刪除行，如果任何一組的值爲空

Question

我有一個DataFrame有很多列，我想刪除行的列的值爲列爲空。我知道如何與一列做到這一點：

df = df[df['Column'] != ''] 

我想用一組列的做到這一點，像這樣：

df = df['' not in [df['Column1'], df['Column2'], df['Column3']]' 

然而，這給出了錯誤：

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

我該怎麼做？

Answer 1

如果值是空字符串創建子集，每行的所有True的add all或any：

df = df[(df[['Column1', 'Column2', 'Column1']] != '').all(axis=1)] 

df = df[~(df[['Column1', 'Column2', 'Column1']] == '').any(axis=1)] 

如果值是NaN S，None小號使用dropna加參數subset：

df = df.dropna(subset=['Column1', 'Column2', 'Column1']) 

樣品：

df = pd.DataFrame({'A':[np.nan,'','p','hh','f'], 
        'B':['',np.nan,'','','o'], 
        'C':['a','s','d','f','g'], 
        'D':['f','g','h','j','k'], 
        'E':['l','i',np.nan,'u','o'], 
        'F':['','','o','i',np.nan]}) 

print (df) 
    A B C D E F 
0 NaN  a f l  
1  NaN s g i  
2 p  d h NaN o 
3 hh  f j u i 
4 f o g k o NaN 

df1 = df.dropna(subset=['A', 'B', 'F']) 
print (df1) 
    A B C D E F 
2 p d h NaN o 
3 hh f j u i 

df2 = df[(df[['A', 'B', 'F']] != '').all(axis=1)] 
print (df2) 
    A B C D E F 
4 f o g k o NaN 

df2 = df[~(df[['A', 'B', 'F']] == '').any(axis=1)] 
print (df2) 
    A B C D E F 
4 f o g k o NaN 

編輯：

爲了比較字符串和一些列的數字得到：

TypeError: Could not compare [''] with block values

有它2個解決方案 - 比較創建numpy的陣列由values或轉換值到string s由astype：

df = pd.DataFrame({'A':[np.nan,7,8,8,8], 
        'B':['',np.nan,'','','o'], 
        'C':['a','s','d','f','g'], 
        'D':['f','g','h','j','k'], 
        'E':['l','i',np.nan,'u','o'], 
        'F':['','','o','i',np.nan]}) 

print (df) 
    A B C D E F 
0 NaN  a f l  
1 7.0 NaN s g i  
2 8.0  d h NaN o 
3 8.0  f j u i 
4 8.0 o g k o NaN 

df2 = df[(df[['A', 'B', 'F']].values != '').all(axis=1)] 
print (df2) 
    A B C D E F 
4 8.0 o g k o NaN 

df2 = df[(df[['A', 'B', 'F']].astype(str) != '').all(axis=1)] 
print (df2) 
    A B C D E F 
4 8.0 o g k o NaN 

Answer 2

df.dropna()是你在找什麼