如何在不使用拆分器的情况下拆分等长字符串并扩展数据帧

Question

我想在没有拆分器的情况下拆分等长字符串，并扩展数据帧。

下面是我使用的测试数据帧：

sample1 = pd.DataFrame({
        'TST': {1: 1535840000000, 2: 1535840000000}, 
        'RCV': {1: 1535840000000, 2: 1535850000000}, 
        'TCU': {1: 358272000000000, 2: 358272000000000}, 
        'SPD': {1: '0', 2: '00000000000000710000007D007C00E2'}
        })

如您所见，SPD列包含不同长度的字符串，没有任何拆分器。

我希望每隔4个字符将SPD列拆分成新的行，然后将它们扩展到数据帧。

             TST            RCV              TCU   SPD
0  1535840000000  1535840000000  358272000000000  0000
1  1535840000000  1535840000000  358272000000000  0000
2  1535840000000  1535840000000  358272000000000  0000
3  1535840000000  1535840000000  358272000000000  0071
4  1535840000000  1535840000000  358272000000000  0000
5  1535840000000  1535840000000  358272000000000  007D
6  1535840000000  1535840000000  358272000000000  007C
7  1535840000000  1535840000000  358272000000000  00E2

我尝试首先使用下面的代码生成一个序列：

pd.concat([pd.Series(re.findall('....', row['SPD'])) for _, row in sample1.iterrows()]).reset_index()

这给了我们

   index     0
0      0  0000
1      1  0000
2      2  0000
3      3  0071
4      4  0000
5      5  007D
6      6  007C
7      7  00E2

但是我不能把它扩展回sample1

Answer 1

您可以使用str.findall拆分每个4字符中的字符串，然后使用unnesting将结果数据帧从链接的解决方案中解套出来：

sample1['SPD'] = sample1.SPD.str.ljust(4, '0').str.findall(r'.{4}?')
unnesting(sample1, ['SPD'])

   SPD            TST            RCV              TCU
1  0000  1535840000000  1535840000000  358272000000000
2  0000  1535840000000  1535850000000  358272000000000
2  0000  1535840000000  1535850000000  358272000000000
2  0000  1535840000000  1535850000000  358272000000000
2  0071  1535840000000  1535850000000  358272000000000
2  0000  1535840000000  1535850000000  358272000000000
2  007D  1535840000000  1535850000000  358272000000000
2  007C  1535840000000  1535850000000  358272000000000
2  00E2  1535840000000  1535850000000  358272000000000

Answer 2

您可以使用str.findall，然后根据来自SPD的4个字符片的数量对行进行repeat。

from itertools import chain

spd4 = df.pop('SPD').str.findall(r'.{4}') 

(pd.DataFrame(df.values.repeat(spd4.str.len(), axis=0), columns=df.columns)
   .assign(SPD=list(chain.from_iterable(spd4))))

             TST            RCV              TCU   SPD
0  1535840000000  1535850000000  358272000000000  0000
1  1535840000000  1535850000000  358272000000000  0000
2  1535840000000  1535850000000  358272000000000  0000
3  1535840000000  1535850000000  358272000000000  0071
4  1535840000000  1535850000000  358272000000000  0000
5  1535840000000  1535850000000  358272000000000  007D
6  1535840000000  1535850000000  358272000000000  007C
7  1535840000000  1535850000000  358272000000000  00E2

Answer 3

使用Series.str.extractall，然后与原始df连接。

sample1.filter(regex='^(?!SPD)').join(
    sample1.SPD.str.extractall('(?P<SPD>.{4})').reset_index(level=1, drop=True)
) 

#             TST            RCV              TCU   SPD
#1  1535840000000  1535840000000  358272000000000   NaN
#2  1535840000000  1535850000000  358272000000000  0000
#2  1535840000000  1535850000000  358272000000000  0000
#2  1535840000000  1535850000000  358272000000000  0000
#2  1535840000000  1535850000000  358272000000000  0071
#2  1535840000000  1535850000000  358272000000000  0000
#2  1535840000000  1535850000000  358272000000000  007D
#2  1535840000000  1535850000000  358272000000000  007C
#2  1535840000000  1535850000000  358272000000000  00E2

使用内部联接(...how='inner')如果您想排除SPD少于4个字符的行。