Question

我在DataFrame中有一列，其中包含一个字符串，我必须使用不同的分隔符从中检索两条信息：

ID  STR
280 11040402-38.58551%;11050101-9.29086%;11070101-52.12363%
351 11130203-35%;11130230-65%
510 11070103-69%
655 11090103-41.63463%;11160102-58.36537%
666 11130205-50.00%;11130207-50%

我一直在尝试将本系列的.apply方法与lambda函数一起使用，一口气进行拆分，无济于事：

df['STR'].apply(lambda x: y.split('-') for y in x.split(';'))

理想情况下，不仅可以一次性拆分字符串，还可以将-的左侧与右侧分开：

ID  STR.LEFT                       STR.RIGHT
280 [11040402, 11050101, 11070101] [38.58551%, 9.29086%, 52.12363%]
351 [11130203, 11130230]           [35%, 65%]
510 [11070103]                     [69%]
655 [11090103, 11160102]           [41.63463%, 58.36537%]
666 [11130205, 11130207]           [50.00%, 50%]

我相信这可以通过.apply和切片来实现，但是任何其他解决方案都欢迎。

Answer 1

您可以尝试拆分几次：

# set ID as index
df.set_index('ID', inplace=True)


new_series = df.STR.str.split(';', expand=True).stack().reset_index(level=-1,drop=True)

new_df = new_series.str.split('-', expand=True)

new_df.groupby('ID').agg(list).reset_index()

输出：

      ID  0                                     1
--  ----  ------------------------------------  --------------------------------------
 0   280  ['11040402', '11050101', '11070101']  ['38.58551%', '9.29086%', '52.12363%']
 1   351  ['11130203', '11130230']              ['35%', '65%']
 2   510  ['11070103']                          ['69%']
 3   655  ['11090103', '11160102']              ['41.63463%', '58.36537%']
 4   666  ['11130205', '11130207']              ['50.00%', '50%']

Answer 2

`var geojesonlayer1 = new google.maps.Data(); var geojesonlayer2 = new google.maps.Data(); geojesonlayer1.loadGeoJson('mygeojson path'); geojesonlayer2.loadGeoJson('mygeojson path'); //layer style geojesonlayer1.setStyle({ strokeColor: 'yellow', strokeWeight: 5 }); geojesonlayer2.setStyle({ strokeColor: 'blue', strokeWeight: 1 }); //add layer to map geojesonlayer1.setMap(map); geojesonlayer2.setMap(map); //remove layer from map geojesonlayer1.setMap(null); geojesonlayer2.setMap(null);`

假设模式始终离开str.split

'l-r;l-r;l-r...'

如果要将这些列表分解成单独的行

s = df.STR.str.split('-|;')
df[['ID']].join(pd.concat({'STR.LEFT': s.str[::2], 'STR.RIGTH': s.str[1::2]}, axis=1))

    ID                        STR.LEFT                         STR.RIGTH
0  280  [11040402, 11050101, 11070101]  [38.58551%, 9.29086%, 52.12363%]
1  351            [11130203, 11130230]                        [35%, 65%]
2  510                      [11070103]                             [69%]
3  655            [11090103, 11160102]            [41.63463%, 58.36537%]
4  666            [11130205, 11130207]                     [50.00%, 50%]

Answer 3

单个str.extractall调用就足以将对提取到单独的列中。然后，您可以使用groupby将它们聚合到列表中。

(df['STR'].str.extractall(r'(.*?)-(.*?)(?=;|$)')
          .groupby(level=0)
          .agg(list)
          .set_axis(['STR.LEFT', 'STR.RIGHT'], axis=1, inplace=False))

                           STR.LEFT                         STR.RIGHT
0  [11040402, ;11050101, ;11070101]  [38.58551%, 9.29086%, 52.12363%]
1  [11130203, ;11130230]             [35%, 65%]                      
2  [11070103]                        [69%]                           
3  [11090103, ;11160102]             [41.63463%, 58.36537%]          
4  [11130205, ;11130207]             [50.00%, 50%]

要加入ID，只需使用join。

(df['STR'].str.extractall(r'(.*?)-(.*?)(?=;|$)')
          .groupby(level=0)
          .agg(list)
          .set_axis(['STR.LEFT', 'STR.RIGHT'], axis=1, inplace=False)
          .join(df['ID'])

                           STR.LEFT                         STR.RIGHT   ID
0  [11040402, ;11050101, ;11070101]  [38.58551%, 9.29086%, 52.12363%]  280
1  [11130203, ;11130230]             [35%, 65%]                        351
2  [11070103]                        [69%]                             510
3  [11090103, ;11160102]             [41.63463%, 58.36537%]            655
4  [11130205, ;11130207]             [50.00%, 50%]                     666

在DataFrame中多次拆分字符串

3 个答案: