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

edited Nov 22 '18 at 17:18

Ali AzG

7131717

asked Nov 22 '18 at 15:51

user8322222

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 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

edited Nov 22 '18 at 17:18

Ali AzG

7131717

asked Nov 22 '18 at 15:51

user8322222

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 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

edited Nov 22 '18 at 17:18

Ali AzG

7131717

asked Nov 22 '18 at 15:51

user8322222

1178

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

user8322222

1178

edited Nov 22 '18 at 17:18

Ali AzG

7131717

asked Nov 22 '18 at 15:51

user8322222

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

asked Nov 22 '18 at 15:51

user8322222

1178

asked Nov 22 '18 at 15:51

user8322222

1178

asked Nov 22 '18 at 15:51

user8322222

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

@.. 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 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

BernardL

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)

signed matches the characters signed literally (case sensitive)

$ asserts position at the end of a line

answered Nov 22 '18 at 18:16

pygo

3,1961721

add a comment |

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3 Answers
3

active

oldest

votes

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 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 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 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 Agarwal

2,32851129

answered Nov 22 '18 at 15:55

Rahul Agarwal

2,32851129

answered Nov 22 '18 at 15:55

Rahul Agarwal

2,32851129

answered Nov 22 '18 at 15:55

Rahul 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

BernardL

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

BernardL

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

BernardL

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

BernardL

2,40811130

answered Nov 22 '18 at 16:16

BernardL

2,40811130

answered Nov 22 '18 at 16:16

BernardL

2,40811130

answered Nov 22 '18 at 16:16

BernardL

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)

signed matches the characters signed literally (case sensitive)

$ asserts position at the end of a line

answered Nov 22 '18 at 18:16

pygo

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)

signed matches the characters signed literally (case sensitive)

$ asserts position at the end of a line

answered Nov 22 '18 at 18:16

pygo

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)

signed matches the characters signed literally (case sensitive)

$ asserts position at the end of a line

answered Nov 22 '18 at 18:16

pygo

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)

signed matches the characters signed literally (case sensitive)

$ asserts position at the end of a line

answered Nov 22 '18 at 18:16

pygo

3,1961721

answered Nov 22 '18 at 18:16

pygo

3,1961721

answered Nov 22 '18 at 18:16

pygo

3,1961721

answered Nov 22 '18 at 18:16

pygo

3,1961721

add a comment |

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