python - 提取数据直到最后，而不是使用多个next(v)[1]

嗨，我有一个python代码，可以在将它们解析为out.csv之前从sample_data.csv中获取数据。

请查看图像以更好地查看sample_data.csv
http://i.imgur.com/wwi4RC7.jpg

我的问题是我如何从最后一个next（v）[1]开始

datetime = next(v)[1],next(v)[1],next(v)[1],next(v)[1],next(v)[1]

并从头开始直到行尾，而不是愚蠢地使用多个next（v）[1]？这是一个问题，因为不同的收据具有不同的行数，因此我无法确定交易的next（v）[1]的固定编号


  transaction = next（v）[1]，next（v）[1]，
  next（v）[1]，next（v）[1]，next（v）[1]，next（v）[1]，next（v）[1]，next（v）[1]

import csv
    from itertools import groupby
    from operator import itemgetter
    import re

    with open("sample_data.csv", "rb") as f, open("out.csv", "wb") as out:
        reader = csv.reader(f)
        next(reader)
        writer = csv.writer(out)
        writer.writerow(["Receipt ID","Name","Address","Date","Time","Items","Amount","Cost","Total"])
        groups = groupby(csv.reader(f), key=itemgetter(0))
        for k, v in groups:
            id_, name = next(v)
            add_date_1, add_date_2 = next(v)[1], next(v)[1]
            combinedaddress = add_date_1+ " " +add_date_2
            datetime = next(v)[1],next(v)[1],next(v)[1],next(v)[1],next(v)[1]
            abcd = str(datetime)
            dateprinter = re.search('(\d\d/\d\d/\d\d\d\d)\s(\d\d:\d\d)', abcd).group(1)
            timeprinter = re.search('(\d\d/\d\d/\d\d\d\d)\s(\d\d:\d\d)', abcd).group(2)

            transaction= next(v)[1], next(v)[1], next(v)[1],next(v)[1],next(v)[1],next(v)[1],next(v)[1],next(v)[1]
            writer.writerow([id_, name, combinedaddress, dateprinter, timeprinter, transaction])

最佳答案

如果我正确理解了这个问题，则可以使用列表推导完成v中的值读取，如下所示：

transaction = [ x[1] for x in v ]

该代码就像抓取所有剩余的next(v)[1]直到v结束。

旁注：始终调用next（v）十分丑陋且不切实际，您可以先将v转换为列表，然后使用简单的列表切片来获得所需的内容：

 v = list(v)
 id_, name = v[0]
 add_date_1, add_date_2 = [x[1] for x in v[1:3]]
 ...
 transaction = [ x[1] for x in v[1234:] ]

关于python - 提取数据直到最后，而不是使用多个next(v)[1]，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/39739093/