,6,8,4,3]
And for a = [0, 1, 0] and b = [None, 2, 3] this will generate
out == [0, 1, 3]
answered Nov 23 '18 at 0:30
lifebalancelifebalance
91921444
Wheres your explanation?
– U9-Forward
Nov 23 '18 at 0:48
The code is self-explanatory, yet added a comment at the risk of stating the obvious
– lifebalance
Nov 23 '18 at 1:30
add a comment |
0
out =
for ea, eb in zip(a, b):
res = ea
# if element in a is 0 and corresponding element in b is not None
if ea == 0 and eb:
res = eb
out.append(res)
assert out == [1,2,3,4,5,4,5,6,3,6,7,5,6,8,4,3]
And for a = [0, 1, 0] and b = [None, 2, 3] this will generate
out == [0, 1, 3]
answered Nov 23 '18 at 0:30
lifebalancelifebalance
91921444
Wheres your explanation?
– U9-Forward
Nov 23 '18 at 0:48
The code is self-explanatory, yet added a comment at the risk of stating the obvious
– lifebalance
Nov 23 '18 at 1:30
add a comment |
0
0
0
out =
for ea, eb in zip(a, b):
res = ea
# if element in a is 0 and corresponding element in b is not None
if ea == 0 and eb:
res = eb
out.append(res)
assert out == [1,2,3,4,5,4,5,6,3,6,7,5,6,8,4,3]
And for a = [0, 1, 0] and b = [None, 2, 3] this will generate
out == [0, 1, 3]
answered Nov 23 '18 at 0:30
lifebalancelifebalance
91921444
out =
for ea, eb in zip(a, b):
res = ea
# if element in a is 0 and corresponding element in b is not None
if ea == 0 and eb:
res = eb
out.append(res)
assert out == [1,2,3,4,5,4,5,6,3,6,7,5,6,8,4,3]
And for a = [0, 1, 0] and b = [None, 2, 3] this will generate
out == [0, 1, 3]
answered Nov 23 '18 at 0:30
lifebalancelifebalance
91921444
edited Nov 23 '18 at 1:25
answered Nov 23 '18 at 0:30
lifebalancelifebalance
91921444
answered Nov 23 '18 at 0:30
lifebalancelifebalance
91921444
answered Nov 23 '18 at 0:30
lifebalancelifebalance
91921444
91921444
Wheres your explanation?
– U9-Forward
Nov 23 '18 at 0:48
The code is self-explanatory, yet added a comment at the risk of stating the obvious
– lifebalance
Nov 23 '18 at 1:30
add a comment |
Wheres your explanation?
– U9-Forward
Nov 23 '18 at 0:48
The code is self-explanatory, yet added a comment at the risk of stating the obvious
– lifebalance
Nov 23 '18 at 1:30
Wheres your explanation?
– U9-Forward
Nov 23 '18 at 0:48
Wheres your explanation?
– U9-Forward
Nov 23 '18 at 0:48
The code is self-explanatory, yet added a comment at the risk of stating the obvious
– lifebalance
Nov 23 '18 at 1:30
The code is self-explanatory, yet added a comment at the risk of stating the obvious
– lifebalance
Nov 23 '18 at 1:30
add a comment |
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2
What have you tried so far?
– Ian Quah
Nov 23 '18 at 0:17
- = dict(zip(a,b)) for k,v in _.items(): if k == 0: a = a.replace(k,v)
– Matt
Nov 23 '18 at 0:21
For
a = [0, 1, 0]andb = [None, 2, 3], what do you expect the result to be?– lifebalance
Nov 23 '18 at 0:58