Question

给出一个列表，说[4,5,6,7]我想编写一个将输出的函数所有对（[4,5,7]，[6]），（[4,5]，[6,7]）子列表的长度不固定

我使用了一个for循环和两个新列表来将第ith个元素追加到列出一个并将其余的附加到列表2中，然后重复该过程，我知道这是一种愚蠢的方法

a = [4,5,6,7]
for x in range(0,len(a)):
     b = []
     c = []
     b.append(a[x])
     for y in range(0,len(a)):
          if y!=x:
               c.append(a[y])
     print(b,c)
for x in range(0,len(a)-1):
     b = []
     c = []
     b.append(a[x])
     b.append(a[x+1])
     for y in range(0,len(a)):
          if y!=x and y!=x+1:
               c.append(a[y])
     print(b,c)

我希望打印所有可能的子列表，但始终会错过不连续的子列表，我会得到（[4，5]，[6,7]）但从不生成（[4,6]，[5,7]）

Answer 1

您可以使用itertools模块来构造所有此类子列表：

import itertools as it

s = {4,5,6,7}
for length in range(len(s)):
    for combination in it.combinations(s, length):
        print(list(combination), list(s - set(combination)))

[] [4, 5, 6, 7]
[4] [5, 6, 7]
[5] [4, 6, 7]
[6] [4, 5, 7]
[7] [4, 5, 6]
[4, 5] [6, 7]
[4, 6] [5, 7]
[4, 7] [5, 6]
[5, 6] [4, 7]
[5, 7] [4, 6]
[6, 7] [4, 5]
[4, 5, 6] [7]
[4, 5, 7] [6]
[4, 6, 7] [5]
[5, 6, 7] [4]

Answer 2

这是不删除重复项的解决方案：

from itertools import combinations

a = [4,4,5,6,7]
result = []

for x in range(len(a) + 1):
    for left in combinations(a, x):
        right = a.copy()
        for ele in left:
            right.remove(ele)
        result.append([list(left), right])

for res in result:
    print(res)

输出：

[[], [4, 4, 5, 6, 7]]
[[4], [4, 5, 6, 7]]
[[4], [4, 5, 6, 7]]
[[5], [4, 4, 6, 7]]
[[6], [4, 4, 5, 7]]
[[7], [4, 4, 5, 6]]
[[4, 4], [5, 6, 7]]
[[4, 5], [4, 6, 7]]
[[4, 6], [4, 5, 7]]
[[4, 7], [4, 5, 6]]
[[4, 5], [4, 6, 7]]
[[4, 6], [4, 5, 7]]
[[4, 7], [4, 5, 6]]
[[5, 6], [4, 4, 7]]
[[5, 7], [4, 4, 6]]
[[6, 7], [4, 4, 5]]
[[4, 4, 5], [6, 7]]
[[4, 4, 6], [5, 7]]
[[4, 4, 7], [5, 6]]
[[4, 5, 6], [4, 7]]
[[4, 5, 7], [4, 6]]
[[4, 6, 7], [4, 5]]
[[4, 5, 6], [4, 7]]
[[4, 5, 7], [4, 6]]
[[4, 6, 7], [4, 5]]
[[5, 6, 7], [4, 4]]
[[4, 4, 5, 6], [7]]
[[4, 4, 5, 7], [6]]
[[4, 4, 6, 7], [5]]
[[4, 5, 6, 7], [4]]
[[4, 5, 6, 7], [4]]
[[4, 4, 5, 6, 7], []]

Answer 3

您可以创建一个随机子列表，然后获取第二个列表中不属于该随机子列表的元素：

# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility

import re

regex = r"identifier:(\s+)?(.*?)(\s+)?\]"

test_str = ("OBNAME[origin:85 copy:1 identifier:TDEP],OBNAME[origin:85 copy:1 identifier:RDEP]\n\n"
    "OBNAME[origin:85 copy:1 identifier: TDEP  ],OBNAME[origin:85 copy:1 identifier:  RDEP  ]")

matches = re.finditer(regex, test_str, re.MULTILINE)

for matchNum, match in enumerate(matches, start=1):

    print ("Match {matchNum} was found at {start}-{end}: {match}".format(matchNum = matchNum, start = match.start(), end = match.end(), match = match.group()))

    for groupNum in range(0, len(match.groups())):
        groupNum = groupNum + 1

        print ("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = match.start(groupNum), end = match.end(groupNum), group = match.group(groupNum)))

# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.

关于如何将原始列表划分为随机块，具体取决于您解决

如何从给定的列表中选择所有可能的子列表对，它们的联合将成为列表？

3 个答案: