将具有多值字段的 CSV 中的 Python 列表转换为 Python 嵌套列表，对嵌套列表值进行排序并导出为 CSV

Question

我使用 Python csv 模块将具有多值字段的 csv 转换为 Python list。输出包含具有多个相关值的字段。

['Route', 'Vehicles', 'Vehicle Class', 'Driver_ID', 'Date', 'Start', 'Arrive']
['ABC', 'ZYG098, AB0134, GF0158', 'A1, B2, C3', 'John Doe, Jane Doe, Abraham Lincoln', '20150301', 'A', 'B']
['AC', 'ZGA123', 'C3', 'George Washington', '20150301', 'A', 'C']
['ABC', 'XAZ012, AB0134, YZ089', 'C1, B2, A2 ', 'John Adams, Jane Doe, Thomas Jefferson', '20150302', 'A', 'B']

我想将 Vehicles、Vehicle Class 和 Driver ID 字段转换为嵌套列表，这样如果我对 Vehicle row[1] 中的每个子列表进行排序，以确保车辆始终按字母顺序出现在子列表中，并且车辆类别和驾驶员保持各自正确的顺序。所以标题和第一行子列表的排列方式如下：

['Route', 'Vehicles', 'Vehicle Class', 'Driver_ID', 'Date', 'Start', 'Arrive']
['ABC', 'AB0134, GF0158, ZYG098', 'B2, C3, A1', 'Jane Doe, Abraham Lincoln, John Doe', '20150301', 'A', 'B']
['AC', 'ZGA123', 'C3', 'George Washington', '20150301', 'A', 'C']
['ABC', 'AB0134, YZ089, XAZ012', 'B2, A2, C1', 'Jane Doe, Thomas Jefferson, John Adams', '20150302', 'A', 'B']

因此，在上面的输出中，车辆的每个子组/列表都按字母顺序排序，并且根据需要重新排列车辆类别和驾驶员 ID，以保持它们与各自车辆的原始关系（即驾驶员 ID - John Doe 驾驶Vehicle - ZYG098 是 Vehicle Class - A1，因此这些项目在其子列表中移动以反映 ZYG098 现在是最后一个，而不是第一个）。如果可以做到这一点，您将如何将生成的嵌套列表导出回带有原始标题的 CSV？

抱歉，如果这很简单或荒谬，我刚刚开始学习 Python。如果嵌套列表不是最佳选择，我愿意接受任何其他解决方案（对于字典，我需要加入字段来创建键，因为没有组合 Route_Date 就没有唯一键）。如果有人有可靠的资源来使用 Python 处理各种 CSV 用例，那么推荐会很棒。

提前感谢您的耐心和帮助。

Answer 1

终于在同一页面上，花了一些工作，但这会做你想要的：

from itertools import chain
import csv


l = [['Route', 'Vehicles', 'Vehicle Class', 'Driver_ID', 'Date', 'Start', 'Arrive'],
     ['ABC', 'ZYG098, AB0134, GF0158', 'A1, B2, C3', 'John Doe, Jane Doe, Abraham Lincoln', '20150301', 'A', 'B'],
     ['AC', 'ZGA123', 'C3', 'George Washington', '20150301', 'A', 'C'],
     ['ABC', 'XAZ012, AB0134, YZ089', 'C1, B2, A2 ', 'John Adams, Jane Doe, Thomas Jefferson', '20150302', 'A', 'B']]
it = map(list,zip(*l))

# transpose original list, row-columns, columns-rows
it =  zip(*l)

# get each column separately, using iter so we can pop first element
# off to get headers efficiently 
route, veh, veh_c, d_id, date, start, arrive = iter(iter(next(it))), iter(next(it)), iter(next(it)), iter(next(it)), iter(next(it)), iter(next(it)), iter(next(it))

# get all headers to write later
headers = next(route), next(veh), next(veh_c), next(d_id), next(date), next(start), next(arrive)

srt_veh = []
key_inds = []

# sort vehicle elements and keep a record of old indexes
# so subelements in Vehicle_class and driver_id can be rearranged to match
for x in veh:
    srt = sorted(x.split(","))
    key_inds.append([x.split(",").index(w) for w in srt])
    srt_veh.append(",".join(srt).strip())

srt_veh_cls = []

# sort vehicle class based on old index of elements in vehicles
# and rejoin split elements
for ind, ele in enumerate(veh_c):
    spl = ele.split(",")
    srt_veh_cls.append(",".join([spl[i].strip() for i in key_inds[ind]]))

srt_dr_id = []

# sort driver_ids  based on old index of elements in vehicle
# and join subelements again after splitting and sorting
for ind, ele in enumerate(d_id):
    spl = ele.split(",")
    srt_dr_id.append(",".join([spl[i].strip() for i in key_inds[ind]]))

 # transpose again for writing
zipped = zip(*(route, srt_veh, srt_veh_cls,
           srt_dr_id, date, start, arrive))

终于用 csv.writerows 写了：

with open("out.csv", "w") as f:
    wr = csv.writer(f)
    wr.writerow(headers)
    wr.writerows(zipped)

输出：

Route,Vehicles,Vehicle Class,Driver_ID,Date,Start,Arrive
ABC,"AB0134, GF0158,ZYG098","B2,C3,A1","Jane Doe,Abraham Lincoln,John Doe",20150301,A,B
AC,ZGA123,C3,George Washington,20150301,A,C
ABC,"AB0134, YZ089,XAZ012","B2,A2,C1","Jane Doe,Thomas Jefferson,John Adams",20150302,A,B

对于 python 2，将 zip 替换为 itertools.izip，将 map 替换为 itertools.imap：

from itertools import izip, imap

您可以压缩更多内容并做一些事情来缩短代码，但我认为这无助于提高可读性。

Answer 2

要转换为您描述的嵌套格式：

nested = zip(*lst)

而 zip 是它自己的逆：

orig = zip(*nested)

但也许你真正想要的是：

import operator

sort = sorted(lst[1:], key=operator.itemgetter(1))

这为您提供了一个按第 1 行排序的新列表。在这种情况下，您没有更改数据的格式，因此您应该能够将其转储为 csv 而无需修改，尽管您需要预先添加来自 lst[0] 的原始标题。