Filter out rows with more than certain number of NaN

Question

In a Pandas dataframe, I would like to filter out all the rows that have more than 2 NaNs.

Essentially, I have 4 columns and I would like to keep only those rows where at least 2 columns have finite values.

Can somebody advise on how to achieve this?

Answer 1

You have phrased 2 slightly different questions here. In the general case, they have different answers.

I would like to keep only those rows where at least 2 columns have finite values.

df = df.dropna(thresh=2)

This keeps rows with 2 or more non-null values.

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I would like to filter out all the rows that have more than 2 NaNs

df = df.dropna(thresh=df.shape[1]-2)

This filters out rows with 2 or more null values.

In your example dataframe of 4 columns, these operations are equivalent, since df.shape[1] - 2 == 2. However, you will notice discrepancies with dataframes which do not have exactly 4 columns.

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Note dropna also has a subset argument should you wish to include only specified columns when applying a threshold. For example:

df = df.dropna(subset=['col1', 'col2', 'col3'], thresh=2)

Answer 2

The following should work

df.dropna(thresh=2)

See the online docs

What we are doing here is dropping any NaN rows, where there are 2 or more non NaN values in a row.

Example:

In [25]:

import pandas as pd

df = pd.DataFrame({'a':[1,2,NaN,4,5], 'b':[NaN,2,NaN,4,5], 'c':[1,2,NaN,NaN,NaN], 'd':[1,2,3,NaN,5]})

df

Out[25]:

    a   b   c   d
0   1 NaN   1   1
1   2   2   2   2
2 NaN NaN NaN   3
3   4   4 NaN NaN
4   5   5 NaN   5

[5 rows x 4 columns]

In [26]:

df.dropna(thresh=2)

Out[26]:

   a   b   c   d
0  1 NaN   1   1
1  2   2   2   2
3  4   4 NaN NaN
4  5   5 NaN   5

[4 rows x 4 columns]

EDIT

For the above example it works but you should note that you would have to know the number of columns and set the thresh value appropriately, I thought originally it meant the number of NaN values but it actually means number of Non NaN values.