dropping infinite values from dataframes in pandas?

Python Programming

Question or problem about Python programming:

what is the quickest/simplest way to drop nan and inf/-inf values from a pandas DataFrame without resetting mode.use_inf_as_null? I’d like to be able to use the subset and how arguments of dropna, except with inf values considered missing, like:

df.dropna(subset=["col1", "col2"], how="all", with_inf=True)

is this possible? Is there a way to tell dropna to include inf in its definition of missing values?

How to solve the problem:

Solution 1:

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

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

and then use the dropna:

df.replace([np.inf, -np.inf], np.nan).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)
0   1
1   2
2 NaN
3 NaN

The same method would work for a Series.

Solution 2:

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.

Solution 3:

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

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

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

Solution 4:

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


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

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

Solution 5:

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')

Hope this helps!