具有重复但不加倍的列表的所有排列 [英] ALL permutations of a list with repetition but not doubles

问题描述

我看到了相似但不相同的内容:这里. 我绝对希望所有列表元素的排列而不是组合.我的是不同的，因为itertools排列的a，b，c返回abc而不是aba(soooo close). 如何获得像aba这样的结果?

I have seen similar but not the same: here. I definitely want the permutations, not combinations, of all list elements. Mine is different because itertools permutation of a,b,c returns abc but not aba (soooo close). How can I get results like aba returned as well?

('a',)     <-excellent
('b',)     <-excellent
('c',)     <-excellent
('a', 'b') <-excellent
('a', 'c') <-excellent
('b', 'a') <-excellent
('b', 'c') <-excellent
('c', 'a') <-excellent
('c', 'b') <-excellent
('a', 'b', 'c') <-- I need a,b,a
('a', 'c', 'b') <-- I need a,c,a
('b', 'a', 'c') <-- I need b,a,b... you get the idea

哦，排列的最大长度(在python.org itertools中为"r")等于len(list)，我不想包含双精度"，例如aab或abb ...或abba:P该列表可以是任意长度.

Oh max length of permutations ("r" in the python.org itertools) is equal to len(list), and I don't want to include 'doubles' such as aab or abb ...or abba :P The list could be any length.

import itertools
from itertools import product
my_list = ["a","b","c"]
#print list(itertools.permutations(my_list, 1))
#print list(itertools.permutations(my_list, 2))
#print list(itertools.permutations(my_list, 3)) <-- this *ALMOST* works

我将以上内容合并为for循环

I combined the above into a for loop

def all_combinations(varsxx):
    repeat = 1
    all_combinations_result = []
    for item in varsxx:
        if repeat <= len(varsxx):
            all_combinations_result.append(list(itertools.permutations(varsxx, repeat)))
        repeat += 1
    return all_combinations_result

作为参考，当我在纸上进行此操作时，我得到了21个结果.

For reference, when I did this on paper I got 21 results.

将字符串列表转换为数字列表也有任何好处.我的想法是，对于排列工具而言，数字将更易于使用.字符串可能是10到50个字符..ish.

Also is there any merit in converting the list of strings into a list of numbers. My thought was that numbers would be easier to work with for the permutations tool. Strings might be 10 to 50 chars..ish.

推荐答案

即使您绝对想要排列"，听起来您也不是想要的，您实际上想要序列的笛卡尔积与1到len(sequence)次，相邻元素相等的结果被滤除.

Even though you "definitely want the permutations", it sounds like you don't really want that, you actually want the Cartesian product of your sequence with itself from 1 to len(sequence) times, with results with neighbouring equal elements filtered out.

类似的东西:

In [16]: from itertools import product

In [17]: def has_doubles(x): return any(i==j for i,j in zip(x, x[1:]))

In [18]: seq = ["a","b","c"]

In [19]: [x for n in range(len(seq)) for x in product(seq, repeat=n+1) 
            if not has_doubles(x)]
Out[19]: 
[('a',),
 ('b',),
 ('c',),
 ('a', 'b'),
 ('a', 'c'),
 ('b', 'a'),
 ('b', 'c'),
 ('c', 'a'),
 ('c', 'b'),
 ('a', 'b', 'a'),
 ('a', 'b', 'c'),
 ('a', 'c', 'a'),
 ('a', 'c', 'b'),
 ('b', 'a', 'b'),
 ('b', 'a', 'c'),
 ('b', 'c', 'a'),
 ('b', 'c', 'b'),
 ('c', 'a', 'b'),
 ('c', 'a', 'c'),
 ('c', 'b', 'a'),
 ('c', 'b', 'c')]

In [20]: len(_)
Out[20]: 21

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