You can use DataFrame.fillna or Series.fillna which will replace the Python object None, not the string 'None'.

import pandas as pd
import numpy as np

For dataframe:

df = df.fillna(value=np.nan)

For column or series:

df.mycol.fillna(value=np.nan, inplace=True)

I understand you want to replace the value 0 in column 'B' with NaN (None) in column 'A'. However, the issue might be arising because the value 0 in column 'B' is of type integer, whereas you are trying to replace it with NaN, which is of type float.

import pandas as pd
import numpy as np

# Sample DataFrame
df = pd.DataFrame({'B': [0, 0, 'A', 'B']})

# Replace 0 with NaN in column 'A'
df['A'] = df['B'].replace(0, np.nan)

# Print DataFrame to check
print(df)

# Save DataFrame to Excel
df.to_excel("ABC.xlsx", index=False)

The simplest thing to do is just:

sala['N10'].replace('None', 'vcv', inplace=True)

that should work.

If they were true NaN values then calling fillna would've worked.

e.g.

sala['N10'].fillna(value='vcv', inplace = True)

also what I suggested in my comment:

sala.loc[sala['N10'].isnull(), 'N10'] = 'vcv'

Updated answer, April 2025:

pd.to_numeric can convert arguments to a numeric type. The option errors='coerce' sets things to NaN. However, it can only work on 1D objects (i.e. scalar, list, tuple, 1-d array, or Series). Therefore, to use it on a DataFrame, we need to use df.apply to convert each column individually. Note that any **kwargs given to apply will be passed onto the function, so we can still set errors='coerce'.

Using pd.to_numeric along with df.apply will set any strings to NaN. If we want to convert those to 0 values, we can then use .fillna(0) on the resulting DataFrame.

For example (and note this also works with the strings suggested by the original question "$-" and "($24)"):

import pandas as pd

df = pd.DataFrame({
    'a': (1, 'sd', 1),
    'b': (2., 2., 'fg'),
    'c': (4, "$-", "($24)")
    })

print(df)

#     a    b  c
# 0   1  2.0  4
# 1  sd  2.0     $-
# 2   1   fg  ($24)

df = df.apply(pd.to_numeric, errors='coerce').fillna(0)

print(df)

#      a    b  c
# 0  1.0  2.0  4.0
# 1  0.0  2.0  0.0
# 2  1.0  0.0  0.0

My original answer from 2015, which is now deprecated

You can use the convert_objects method of the DataFrame, with convert_numeric=True to change the strings to NaNs

From the docs:

convert_numeric: If True, attempt to coerce to numbers (including strings), with unconvertible values becoming NaN.

In [17]: df
Out[17]: 
    a   b  c
0  1.  2.  4
1  sd  2.  4
2  1.  fg  5

In [18]: df2 = df.convert_objects(convert_numeric=True)

In [19]: df2
Out[19]: 
    a   b  c
0   1   2  4
1 NaN   2  4
2   1 NaN  5

Finally, if you want to convert those NaNs to 0's, you can use df.replace

In [20]: df2.replace('NaN',0)
Out[20]: 
   a  b  c
0  1  2  4
1  0  2  4
2  1  0  5

This is sufficient

df.fillna("",inplace=True)

df
Out[142]: 
   A                    B  C  D  E
0  A  2014-01-02 02:00:00     A  1
1  B  2014-01-02 03:00:00  B  B  2
2     2014-01-02 04:00:00  C  C   
3  C                       C     4

edit 2021-07-26 complete response following @dWitty's comment

If you really want to keep Nat and NaN values on other than text, you just need fill Na for your text column In your exemple this is A, C, D

You just send a dict of replacement value for your columns. value can be differents for each column. For your case you just need construct the dict

# default values to replace NA (None)
# values = {"A": "", "C": "", "D": ""}
values = (dict([[e,""] for e in ['A','C','D']]))
df.fillna(value=values, inplace=True)

df
Out[142]: 
   A                   B  C  D    E
0  A 2014-01-02 02:00:00     A  1.0
1  B 2014-01-02 03:00:00  B  B  2.0
2    2014-01-02 04:00:00  C  C  NaN
3  C                 NaT  C     4.0