'generator' object is not callable when if inside a for loop in list compression
I am getting the following error when i ran this:
df['initial_referrer'].apply(lambda x: value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan for value in x.split('&'))
TypeError: 'generator' object is not callable
I am not sure the meaning of the error and how to modify my this to get rid of this. I have read couple of similar questions here but could not figure our what could be the issue.
So I have values in df['initial_referrer'] be like:
df['initial_referrer'].head()
0 /login/index.php
1 /login/index.php?utm_source=INTERNAL&utm_medium=EMAIL&utm_campaign=login-day1
2 /login/index.php
3 /login/index.php?utm_source=INTERNAL&utm_medium=EMAIL&utm_campaign=login-day1
4 /login/index.php
And in this, I wanted to extract the value of utm_campaign which is login-day1 thats why I was using the for loop and then if statement it was taking a lot of time/days to process 20mil rows. Therefore I wanted to use generator expression or list compression to process it faster.
python pandas performance
edited Nov 12 '18 at 11:31
jpp
91.9k2052102
asked Nov 12 '18 at 10:23
Gagan
392520
|
show 1 more comment
I am getting the following error when i ran this:
df['initial_referrer'].apply(lambda x: value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan for value in x.split('&'))
TypeError: 'generator' object is not callable
I am not sure the meaning of the error and how to modify my this to get rid of this. I have read couple of similar questions here but could not figure our what could be the issue.
So I have values in df['initial_referrer'] be like:
df['initial_referrer'].head()
0 /login/index.php
1 /login/index.php?utm_source=INTERNAL&utm_medium=EMAIL&utm_campaign=login-day1
2 /login/index.php
3 /login/index.php?utm_source=INTERNAL&utm_medium=EMAIL&utm_campaign=login-day1
4 /login/index.php
And in this, I wanted to extract the value of utm_campaign which is login-day1 thats why I was using the for loop and then if statement it was taking a lot of time/days to process 20mil rows. Therefore I wanted to use generator expression or list compression to process it faster.
python pandas performance
edited Nov 12 '18 at 11:31
jpp
91.9k2052102
asked Nov 12 '18 at 10:23
Gagan
392520
1
you're not using a list-comprehension, you are using a generator expression
– juanpa.arrivillaga
Nov 12 '18 at 10:25
Basically I was doing this using itterrows earlier but it was taking a lot of time so I thought of doing this using apply
– Gagan
Nov 12 '18 at 10:28
use itertuples. But.applyisn't going to be much faster. It's basically a plain python for-loop underneath the hood
– juanpa.arrivillaga
Nov 12 '18 at 10:28
If you'd extract a Minimal, Complete, and Verifiable example, you might find the error yourself. Also, it would make this question much more valuable to others.
– Ulrich Eckhardt
Nov 12 '18 at 10:39
@UlrichEckhardt let me know if question make sense now. I dont think downvote would solve the purpose though but anyways.
– Gagan
Nov 12 '18 at 10:44
|
show 1 more comment
I am getting the following error when i ran this:
df['initial_referrer'].apply(lambda x: value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan for value in x.split('&'))
TypeError: 'generator' object is not callable
I am not sure the meaning of the error and how to modify my this to get rid of this. I have read couple of similar questions here but could not figure our what could be the issue.
So I have values in df['initial_referrer'] be like:
df['initial_referrer'].head()
0 /login/index.php
1 /login/index.php?utm_source=INTERNAL&utm_medium=EMAIL&utm_campaign=login-day1
2 /login/index.php
3 /login/index.php?utm_source=INTERNAL&utm_medium=EMAIL&utm_campaign=login-day1
4 /login/index.php
And in this, I wanted to extract the value of utm_campaign which is login-day1 thats why I was using the for loop and then if statement it was taking a lot of time/days to process 20mil rows. Therefore I wanted to use generator expression or list compression to process it faster.
python pandas performance
edited Nov 12 '18 at 11:31
jpp
91.9k2052102
asked Nov 12 '18 at 10:23
Gagan
392520
I am getting the following error when i ran this:
df['initial_referrer'].apply(lambda x: value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan for value in x.split('&'))
TypeError: 'generator' object is not callable
I am not sure the meaning of the error and how to modify my this to get rid of this. I have read couple of similar questions here but could not figure our what could be the issue.
So I have values in df['initial_referrer'] be like:
df['initial_referrer'].head()
0 /login/index.php
1 /login/index.php?utm_source=INTERNAL&utm_medium=EMAIL&utm_campaign=login-day1
2 /login/index.php
3 /login/index.php?utm_source=INTERNAL&utm_medium=EMAIL&utm_campaign=login-day1
4 /login/index.php
And in this, I wanted to extract the value of utm_campaign which is login-day1 thats why I was using the for loop and then if statement it was taking a lot of time/days to process 20mil rows. Therefore I wanted to use generator expression or list compression to process it faster.
python pandas performance
python pandas performance
edited Nov 12 '18 at 11:31
jpp
91.9k2052102
asked Nov 12 '18 at 10:23
Gagan
392520
edited Nov 12 '18 at 11:31
jpp
91.9k2052102
asked Nov 12 '18 at 10:23
Gagan
392520
edited Nov 12 '18 at 11:31
jpp
91.9k2052102
edited Nov 12 '18 at 11:31
jpp
91.9k2052102
edited Nov 12 '18 at 11:31
jpp
91.9k2052102
91.9k2052102
asked Nov 12 '18 at 10:23
Gagan
392520
asked Nov 12 '18 at 10:23
Gagan
392520
asked Nov 12 '18 at 10:23
Gagan
392520
392520
1
you're not using a list-comprehension, you are using a generator expression
– juanpa.arrivillaga
Nov 12 '18 at 10:25
Basically I was doing this using itterrows earlier but it was taking a lot of time so I thought of doing this using apply
– Gagan
Nov 12 '18 at 10:28
use itertuples. But.applyisn't going to be much faster. It's basically a plain python for-loop underneath the hood
– juanpa.arrivillaga
Nov 12 '18 at 10:28
If you'd extract a Minimal, Complete, and Verifiable example, you might find the error yourself. Also, it would make this question much more valuable to others.
– Ulrich Eckhardt
Nov 12 '18 at 10:39
@UlrichEckhardt let me know if question make sense now. I dont think downvote would solve the purpose though but anyways.
– Gagan
Nov 12 '18 at 10:44
|
show 1 more comment
1
you're not using a list-comprehension, you are using a generator expression
– juanpa.arrivillaga
Nov 12 '18 at 10:25
Basically I was doing this using itterrows earlier but it was taking a lot of time so I thought of doing this using apply
– Gagan
Nov 12 '18 at 10:28
use itertuples. But.applyisn't going to be much faster. It's basically a plain python for-loop underneath the hood
– juanpa.arrivillaga
Nov 12 '18 at 10:28
If you'd extract a Minimal, Complete, and Verifiable example, you might find the error yourself. Also, it would make this question much more valuable to others.
– Ulrich Eckhardt
Nov 12 '18 at 10:39
@UlrichEckhardt let me know if question make sense now. I dont think downvote would solve the purpose though but anyways.
– Gagan
Nov 12 '18 at 10:44
1
1
you're not using a list-comprehension, you are using a generator expression
– juanpa.arrivillaga
Nov 12 '18 at 10:25
you're not using a list-comprehension, you are using a generator expression
– juanpa.arrivillaga
Nov 12 '18 at 10:25
Basically I was doing this using itterrows earlier but it was taking a lot of time so I thought of doing this using apply
– Gagan
Nov 12 '18 at 10:28
Basically I was doing this using itterrows earlier but it was taking a lot of time so I thought of doing this using apply
– Gagan
Nov 12 '18 at 10:28
use itertuples. But
.apply isn't going to be much faster. It's basically a plain python for-loop underneath the hood– juanpa.arrivillaga
Nov 12 '18 at 10:28
use itertuples. But
.apply isn't going to be much faster. It's basically a plain python for-loop underneath the hood– juanpa.arrivillaga
Nov 12 '18 at 10:28
If you'd extract a Minimal, Complete, and Verifiable example, you might find the error yourself. Also, it would make this question much more valuable to others.
– Ulrich Eckhardt
Nov 12 '18 at 10:39
If you'd extract a Minimal, Complete, and Verifiable example, you might find the error yourself. Also, it would make this question much more valuable to others.
– Ulrich Eckhardt
Nov 12 '18 at 10:39
@UlrichEckhardt let me know if question make sense now. I dont think downvote would solve the purpose though but anyways.
– Gagan
Nov 12 '18 at 10:44
@UlrichEckhardt let me know if question make sense now. I dont think downvote would solve the purpose though but anyways.
– Gagan
Nov 12 '18 at 10:44
|
show 1 more comment
1 Answer
1
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oldest
votes
It's instructive to first use apply with a regular function:
def func(x):
return [value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan
for value in x.split('&')]
df['initial_referrer'].apply(func)
Notice the square brackets representing the list comprehension. You need to translate this to your lambda function:
df['initial_referrer'].apply(lambda x: [value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan for value in x.split('&')])
But the latter is unreadable. You are better off writing a regular function.
Note pd.Series.apply is a Python-level loop. You can use map instead and will likely see a performance improvement:
df['initial_referrer'] = list(map(func, df['initial_referrer'].values))
Or even a list comprehension:
df['initial_referrer'] = [func(x) for x in df['initial_referrer'].values]
answered Nov 12 '18 at 10:28
jpp
91.9k2052102
Thanks jpp. any particular idea which way would be faster? I have 20mil rows to process
– Gagan
Nov 12 '18 at 10:32
@Gagan, See my last 2 suggestions. Test them and time to see what's most efficient.
– jpp
Nov 12 '18 at 10:33
Thanks @jpp, I changed the function to this: def func(x): return [value.split("utm_campaign=",1)[1] for value in x.split('&') if 'utm_campaign' in value][0] And statement to this: df['initial_referrer_1'] = [func(x) if 'utm_campaign' in x else np.nan for x in df['initial_referrer'].values]. And it worked.
– Gagan
Nov 12 '18 at 11:12
add a comment |
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It's instructive to first use apply with a regular function:
def func(x):
return [value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan
for value in x.split('&')]
df['initial_referrer'].apply(func)
Notice the square brackets representing the list comprehension. You need to translate this to your lambda function:
df['initial_referrer'].apply(lambda x: [value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan for value in x.split('&')])
But the latter is unreadable. You are better off writing a regular function.
Note pd.Series.apply is a Python-level loop. You can use map instead and will likely see a performance improvement:
df['initial_referrer'] = list(map(func, df['initial_referrer'].values))
Or even a list comprehension:
df['initial_referrer'] = [func(x) for x in df['initial_referrer'].values]
answered Nov 12 '18 at 10:28
jpp
91.9k2052102
Thanks jpp. any particular idea which way would be faster? I have 20mil rows to process
– Gagan
Nov 12 '18 at 10:32
@Gagan, See my last 2 suggestions. Test them and time to see what's most efficient.
– jpp
Nov 12 '18 at 10:33
Thanks @jpp, I changed the function to this: def func(x): return [value.split("utm_campaign=",1)[1] for value in x.split('&') if 'utm_campaign' in value][0] And statement to this: df['initial_referrer_1'] = [func(x) if 'utm_campaign' in x else np.nan for x in df['initial_referrer'].values]. And it worked.
– Gagan
Nov 12 '18 at 11:12
add a comment |
It's instructive to first use apply with a regular function:
def func(x):
return [value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan
for value in x.split('&')]
df['initial_referrer'].apply(func)
Notice the square brackets representing the list comprehension. You need to translate this to your lambda function:
df['initial_referrer'].apply(lambda x: [value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan for value in x.split('&')])
But the latter is unreadable. You are better off writing a regular function.
Note pd.Series.apply is a Python-level loop. You can use map instead and will likely see a performance improvement:
df['initial_referrer'] = list(map(func, df['initial_referrer'].values))
Or even a list comprehension:
df['initial_referrer'] = [func(x) for x in df['initial_referrer'].values]
answered Nov 12 '18 at 10:28
jpp
91.9k2052102
Thanks jpp. any particular idea which way would be faster? I have 20mil rows to process
– Gagan
Nov 12 '18 at 10:32
@Gagan, See my last 2 suggestions. Test them and time to see what's most efficient.
– jpp
Nov 12 '18 at 10:33
Thanks @jpp, I changed the function to this: def func(x): return [value.split("utm_campaign=",1)[1] for value in x.split('&') if 'utm_campaign' in value][0] And statement to this: df['initial_referrer_1'] = [func(x) if 'utm_campaign' in x else np.nan for x in df['initial_referrer'].values]. And it worked.
– Gagan
Nov 12 '18 at 11:12
add a comment |
It's instructive to first use apply with a regular function:
def func(x):
return [value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan
for value in x.split('&')]
df['initial_referrer'].apply(func)
Notice the square brackets representing the list comprehension. You need to translate this to your lambda function:
df['initial_referrer'].apply(lambda x: [value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan for value in x.split('&')])
But the latter is unreadable. You are better off writing a regular function.
Note pd.Series.apply is a Python-level loop. You can use map instead and will likely see a performance improvement:
df['initial_referrer'] = list(map(func, df['initial_referrer'].values))
Or even a list comprehension:
df['initial_referrer'] = [func(x) for x in df['initial_referrer'].values]
answered Nov 12 '18 at 10:28
jpp
91.9k2052102
It's instructive to first use apply with a regular function:
def func(x):
return [value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan
for value in x.split('&')]
df['initial_referrer'].apply(func)
Notice the square brackets representing the list comprehension. You need to translate this to your lambda function:
df['initial_referrer'].apply(lambda x: [value.split("utm_campaign=",1)[1] if 'utm_campaign' in value else np.nan for value in x.split('&')])
But the latter is unreadable. You are better off writing a regular function.
Note pd.Series.apply is a Python-level loop. You can use map instead and will likely see a performance improvement:
df['initial_referrer'] = list(map(func, df['initial_referrer'].values))
Or even a list comprehension:
df['initial_referrer'] = [func(x) for x in df['initial_referrer'].values]
answered Nov 12 '18 at 10:28
jpp
91.9k2052102
edited Nov 12 '18 at 10:33
answered Nov 12 '18 at 10:28
jpp
91.9k2052102
answered Nov 12 '18 at 10:28
jpp
91.9k2052102
answered Nov 12 '18 at 10:28
jpp
91.9k2052102
91.9k2052102
Thanks jpp. any particular idea which way would be faster? I have 20mil rows to process
– Gagan
Nov 12 '18 at 10:32
@Gagan, See my last 2 suggestions. Test them and time to see what's most efficient.
– jpp
Nov 12 '18 at 10:33
Thanks @jpp, I changed the function to this: def func(x): return [value.split("utm_campaign=",1)[1] for value in x.split('&') if 'utm_campaign' in value][0] And statement to this: df['initial_referrer_1'] = [func(x) if 'utm_campaign' in x else np.nan for x in df['initial_referrer'].values]. And it worked.
– Gagan
Nov 12 '18 at 11:12
add a comment |
Thanks jpp. any particular idea which way would be faster? I have 20mil rows to process
– Gagan
Nov 12 '18 at 10:32
@Gagan, See my last 2 suggestions. Test them and time to see what's most efficient.
– jpp
Nov 12 '18 at 10:33
Thanks @jpp, I changed the function to this: def func(x): return [value.split("utm_campaign=",1)[1] for value in x.split('&') if 'utm_campaign' in value][0] And statement to this: df['initial_referrer_1'] = [func(x) if 'utm_campaign' in x else np.nan for x in df['initial_referrer'].values]. And it worked.
– Gagan
Nov 12 '18 at 11:12
Thanks jpp. any particular idea which way would be faster? I have 20mil rows to process
– Gagan
Nov 12 '18 at 10:32
Thanks jpp. any particular idea which way would be faster? I have 20mil rows to process
– Gagan
Nov 12 '18 at 10:32
@Gagan, See my last 2 suggestions. Test them and time to see what's most efficient.
– jpp
Nov 12 '18 at 10:33
@Gagan, See my last 2 suggestions. Test them and time to see what's most efficient.
– jpp
Nov 12 '18 at 10:33
Thanks @jpp, I changed the function to this: def func(x): return [value.split("utm_campaign=",1)[1] for value in x.split('&') if 'utm_campaign' in value][0] And statement to this: df['initial_referrer_1'] = [func(x) if 'utm_campaign' in x else np.nan for x in df['initial_referrer'].values]. And it worked.
– Gagan
Nov 12 '18 at 11:12
Thanks @jpp, I changed the function to this: def func(x): return [value.split("utm_campaign=",1)[1] for value in x.split('&') if 'utm_campaign' in value][0] And statement to this: df['initial_referrer_1'] = [func(x) if 'utm_campaign' in x else np.nan for x in df['initial_referrer'].values]. And it worked.
– Gagan
Nov 12 '18 at 11:12
add a comment |
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1
you're not using a list-comprehension, you are using a generator expression
– juanpa.arrivillaga
Nov 12 '18 at 10:25
Basically I was doing this using itterrows earlier but it was taking a lot of time so I thought of doing this using apply
– Gagan
Nov 12 '18 at 10:28
use itertuples. But
.applyisn't going to be much faster. It's basically a plain python for-loop underneath the hood– juanpa.arrivillaga
Nov 12 '18 at 10:28
If you'd extract a Minimal, Complete, and Verifiable example, you might find the error yourself. Also, it would make this question much more valuable to others.
– Ulrich Eckhardt
Nov 12 '18 at 10:39
@UlrichEckhardt let me know if question make sense now. I dont think downvote would solve the purpose though but anyways.
– Gagan
Nov 12 '18 at 10:44