使用pandas中其他两列的匹配项创建新列

Question

下面是pandas data frame我的一部分

           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

我试图得到一个新的列(comb_matches),它拉出name_matches和dist_matches列中的匹配值.有时,列中将有一个或多个值以逗号分隔.我希望获得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

有没有简单的方法来获得上述内容？

Answer 1

没有简单的方法.熊猫不是为这种任务而设计的,它不是可矢量化的.您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)

绩效基准

为了证明Pandas str方法不是真正的矢量化:

# 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