Replacing multiple value names in a panda dataframe
I have a column colA that has multiple values in a pandas dataframe. I want every value that starts with spare1 in this column to be replaced with the words email_petition. e.g. spare1signed, spare1not signed yet' etc. will all be converted to just email_petition.
I am using the following code:
petition = df.colA.str.startswith('spare1')
if df.colA == petition:
df.colA.replace(petition, 'email_petition', inplace=True)
but I get the following error:
The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
I also tried the following code which doesn't give me an error but doesn't seem to work as the values don't change:
petition = df.colA.str.startswith('spare1')
if df.colA is petition:
df.colA.replace(petition, 'email_petition', inplace=True)
would love some advice on this!
thanks
python pandas
edited Nov 22 '18 at 17:18
Ali AzG
7131717
asked Nov 22 '18 at 15:51
user8322222user8322222
1178
add a comment |
I have a column colA that has multiple values in a pandas dataframe. I want every value that starts with spare1 in this column to be replaced with the words email_petition. e.g. spare1signed, spare1not signed yet' etc. will all be converted to just email_petition.
I am using the following code:
petition = df.colA.str.startswith('spare1')
if df.colA == petition:
df.colA.replace(petition, 'email_petition', inplace=True)
but I get the following error:
The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
I also tried the following code which doesn't give me an error but doesn't seem to work as the values don't change:
petition = df.colA.str.startswith('spare1')
if df.colA is petition:
df.colA.replace(petition, 'email_petition', inplace=True)
would love some advice on this!
thanks
python pandas
edited Nov 22 '18 at 17:18
Ali AzG
7131717
asked Nov 22 '18 at 15:51
user8322222user8322222
1178
Does my answer helped you ?
– Rahul Agarwal
Nov 22 '18 at 16:45
@.. user8322222 what i should do when someone answers my question must learn!
– pygo
Nov 23 '18 at 19:18
add a comment |
I have a column colA that has multiple values in a pandas dataframe. I want every value that starts with spare1 in this column to be replaced with the words email_petition. e.g. spare1signed, spare1not signed yet' etc. will all be converted to just email_petition.
I am using the following code:
petition = df.colA.str.startswith('spare1')
if df.colA == petition:
df.colA.replace(petition, 'email_petition', inplace=True)
but I get the following error:
The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
I also tried the following code which doesn't give me an error but doesn't seem to work as the values don't change:
petition = df.colA.str.startswith('spare1')
if df.colA is petition:
df.colA.replace(petition, 'email_petition', inplace=True)
would love some advice on this!
thanks
python pandas
edited Nov 22 '18 at 17:18
Ali AzG
7131717
asked Nov 22 '18 at 15:51
user8322222user8322222
1178
I have a column colA that has multiple values in a pandas dataframe. I want every value that starts with spare1 in this column to be replaced with the words email_petition. e.g. spare1signed, spare1not signed yet' etc. will all be converted to just email_petition.
I am using the following code:
petition = df.colA.str.startswith('spare1')
if df.colA == petition:
df.colA.replace(petition, 'email_petition', inplace=True)
but I get the following error:
The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
I also tried the following code which doesn't give me an error but doesn't seem to work as the values don't change:
petition = df.colA.str.startswith('spare1')
if df.colA is petition:
df.colA.replace(petition, 'email_petition', inplace=True)
would love some advice on this!
thanks
python pandas
python pandas
edited Nov 22 '18 at 17:18
Ali AzG
7131717
asked Nov 22 '18 at 15:51
user8322222user8322222
1178
edited Nov 22 '18 at 17:18
Ali AzG
7131717
asked Nov 22 '18 at 15:51
user8322222user8322222
1178
edited Nov 22 '18 at 17:18
Ali AzG
7131717
edited Nov 22 '18 at 17:18
Ali AzG
7131717
edited Nov 22 '18 at 17:18
Ali AzG
7131717
7131717
asked Nov 22 '18 at 15:51
user8322222user8322222
1178
asked Nov 22 '18 at 15:51
user8322222user8322222
1178
asked Nov 22 '18 at 15:51
user8322222user8322222
1178
1178
Does my answer helped you ?
– Rahul Agarwal
Nov 22 '18 at 16:45
@.. user8322222 what i should do when someone answers my question must learn!
– pygo
Nov 23 '18 at 19:18
add a comment |
Does my answer helped you ?
– Rahul Agarwal
Nov 22 '18 at 16:45
@.. user8322222 what i should do when someone answers my question must learn!
– pygo
Nov 23 '18 at 19:18
Does my answer helped you ?
– Rahul Agarwal
Nov 22 '18 at 16:45
Does my answer helped you ?
– Rahul Agarwal
Nov 22 '18 at 16:45
@.. user8322222 what i should do when someone answers my question must learn!
– pygo
Nov 23 '18 at 19:18
@.. user8322222 what i should do when someone answers my question must learn!
– pygo
Nov 23 '18 at 19:18
add a comment |
3 Answers
3
active
oldest
votes
Try this:
df.colA.replace({'spare1':'email_petition'}, regex=True)
For complete removal:
df['colA'].replace({'spare1signed':'email_petition','spare1notsigned':'email_petition'})
answered Nov 22 '18 at 15:55
Rahul AgarwalRahul Agarwal
2,32851129
add a comment |
When possible always vectorize your operations on a dataframe. In your case the for loop is not required, you can simply apply a function to the whole column.
df = pd.DataFrame({'colA':['spare1signed','spare1not signed','no action']})
df.colA = df.colA.apply(lambda x: 'email_sent' if 'spare1' in x else x)
df
>>
colA
0 email_sent
1 email_sent
2 no action
Here we assign the column with a lambda function that replaces any value in the column if spare1 is found with email_sent.
answered Nov 22 '18 at 16:16
BernardLBernardL
2,40811130
add a comment |
This can be simplified with replace using regex pattern:
Borrowed the data from @BernardL
Example DataFrame with column name colA :
>>> df
colA
0 spare1signed
1 spare1not signed
2 no action
Applying regex method which says whatever ends with signed$ Just replace them to email_sent :
Result:
>>> df['colA'] = df.colA.replace(r'.*signed$', 'email_sent', regex=True)
>>> df
colA
0 email_sent
1 email_sent
2 no action
Regex Meaning:
.*matches any character (except for line terminators)
*Quantifier — Matches between zero and unlimited times, as many
times as possible, giving back as needed (greedy)
signedmatches the characters signed literally (case sensitive)
$asserts position at the end of a line
answered Nov 22 '18 at 18:16
pygopygo
3,1961721
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Try this:
df.colA.replace({'spare1':'email_petition'}, regex=True)
For complete removal:
df['colA'].replace({'spare1signed':'email_petition','spare1notsigned':'email_petition'})
answered Nov 22 '18 at 15:55
Rahul AgarwalRahul Agarwal
2,32851129
add a comment |
Try this:
df.colA.replace({'spare1':'email_petition'}, regex=True)
For complete removal:
df['colA'].replace({'spare1signed':'email_petition','spare1notsigned':'email_petition'})
answered Nov 22 '18 at 15:55
Rahul AgarwalRahul Agarwal
2,32851129
add a comment |
Try this:
df.colA.replace({'spare1':'email_petition'}, regex=True)
For complete removal:
df['colA'].replace({'spare1signed':'email_petition','spare1notsigned':'email_petition'})
answered Nov 22 '18 at 15:55
Rahul AgarwalRahul Agarwal
2,32851129
Try this:
df.colA.replace({'spare1':'email_petition'}, regex=True)
For complete removal:
df['colA'].replace({'spare1signed':'email_petition','spare1notsigned':'email_petition'})
answered Nov 22 '18 at 15:55
Rahul AgarwalRahul Agarwal
2,32851129
answered Nov 22 '18 at 15:55
Rahul AgarwalRahul Agarwal
2,32851129
answered Nov 22 '18 at 15:55
Rahul AgarwalRahul Agarwal
2,32851129
answered Nov 22 '18 at 15:55
Rahul AgarwalRahul Agarwal
2,32851129
2,32851129
add a comment |
add a comment |
When possible always vectorize your operations on a dataframe. In your case the for loop is not required, you can simply apply a function to the whole column.
df = pd.DataFrame({'colA':['spare1signed','spare1not signed','no action']})
df.colA = df.colA.apply(lambda x: 'email_sent' if 'spare1' in x else x)
df
>>
colA
0 email_sent
1 email_sent
2 no action
Here we assign the column with a lambda function that replaces any value in the column if spare1 is found with email_sent.
answered Nov 22 '18 at 16:16
BernardLBernardL
2,40811130
add a comment |
When possible always vectorize your operations on a dataframe. In your case the for loop is not required, you can simply apply a function to the whole column.
df = pd.DataFrame({'colA':['spare1signed','spare1not signed','no action']})
df.colA = df.colA.apply(lambda x: 'email_sent' if 'spare1' in x else x)
df
>>
colA
0 email_sent
1 email_sent
2 no action
Here we assign the column with a lambda function that replaces any value in the column if spare1 is found with email_sent.
answered Nov 22 '18 at 16:16
BernardLBernardL
2,40811130
add a comment |
When possible always vectorize your operations on a dataframe. In your case the for loop is not required, you can simply apply a function to the whole column.
df = pd.DataFrame({'colA':['spare1signed','spare1not signed','no action']})
df.colA = df.colA.apply(lambda x: 'email_sent' if 'spare1' in x else x)
df
>>
colA
0 email_sent
1 email_sent
2 no action
Here we assign the column with a lambda function that replaces any value in the column if spare1 is found with email_sent.
answered Nov 22 '18 at 16:16
BernardLBernardL
2,40811130
When possible always vectorize your operations on a dataframe. In your case the for loop is not required, you can simply apply a function to the whole column.
df = pd.DataFrame({'colA':['spare1signed','spare1not signed','no action']})
df.colA = df.colA.apply(lambda x: 'email_sent' if 'spare1' in x else x)
df
>>
colA
0 email_sent
1 email_sent
2 no action
Here we assign the column with a lambda function that replaces any value in the column if spare1 is found with email_sent.
answered Nov 22 '18 at 16:16
BernardLBernardL
2,40811130
answered Nov 22 '18 at 16:16
BernardLBernardL
2,40811130
answered Nov 22 '18 at 16:16
BernardLBernardL
2,40811130
answered Nov 22 '18 at 16:16
BernardLBernardL
2,40811130
2,40811130
add a comment |
add a comment |
This can be simplified with replace using regex pattern:
Borrowed the data from @BernardL
Example DataFrame with column name colA :
>>> df
colA
0 spare1signed
1 spare1not signed
2 no action
Applying regex method which says whatever ends with signed$ Just replace them to email_sent :
Result:
>>> df['colA'] = df.colA.replace(r'.*signed$', 'email_sent', regex=True)
>>> df
colA
0 email_sent
1 email_sent
2 no action
Regex Meaning:
.*matches any character (except for line terminators)
*Quantifier — Matches between zero and unlimited times, as many
times as possible, giving back as needed (greedy)
signedmatches the characters signed literally (case sensitive)
$asserts position at the end of a line
answered Nov 22 '18 at 18:16
pygopygo
3,1961721
add a comment |
This can be simplified with replace using regex pattern:
Borrowed the data from @BernardL
Example DataFrame with column name colA :
>>> df
colA
0 spare1signed
1 spare1not signed
2 no action
Applying regex method which says whatever ends with signed$ Just replace them to email_sent :
Result:
>>> df['colA'] = df.colA.replace(r'.*signed$', 'email_sent', regex=True)
>>> df
colA
0 email_sent
1 email_sent
2 no action
Regex Meaning:
.*matches any character (except for line terminators)
*Quantifier — Matches between zero and unlimited times, as many
times as possible, giving back as needed (greedy)
signedmatches the characters signed literally (case sensitive)
$asserts position at the end of a line
answered Nov 22 '18 at 18:16
pygopygo
3,1961721
add a comment |
This can be simplified with replace using regex pattern:
Borrowed the data from @BernardL
Example DataFrame with column name colA :
>>> df
colA
0 spare1signed
1 spare1not signed
2 no action
Applying regex method which says whatever ends with signed$ Just replace them to email_sent :
Result:
>>> df['colA'] = df.colA.replace(r'.*signed$', 'email_sent', regex=True)
>>> df
colA
0 email_sent
1 email_sent
2 no action
Regex Meaning:
.*matches any character (except for line terminators)
*Quantifier — Matches between zero and unlimited times, as many
times as possible, giving back as needed (greedy)
signedmatches the characters signed literally (case sensitive)
$asserts position at the end of a line
answered Nov 22 '18 at 18:16
pygopygo
3,1961721
This can be simplified with replace using regex pattern:
Borrowed the data from @BernardL
Example DataFrame with column name colA :
>>> df
colA
0 spare1signed
1 spare1not signed
2 no action
Applying regex method which says whatever ends with signed$ Just replace them to email_sent :
Result:
>>> df['colA'] = df.colA.replace(r'.*signed$', 'email_sent', regex=True)
>>> df
colA
0 email_sent
1 email_sent
2 no action
Regex Meaning:
.*matches any character (except for line terminators)
*Quantifier — Matches between zero and unlimited times, as many
times as possible, giving back as needed (greedy)
signedmatches the characters signed literally (case sensitive)
$asserts position at the end of a line
answered Nov 22 '18 at 18:16
pygopygo
3,1961721
answered Nov 22 '18 at 18:16
pygopygo
3,1961721
answered Nov 22 '18 at 18:16
pygopygo
3,1961721
answered Nov 22 '18 at 18:16
pygopygo
3,1961721
3,1961721
add a comment |
add a comment |
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Does my answer helped you ?
– Rahul Agarwal
Nov 22 '18 at 16:45
@.. user8322222 what i should do when someone answers my question must learn!
– pygo
Nov 23 '18 at 19:18