Replace All Occurrences of a String in a Pandas Dataframe (Python)

Replace all occurrences of a string in a pandas dataframe (Python)

You can use replace and pass the strings to find/replace as dictionary keys/items:

df.replace({'\n': '<br>'}, regex=True)

For example:

>>> df = pd.DataFrame({'a': ['1\n', '2\n', '3'], 'b': ['4\n', '5', '6\n']})
>>> df
   a    b
0  1\n  4\n
1  2\n  5
2  3    6\n

>>> df.replace({'\n': '<br>'}, regex=True)
   a      b
0  1<br>  4<br>
1  2<br>  5
2  3      6<br>

Note that this method returns a new DataFrame instance by default (it does not modify the original), so you'll need to either reassign the output:

df = df.replace({'\n': '<br>'}, regex=True)

or specify inplace=True:

df.replace({'\n': '<br>'}, regex=True, inplace=True)

How do I replace all the instances of a certain character in a dataframe?

Consider the dataframe df

df = pd.DataFrame([['?', 1], [2, '?']])

print(df)

   0  1
0  ?  1
1  2  ?

replace

df.replace('?', 0)

   0  1
0  0  1
1  2  0

mask or where

df.mask(df == '?', 0)
# df.where(df != '?', 0)

   0  1
0  0  1
1  2  0

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However, imagine your dataframe has ? within longer strings.

df = pd.DataFrame([['a?', 1], [2, '?b']])

print(df)

    0   1
0  a?   1
1   2  ?b

replace with regex=True

df.replace('\?', '0', regex=True)

    0   1
0  a0   1
1   2  0b

Pandas: Search and replace a string across all columns in a dataframe

Try this

df = pd.DataFrame({'A': ['None', 'NONE', '565'], 
                   'B': ['234', 'NONE', '347']})
# replace NONE by None
df = df.replace('NONE', 'None')
print(df)

      A     B
0  None   234
1  None  None
2   565   347

Is there a way to replace all substrings in a pandas dataframe?

Try:

df = df.astype(str).replace("\[xReturn\]", "", regex=True)

How to replace text in a string column of a Pandas dataframe?

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.

Pandas: replace substring in string

Use replace with dict for replacing and regex=True:

df['url'] = df['url'].replace({'icashier.alipay.com': 'aliexpress.com'}, regex=True)
print (df)
                                          url
0  aliexpress.com/catalog/2758186/detail.aspx
1  aliexpress.com/catalog/2758186/detail.aspx
2  aliexpress.com/catalog/2758186/detail.aspx
3                                      vk.com

How to replace exact string to other using replace() of Panda.DataFrame?

Try:

df['tumor-size'] = df['tumor-size'].replace("^'0-4'$", "'00-04'")

replace part of an int or string in a pandas dataframe column upon condition

You can use .str.replace with regex

df['New_date'] = df['Original_date'].astype(str).str.replace('(\d{4})(13|14)(\d{2})', r'\g<1>12\3', regex=True)

print(df)

   Original_date  New_date
0       20190101  20190101
1       20191301  20191201
2       20191401  20191201

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