Question

下面是我拥有的pandas data frame的子集

           index             name_matches dist_matches
38  PO1000000345                  M-00346      M-00346
39  PO1000000352                               M-00804
40  PO1000000354                  M-00196      M-00196
41  PO1000000355                  M-00514      M-00514
42  PO1000000382          M-00353,M-00354      M-00354
43  PO1000000411                                      
44  PO1000000451                                      
45  PO1000000512                               M-00680
46  PO1000000530                  M-00089             
47  PO1000000531                  M-00087      M-00087
48  PO1000000553  M-00917,M-00920,M-00922      M-00920

我正在尝试获取一个新列（comb_matches），该列会提取name_matches和dist_matches列中的匹配值。有时，列中会有一个或多个用逗号分隔的值。我想要获取的输出示例如下所示。

           index             name_matches dist_matches  comb_matches
38  PO1000000345                  M-00346      M-00346       M-00346
39  PO1000000352                               M-00804
40  PO1000000354                  M-00196      M-00196       M-00196
41  PO1000000355                  M-00514      M-00514       M-00514
42  PO1000000382          M-00353,M-00354      M-00354       M-00354
43  PO1000000411                                      
44  PO1000000451                                      
45  PO1000000512                               M-00680
46  PO1000000530                  M-00089             
47  PO1000000531                  M-00087      M-00087       M-00087
48  PO1000000553  M-00917,M-00920,M-00922      M-00920       M-00920

有什么简单的方法可以达到上述要求？

Answer 1

没有简单方式。熊猫不是为此类任务而设计的，它不是矢量化的。最好的选择可能是列表理解：

s1 = df['dist_matches'].astype(str)
s2 = df['name_matches'].astype(str).str.split(',')
mask = [i in j for i, j in zip(s1, s2)]

df['comb_match'] = np.where(mask, df['dist_matches'], np.nan)

性能基准测试

要证明Pandas str方法并不是真正的矢量化事实，

# Python 3.6.5, Pandas 0.23.0

def wen(df):
    Bool = df.name_matches.str.split(',',expand=True).isin(df.dist_matches).any(1)    
    df['comb_match'] = np.where(Bool, df.dist_matches, '')
    return df

def jpp(df):
    s1 = df['dist_matches'].astype(str)
    s2 = df['name_matches'].astype(str).str.split(',')
    mask = [i in j for i, j in zip(s1, s2)]
    df['comb_match'] = np.where(mask, df['dist_matches'], np.nan)
    return df

df = pd.concat([df]*1000, ignore_index=True)

assert jpp(df).equals(wen(df))

%timeit jpp(df)  # 12.2 ms
%timeit wen(df)  # 32.7 ms

Answer 2

在str.split之前使用isin。然后我们将布尔值实现为np.where

Bool=df.name_matches.str.split(',',expand=True).isin(df.dist_matches).any(1)    
df['comb_match']=np.where(Bool,df.dist_matches,'')
df
Out[520]: 
           index             name_matches dist_matches comb_match
38  PO1000000345                  M-00346      M-00346    M-00346
39  PO1000000352                               M-00804           
40  PO1000000354                  M-00196      M-00196    M-00196
41  PO1000000355                  M-00514      M-00514    M-00514
42  PO1000000382          M-00353,M-00354      M-00354    M-00354
43  PO1000000411                                                 
44  PO1000000451                                                 
45  PO1000000512                               M-00680           
46  PO1000000530                  M-00089                        
47  PO1000000531                  M-00087      M-00087    M-00087
48  PO1000000553  M-00917,M-00920,M-00922      M-00920    M-00920

根据熊猫中其他两个列的匹配值创建新列

2 个答案:

性能基准测试