Dropping infinite values from dataframes in pandas?

Question

The simplest way would be to first replace() infs to NaN:

df.replace([np.inf, -np.inf], np.nan, inplace=True)

and then use the dropna():

df.replace([np.inf, -np.inf], np.nan, inplace=True) \
    .dropna(subset=["col1", "col2"], how="all")

For example:

In [11]: df = pd.DataFrame([1, 2, np.inf, -np.inf])

In [12]: df.replace([np.inf, -np.inf], np.nan, inplace=True)
Out[12]:
    0
0   1
1   2
2 NaN
3 NaN

The same method would work for a Series.

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.

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

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