How do I drop nan, inf, and -inf values from a DataFrame without resetting mode.use_inf_as_null?
Can I tell dropna to include inf in its definition of missing values so that the following works?
df.dropna(subset=["col1", "col2"], how="all")
First replace() infs with NaN:
df.replace([np.inf, -np.inf], np.nan, inplace=True)
and then drop NaNs via dropna():
df.dropna(subset=["col1", "col2"], how="all", inplace=True)
For example:
>>> df = pd.DataFrame({"col1": [1, np.inf, -np.inf], "col2": [2, 3, np.nan]})
>>> df
col1 col2
0 1.0 2.0
1 inf 3.0
2 -inf NaN
>>> df.replace([np.inf, -np.inf], np.nan, inplace=True)
>>> df
col1 col2
0 1.0 2.0
1 NaN 3.0
2 NaN NaN
>>> df.dropna(subset=["col1", "col2"], how="all", inplace=True)
>>> df
col1 col2
0 1.0 2.0
1 NaN 3.0
The same method also works for Series.
inf values to a predefined int such as 0, in a certain column?
.replace(..., 0). To just do on columns you update those columns i.e. df[cols] = df[cols].replace(..., 0)
replace does not work in-place, so a new DataFrame is returnedDEPRECATED
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_inf_as_null had been deprecated and will be removed in a future version. Use use_inf_as_na instead. Add to/update answer?
inf as nulls at the global setting levels instead of operational level. This could potentially saves time imputing the values first.
FutureWarning: use_inf_as_na option is deprecated and will be removed in a future version. Convert inf values to NaN before operating instead.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
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)
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]
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
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
Unlike other answers here, this one line code worked for me.
import numpy as np
df= df[df['required_column_name']!= np.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.
