python - 解码列并使用新数据创建新数据框

Question

我有一个具有以下结构的数据框:

Timestamp         Payload                                                                                                     
2022-03-16 18:00  3c313034393536353534303b31332e32333534313b303b302e303637313b302e303138313b332e33353b36322e31353338323e
2022-03-16 18:10  3c313034393536363134303b31332e32313134383b303b302e303637323b302e3031383b342e3632353b37332e34353133313e
2022-03-16 18:20  3c313034393536363734303b31332e31393630313b303b302e303637323b302e3031383b342e303132353b36322e34343238353e     
2022-03-16 18:30  3c313034393537353734303b31332e31303137343b303b302e3036373b302e303137393b342e383132353b37352e343931393e     
2022-03-16 18:40  3c313034393538323934303b31332e30353934353b303b302e303636373b302e303137393b332e313337353b37352e37313830393e  

我想要的是在这两个字段出现的地方创建一个新的数据帧，以及有效负载列的解码。我遇到的困难是对那组字符的解码会生成一组数据。例如，如果我们解码以下行，

3c313034393536353534303b31332e32333534313b303b302e303637313b302e303138313b332e33353b36322e31353338323e

我们得到，

<1049565540;13.23541;0;0.0671;0.0181;3.35;62.15382>

其中字符< , >和 ;是分隔符。所以我的目标是获得类似于以下数据框的内容，

2022-03-16 18:00  1049565540  13.23541  0  0.0671  0.0181  3.3500  62.15382  3c313034393536353534303b31332e32333534313b303b302e303637313b302e303138313b332e33353b36322e31353338323e
2022-03-16 18:10  1049566140  13.21148  0  0.0672  0.0180  4.6250  73.45131  3c313034393536363134303b31332e32313134383b303b302e303637323b302e3031383b342e3632353b37332e34353133313e
2022-03-16 18:20  1049566740  13.19601  0  0.0672  0.0180  4.0125  62.44285  3c313034393536363734303b31332e31393630313b303b302e303637323b302e3031383b342e303132353b36322e34343238353e     
2022-03-16 18:30  1049575740  13.10174  0  0.0670  0.0179  4.8125  75.49190  3c313034393537353734303b31332e31303137343b303b302e3036373b302e303137393b342e383132353b37352e343931393e     
2022-03-16 18:40  1049582940  13.05945  0  0.0667  0.0179  3.1375  75.71809  3c313034393538323934303b31332e30353934353b303b302e303636373b302e303137393b332e313337353b37352e37313830393e  

实际上我正在执行以下代码，没关系，但我确定我没有利用 Pandas 的潜力图书馆。

def myFunc(pay):
    decode = bytearray.fromhex(pay[2:-2]).decode()
    return decode.split(";")
  
output2['datos'] = output2['payload'].apply(myFunc)
payloads = output2['datos'].values
df = pd.DataFrame(payloads.tolist(), columns=None)

能否以更高效的方式完成？

非常感谢!

Answer 1

如果想要通过特殊功能工作是可能的，请使用 .apply 作为您的解决方案或列表理解:

L = [bytearray.fromhex(pay[2:-2]).decode().split(";") for pay in output2.pop('Payload')]
df = pd.DataFrame(L, index=output2.index)

上次使用加入:

df = output2.join(df)
print (df)
          Timestamp           0         1  2       3       4       5         6
0  2022-03-16 18:10  1049566140  13.21148  0  0.0672   0.018   4.625  73.45131
1  2022-03-16 18:20  1049566740  13.19601  0  0.0672   0.018  4.0125  62.44285
2  2022-03-16 18:30  1049575740  13.10174  0   0.067  0.0179  4.8125   75.4919
3  2022-03-16 18:40  1049582940  13.05945  0  0.0667  0.0179  3.1375  75.71809

如果还需要原创专栏:

L = [bytearray.fromhex(pay[2:-2]).decode().split(";") for pay in output2['Payload']]
df = pd.DataFrame(L, index=output2.index)
df = output2.join(df)
print (df)
          Timestamp                                            Payload  \
0  2022-03-16 18:10  3c313034393536363134303b31332e32313134383b303b...   
1  2022-03-16 18:20  3c313034393536363734303b31332e31393630313b303b...   
2  2022-03-16 18:30  3c313034393537353734303b31332e31303137343b303b...   
3  2022-03-16 18:40  3c313034393538323934303b31332e30353934353b303b...   

            0         1  2       3       4       5         6  
0  1049566140  13.21148  0  0.0672   0.018   4.625  73.45131  
1  1049566740  13.19601  0  0.0672   0.018  4.0125  62.44285  
2  1049575740  13.10174  0   0.067  0.0179  4.8125   75.4919  
3  1049582940  13.05945  0  0.0667  0.0179  3.1375  75.71809