Question

我在list中有以下示例python：

room_list = [['Room A','ok'],['Room B','ok'],['Room C','ok'],['Room D','ok'],['Room E','ok'],['Room F','ok'],['Room G','ok'],['Room H','ok'],['Room I','ok']]

我正在尝试来自list list的另一个['%d. Room number: %s']，%d其中%s的数字越来越多，5是房间号值从上面的列表（每个内部列表中的第1项）。但我想要做的是跳过%d的{{1}}号并将其称为6 - 意味着以下应该是上面示例列表的新列表的输出：

new_room_list = ['1. Room number: Room A','2. Room number: Room B','3. Room number:  Room C','4. Room number: Room D','6. Room number: Room E','7. Room number: Room F','8. Room number: Room G','9. Room number: Room H','10. Room number: Room I']

正如您所看到的，我不想跳过任何数据，只是跳过号码5并将其称为6（上面的输出Room E应该是5但被命名为6）。我为代码编写了这一行，但它没有跳过第5行：

new_room_list = ["%d. Room number: %s" % (i,*(x[:-1])) for i, x in enumerate(room_list, 1)]

如何跳过5并将其重命名为6？

Answer 1

new_room_list = ["%d. Room number: %s" % (i+1 if i > 4 else i,*(x[:-1])) for i, x in enumerate(room_list, 1)]

输出：

['1. Room number: Room A', '2. Room number: Room B', '3. Room number: Room C', '4. Room number: Room D', '6. Room number: Room E', '7. Room number: Room F', '8. Room number: Room G', '9. Room number: Room H', '10. Room number: Room I']

我所做的唯一更改是将(i,*(x[:-1]))替换为(i+1 if i > 4 else i,*(x[:-1]))

我使用了python的三元运算符版本，其使用方法如下：

a if condition else b

Answer 2

只需使用自定义迭代器，itertools通常可以帮助您解决此问题：

In [7]: from itertools import chain, count

In [8]: it = zip(chain(range(1, 5), count(6)), room_list)

In [9]: ['%d. Room number: %s'%(i, rm) for i, (rm,  _) in it]
Out[9]:
['1. Room number: Room A',
 '2. Room number: Room B',
 '3. Room number: Room C',
 '4. Room number: Room D',
 '6. Room number: Room E',
 '7. Room number: Room F',
 '8. Room number: Room G',
 '9. Room number: Room H',
 '10. Room number: Room I']

小心，chain(range(1, 6), count(7))是一个无限的迭代器。当最短的迭代器停止时zip停止，但是，例如，如果你使用itertools.zip_longest，则会爆炸。

Answer 3

您可以使用next：

room_list = [['Room A','ok'],['Room B','ok'],['Room C','ok'],['Room D','ok'],['Room E','ok'],['Room F','ok'],['Room G','ok'],['Room H','ok'],['Room I','ok']]
l = iter(range(1, len(room_list)+2))
final_rooms = ["{}. Room number: {}".format(next(l) if i != 4 else [next(l), next(l)][0], a) for i, (a, _) in enumerate(room_list, start = 1)]

输出：

['1. Room number: Room A', '2. Room number: Room B', '3. Room number: Room C', '4. Room number: Room D', '6. Room number: Room E', '7. Room number: Room F', '8. Room number: Room G', '9. Room number: Room H', '10. Room number: Room I']

Answer 4

这是一种可视化的简单方法;解压缩你拥有的单例并将其分解为其组件。进行正常追加for循环，并根据需要重新创建单例。

#! /usr/bin/env python3
from pprint import PrettyPrinter

pp = PrettyPrinter(indent=2)

rooms_list = [
    ["Room A", "ok"],
    ["Room B", "ok"],
    ["Room C", "ok"],
    ["Room D", "ok"],
    ["Room E", "ok"],
    ["Room F", "ok"],
    ["Room G", "ok"],
    ["Room H", "ok"],
    ["Room I", "ok"]
]

print("Rooms Lists")
pp.pprint(rooms_list)


new_rooms_list = []

for (index, (room, status)) in enumerate(rooms_list):
    new_rooms_list.append("{}. Room number: {}".format(
        index if index < 5 else index + 1,
        room
    )
)


print("Formatted Rooms")
pp.pprint(new_rooms_list)

输出如下：

Rooms Lists
[ ['Room A', 'ok'],
  ['Room B', 'ok'],
  ['Room C', 'ok'],
  ['Room D', 'ok'],
  ['Room E', 'ok'],
  ['Room F', 'ok'],
  ['Room G', 'ok'],
  ['Room H', 'ok'],
  ['Room I', 'ok']]
Formatted Rooms
[ '0. Room number: Room A',
  '1. Room number: Room B',
  '2. Room number: Room C',
  '3. Room number: Room D',
  '4. Room number: Room E',
  '6. Room number: Room F',
  '7. Room number: Room G',
  '8. Room number: Room H',
  '9. Room number: Room I']

现在，您可以将中间部分作为单例/生成器返回：

new_rooms_list = [
    ("{}. Room number: {}".format(
        index if index < 5 else index + 1,
        room
    )) for index, (room, status) in enumerate(rooms_list)
]

从现有列表的第二个列表中使用枚举时跳过索引的问题

4 个答案: