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

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)

You can use the to_numeric method, but it's not changing the value in place. You need to set the column to the new values:

training_data['usagequantity'] = (
    pd.to_numeric(training_data['usagequantity'],
                  errors='coerce')
      .fillna(0)
    )

to_numeric sets the non-numeric values to NaNs, and then the chained fillna method replaces the NaNs with zeros.

I think you need replace by dict:

cols = ["Weight","Height","BootSize","SuitSize","Type"]
df2[cols] = df2[cols].replace({'0':np.nan, 0:np.nan})

Use mask:

df['X'] = df['X'].mask(df.Y == 1, 0)
df[['X', 'Y']] = df[['X', 'Y']].mask(df.Z == 1, 0)

Another solution with DataFrame.loc:

df.loc[df.Y == 1, 'X'] = 0
df.loc[df.Z == 1, ['X', 'Y']] = 0

print (df)
   X  Y  Z
0  0  0  1
1  0  1  0
2  0  0  1