Dropping infinite values from dataframes in pandas?

The above solution will modify the infs that are not in the target columns. To remedy that,

lst = [np.inf, -np.inf]
to_replace = {v: lst for v in ['col1', 'col2']}
df.replace(to_replace, np.nan)

Here is another method using .loc to replace inf with nan on a Series:

s.loc[(~np.isfinite(s)) & s.notnull()] = np.nan

So, in response to the original question:

df = pd.DataFrame(np.ones((3, 3)), columns=list('ABC'))

for i in range(3): 
    df.iat[i, i] = np.inf

df
          A         B         C
0       inf  1.000000  1.000000
1  1.000000       inf  1.000000
2  1.000000  1.000000       inf

df.sum()
A    inf
B    inf
C    inf
dtype: float64

df.apply(lambda s: s[np.isfinite(s)].dropna()).sum()
A    2
B    2
C    2
dtype: float64

With option context, this is possible without permanently setting use_inf_as_na. For example:

with pd.option_context('mode.use_inf_as_na', True):
    df = df.dropna(subset=['col1', 'col2'], how='all')

Of course it can be set to treat inf as NaN permanently with

pd.set_option('use_inf_as_na', True)

For older versions, replace use_inf_as_na with use_inf_as_null.

Yet another solution would be to use the isin method. Use it to determine whether each value is infinite or missing and then chain the all method to determine if all the values in the rows are infinite or missing.

Finally, use the negation of that result to select the rows that don’t have all infinite or missing values via boolean indexing.

all_inf_or_nan = df.isin([np.inf, -np.inf, np.nan]).all(axis='columns')
df[~all_inf_or_nan]

You can use pd.DataFrame.mask with np.isinf. You should ensure first your dataframe series are all of type float. Then use dropna with your existing logic.

print(df)

       col1      col2
0 -0.441406       inf
1 -0.321105      -inf
2 -0.412857  2.223047
3 -0.356610  2.513048

df = df.mask(np.isinf)

print(df)

       col1      col2
0 -0.441406       NaN
1 -0.321105       NaN
2 -0.412857  2.223047
3 -0.356610  2.513048

Use (fast and simple):

df = df[np.isfinite(df).all(1)]

This answer is based on DougR’s answer in an other question.
Here an example code:

import pandas as pd
import numpy as np
df=pd.DataFrame([1,2,3,np.nan,4,np.inf,5,-np.inf,6])
print('Input:n',df,sep='')
df = df[np.isfinite(df).all(1)]
print('nDropped:n',df,sep='')

Result:

Input:
    0
0  1.0000
1  2.0000
2  3.0000
3     NaN
4  4.0000
5     inf
6  5.0000
7    -inf
8  6.0000

Dropped:
     0
0  1.0
1  2.0
2  3.0
4  4.0
6  5.0
8  6.0

To remove both Nan, and inf using a single command use

df = df[ np.isfinite( df ).all( axis = 1) ]

If for some reason the above doesn’t work for you, please try the following 2 steps:

df = df[ ~( df.isnull().any( axis = 1 ) ) ] #to remove nan
df = df[ ~( df.isin( [np.inf, -np.inf]).any(axis =1) )] #to remove inf

Just stumbled upon this one and I found a one line without replace or numpy:

df = pd.DataFrame(
    [[1, np.inf],
     [1, -np.inf],
     [1, 2]],
    columns=['a', 'b']
)
df.query("b not in [inf, -inf]")
>>> a  b
 2  1  2.0

For some version of pandas, one might need to use back ` around the name of the column b.

Unlike other answers here, this one line code worked for me.

import numpy as np
df= df[df['required_column_name']!= np.inf]