drops a column if it exceeds a specific number of NA values

Question

i want to write a program that drops a column if it exceeds a specific number of NA values .This is what i did.

def check(x):
for column in df:
    if df.column.isnull().sum() > 2:
        df.drop(column,axis=1)

there is no error in executing the above code , but while doing df.apply(check), there are a ton of errors.

P.S:I know about the thresh arguement in df.dropna(thresh,axis)

Any tips?Why isnt my code working?

Thanks

Answer 1

Although jezrael's answer works that is not the approach you should do. Instead, create a mask: ~df.isnull().sum().gt(2) and apply it with .loc[:,m] to access columns.

Full example:

import pandas as pd
import numpy as np

df = pd.DataFrame({
    'A':list('abcdef'),
    'B':[np.nan,np.nan,np.nan,5,5,np.nan],
    'C':[np.nan,8,np.nan,np.nan,2,3],
    'D':[1,3,5,7,1,0],
    'E':[5,3,6,9,2,np.nan],
    'F':list('aaabbb')
})

m = ~df.isnull().sum().gt(2)
df = df.loc[:,m]

print(df)

Returns:

   A  D    E  F
0  a  1  5.0  a
1  b  3  3.0  a
2  c  5  6.0  a
3  d  7  9.0  b
4  e  1  2.0  b
5  f  0  NaN  b

Explanation

Assume we print the columns and the mask before applying it.

print(df.columns.tolist())
print(m.tolist())

It would return this:

['A', 'B', 'C', 'D', 'E', 'F']
[True, False, False, True, True, True]

Columns B and C are unwanted (False). They are removed when the mask is applied.

Answer 2

I think best here is use dropna with parameter thresh:

thresh : int, optional

Require that many non-NA values.

So for vectorize solution subtract it from length of DataFrame:

N = 2
df = df.dropna(thresh=len(df)-N, axis=1)
print (df)
   A  D    E  F
0  a  1  5.0  a
1  b  3  3.0  a
2  c  5  6.0  a
3  d  7  9.0  b
4  e  1  2.0  b
5  f  0  NaN  b

---

I suggest use DataFrame.pipe for apply function for input DataFrame with change df.column to df[column], because dot notation with dynamic column names from variable failed (it try select column name column):

df = pd.DataFrame({'A':list('abcdef'),
                   'B':[np.nan,np.nan,np.nan,5,5,np.nan],
                   'C':[np.nan,8,np.nan,np.nan,2,3],
                   'D':[1,3,5,7,1,0],
                   'E':[5,3,6,9,2,np.nan],
                   'F':list('aaabbb')})

print (df)
   A    B    C  D    E  F
0  a  NaN  NaN  1  5.0  a
1  b  NaN  8.0  3  3.0  a
2  c  NaN  NaN  5  6.0  a
3  d  5.0  NaN  7  9.0  b
4  e  5.0  2.0  1  2.0  b
5  f  NaN  3.0  0  NaN  b

def check(df):
    for column in df:
        if df[column].isnull().sum() > 2:
            df.drop(column,axis=1, inplace=True)
    return df
            
print (df.pipe(check))
   A  D    E  F
0  a  1  5.0  a
1  b  3  3.0  a
2  c  5  6.0  a
3  d  7  9.0  b
4  e  1  2.0  b
5  f  0  NaN  b
        
        

Answer 3

Alternatively, you can use count which counts non-null values

In [23]: df.loc[:, df.count().gt(len(df.index) - 2)]
Out[23]:
   A  D    E  F
0  a  1  5.0  a
1  b  3  3.0  a
2  c  5  6.0  a
3  d  7  9.0  b
4  e  1  2.0  b
5  f  0  NaN  b