如何在列表（字符串）中进行排列？你能用一套吗？

Question

我想要一份

nums = (("2q","3q","4q","3q"),("1q"), ("1q","2q"))

导致：

pairs = (("2q", "3q"), ("2q", "1q"),("1q","2q")).......

可能吗

如果我有一套更像：

nums = (("2q,3q,4q,3q"),("1q"), ("1q,2q"))

然后用劈叉

Answer 1

我想你的nums有一个打字错误。。。如果希望它们都是元组，("1q")必须有一个逗号，如("1q",)：

nums = (("2q","3q","4q","3q"),("1q",), ("1q","2q"))

如果输入是元组的元组（或列表的列表等），则假设可以选择两次值，则以下给出所有对：

import itertools
from more_itertools import distinct_permutations

out = list(distinct_permutations(itertools.chain.from_iterable(nums), r=2))

print(out)
[('1q', '1q'), ('1q', '2q'), ('1q', '3q'), ('1q', '4q'), ('2q', '1q'), ('2q', '2q'), ('2q', '3q'), ('2q', '4q'), ('3q', '1q'), ('3q', '2q'), ('3q', '3q'), ('3q', '4q'), ('4q', '1q'), ('4q', '2q'), ('4q', '3q')]

如果不希望成对出现重复项，可以在运行之前使用set消除它们：

out_no_duplicates = list(distinct_permutations(set(itertools.chain.from_iterable(nums)), r=2))

print(out_no_duplicates)
[('1q', '2q'), ('1q', '3q'), ('1q', '4q'), ('2q', '1q'), ('2q', '3q'), ('2q', '4q'), ('3q', '1q'), ('3q', '2q'), ('3q', '4q'), ('4q', '1q'), ('4q', '2q'), ('4q', '3q')]

Answer 2

如果要查找文字置换，请查看itertools.permutationshttps://docs.python.org/3/library/itertools.html

示例来自： https://www.geeksforgeeks.org/python-itertools-permutations/

from itertools import permutations  

a = 'string value here.' # list or string value.
a = ['2q', '3q', '4q'] #... Add other items here.
p = permutations(a,2)   #for pairs of len 2

# Print the obtained permutations  
for j in list(p):  
    print(j)  

#For multiple length clusters..
for c in range(min_cluster_len, max_cluster_len): 
    for j in list(permutations(a, c)):
        print(j)

#Output: 
('2q', '3q')
('2q', '4q')
('2q', '5q')
('3q', '2q')
('3q', '4q')
('3q', '5q')
('4q', '2q')
('4q', '3q')
('4q', '5q')
('5q', '2q')
('5q', '3q')
('5q', '4q')
('2q', '3q', '4q')
('2q', '3q', '5q')
('2q', '4q', '3q')
... 

如果顺序不重要，即（2q，3q）=（3q，2q），则使用itertools.compositions

对于您的实际问题数据，如果任何内容的长度大于或等于；x、 我不确定你是如何将这些进行分组的，所以不完全确定。也可以使用正则表达式，但并不理想