如何在任何行在 Python pandas dataframe 中具有 NaN 值后删除列

Question

Toy example code

Let's say I have following DataFrame :

import pandas as pd
import numpy as np
df = pd.DataFrame({"A":[11,21,31], "B":[12,22,32], "C":[np.nan,23,33], "D":[np.nan,24,34], "E":[15,25,35]})

Which would return:

>>> df
    A   B     C     D   E
0  11  12   NaN   NaN  15
1  21  22  23.0  24.0  25
2  31  32  33.0  34.0  35

Remove all columns with nan values

I know how to remove all the columns which have any row with a nan value like this:

out1 = df.dropna(axis=1, how="any")

Which returns:

>>> out1
    A   B   E
0  11  12  15
1  21  22  25
2  31  32  35

Expected output

However what I expect is to remove all columns after a nan value is found. In the toy example code the expected output would be:

    A   B
0  11  12
1  21  22
2  31  32

Question

How can I remove all columns after a nan is found within any row in a pandas DataFrame ?

Answer 1

What I would do:

1. check every element for being null/not null
2. cumulative sum every row across the columns
3. check any for every column, across the rows
4. use that result as an indexer:

df.loc[:, ~df.isna().cumsum(axis=1).any(axis=0)]

Give me:

    A   B
0  11  12
1  21  22
2  31  32

Answer 2

I could find a way as follows to get the expected output:

colFirstNaN = df.isna().any(axis=0).idxmax() # Find column that has first NaN element in any row
indexColLastValue = df.columns.tolist().index(colFirstNaN) -1
ColLastValue = df.columns[indexColLastValue]
out2 = df.loc[:, :ColLastValue]

And the output would be then:

>>> out2
    A   B
0  11  12
1  21  22
2  31  32