python - 如果列具有一定百分比的 NaN 值，则返回列 (Python)

Question

希望仅返回至少有 25% NaN 值的列作为新的 df

I'm thinking either a conditional statement using .loc, .isnull, or count, but I'm not certain what the most efficient method is. Appreciate any and all assistance.

DF:

df1:
(axis 1 = A,B,C = series)

    A    B    C
1   1    2    1
2   NaN  NaN  3
3   4    NaN  1
4   2    NaN  4 

思考:

df.loc[df['series'] == nan >= 25% ]

或者类似的东西:

if count(nan) for column(x) in 'series' is >= (.25 * (count(x)))
    return loc[x]

返回新数据框:

df2:

    A    B    
1   1    2    
2   NaN  NaN  
3   4    NaN  
4   2    NaN

返回 A 和 B，因为它们中至少有 25% 的列条目为 NaN(缺失)

Answer 1

基于 https://datascience.stackexchange.com/q/12645 的回复.

na_count_mask = df.isna().sum(axis=0) >= (col_count // 4)

res_df = df.loc[na_count_mask]