解析文本文件，根据其索引位置提取值

Question

嗨，伙计们，你好吗？我希望你没事！ 如何解析使用索引位置提取特定值的文本文件，将值附加到列表，然后将其转换为dataframe。到目前为止，我能够编写以下代码： 文本示例：

header:0RCPF049100000084220210407
body:1927907801100032G 00sucess
1067697546140032G 00sucess
1053756666000032G 00sucess
1321723368900032G 00sucess
1037673956810032G 00sucess

例如，第一行是标题，从中，我只需要位于以下索引位置的日期： date_from_header=linhas[0][18:26] 其余的值在body中

import csv
import pandas as pd

headers = ["data_mov", "chave_detalhe", "cpf_cliente", "cd_clube",
           "cd_operacao","filler","cd_retorno","tc_recusa"]

# This is the actual code
with open('RCPF0491.20210407.1609.txt', "r")as f:
  linhas = [linha.rstrip() for linha in f.readlines()]
  for i in range(0,len(linhas)):
     data_mov = linhas[0][18:26]
     chave_detalhe=linhas[1][0:1]
     cpf_cliente=linhas[1][1:12]
     cd_clube=linhas[1][12:16]
     cd_operacao=linhas[1][16:17]
     filler=linhas[1][17:40]
     cd_retorno=linhas[1][40:42]
     tx_recusa=linhas[1][42:100]
data = [data_mov,chave_detalhe,cpf_cliente,cd_clube,cd_operacao","filler,cd_retorno,tc_recusa]

预期结果如下所示：

data_mov chave_detalhe cpf_cliente cd_clube cd_operacao filler cd_retorno tx_recusa
'20210407' '1'         92790780110 '0032'   'G'        'blank space' '00'   'sucesso'
'20210407' '1'         92790780110 '0032'   'G'        'blank space' '00'   'sucesso'
'20210407' '1'         92790780110 '0032'   'G'        'blank space' '00'   'sucesso'

Answer 1

我感谢SamBob的帮助，以下是最终解决方案，以防任何人需要：

import itertools
import pandas as pd

pd.options.display.width = 0

def parse_file(filename):
    indices=[0,1,12,16,17,18,42]  # list of indexes
    parsed_data = [] # return a list
    with open(filename) as f:
        header = next(f) 
        data_mov = header[18:26]
        for line in itertools.islice(f,1,100): 
            # dividr de acordo com os índices.
            parts = [data_mov] + [line.rstrip()[i:j] for i,j in zip(indices, indices[1:]+[None])]
            parsed_data.append(parts)
            
            # convert to dataframe
            cols = ['data_mov', 'chave_detalhe', 'cpf_cliente','cd_clube','cd_operacao','filler','cd_retorno','tx_recusa']
            df = pd.DataFrame(parsed_data, columns=cols)

    return df


df = (parse_file("filename.txt"))

Answer 2

使用stackoverflow.com/a/10851479/1581658

def parse_file(filename):
    indices = [0,1,12,16,17,18,20] # list the indices to split on
    parsed_data = [] # returned array by line
    with open(filename) as f:
        header = next(f) #skip the header
        data_mov = header[18:26] # and get data_mov from header
        for line in f: #loop through lines
            #split each line by the indices
            parts = [data_mov] + [line.rstrip()[i:j] for i,j in zip(indices, indices[1:]+[None])]
            parsed_data.append(parts)
    return parsed_data

print(parse_file("filename.txt"))