264

I have a column in my dataframe like this:

range
"(2,30)"
"(50,290)"
"(400,1000)"
... 

and I want to replace the , comma with - dash. I'm currently using this method but nothing is changed.

org_info_exc['range'].replace(',', '-', inplace=True)

Can anybody help?

7 Answers 7

510

Use the vectorised str method replace:

df['range'] = df['range'].str.replace(',','-')

df
      range
0    (2-30)
1  (50-290)

EDIT: so if we look at what you tried and why it didn't work:

df['range'].replace(',','-',inplace=True)

from the docs we see this description:

str or regex: str: string exactly matching to_replace will be replaced with value

So because the str values do not match, no replacement occurs, compare with the following:

df = pd.DataFrame({'range':['(2,30)',',']})
df['range'].replace(',','-', inplace=True)

df['range']

0    (2,30)
1         -
Name: range, dtype: object

here we get an exact match on the second row and the replacement occurs.

Sign up to request clarification or add additional context in comments.

2 Comments

I wonder why the warning occurs: A value is trying to be set on a copy of a slice from a DataFrame. Try using .loc[row_indexer,col_indexer] = value instead See the caveats in the documentation: pandas.pydata.org/pandas-docs/stable/user_guide/…
It's because of changes to Copy-On-Write semantics in pandas 2, which will be the only copy mechanic in pandas 3 pandas.pydata.org/docs/user_guide/…
134

For anyone else arriving here from Google search on how to do a string replacement on all columns (for example, if one has multiple columns like the OP's 'range' column): Pandas has a built in replace method available on a dataframe object.

df.replace(',', '-', regex=True)

1 Comment

me too, in my example: price_df['Mid-price (p)'].replace(',','',regex=True,inplace=True) worked. Withoutregex=True, it didn't.
29

If you only need to replace characters in one specific column, somehow regex=True and in place=True all failed, I think this way will work:

data["column_name"] = data["column_name"].apply(lambda x: x.replace("characters_need_to_replace", "new_characters"))

lambda is more like a function that works like a for loop in this scenario. x here represents every one of the entries in the current column.

The only thing you need to do is to change the "column_name", "characters_need_to_replace" and "new_characters".

2 Comments

This is what I ended up doing to replace, but I get the "A value is trying to be set on a copy of a slice from a DataFrame." warning. The resulting dataframe is exactly what I want, but the warning is off-putting. Rather than just turning the warning off, I'm in search of a non-warning-inducing way.
Thanks! Also you need to do some conversion in parameter of lambda function, like all x to string using str(x)
10

Replace all commas with underscore in the column names

data.columns= data.columns.str.replace(' ','_',regex=True)

Comments

8

In addition, for those looking to replace more than one character in a column, you can do it using regular expressions:

import re
chars_to_remove = ['.', '-', '(', ')', '']
regular_expression = '[' + re.escape (''. join (chars_to_remove)) + ']'

df['string_col'].str.replace(regular_expression, '', regex=True)

Comments

5

Almost similar to the answer by Nancy K, this works for me:

data["column_name"] = data["column_name"].apply(lambda x: x.str.replace("characters_need_to_replace", "new_characters"))

Comments

2

If you want to remove two or more elements from a string, example the characters '$' and ',' :

Column_Name
===========
$100,000
$1,100,000

... then use:

data.Column_Name.str.replace("[$,]", "", regex=True)

=> [ 100000, 1100000 ]

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.