[英]Get all possible combination of values in a list — Python
I have a list containing ['a', 'bill', 'smith']
and I would like to write a python code in order to obtain all possible combinations applying a certain criteria.我有一个包含['a', 'bill', 'smith']
的列表,我想编写一个 python 代码,以获得应用特定标准的所有可能组合。 To be more precise, I would like to obtain combination of those three element in the list plus the first letter of each element if that element isn't yet present in the output list.更准确地说,如果 output 列表中尚不存在该元素,我想获得列表中这三个元素的组合加上每个元素的第一个字母。 For example, given the list ['a', 'bill', 'smith']
, one part of the expected output would be: ['a', 'bill', 'smith'], ['bill', 'smith'], ['a', 'smith']
but as well, ['a', 'b, 'smith'], ['bill, 's'], ['a', 's']
.例如,给定列表['a', 'bill', 'smith']
,预期 output 的一部分将是: ['a', 'bill', 'smith'], ['bill', 'smith'], ['a', 'smith']
但也有['a', 'b, 'smith'], ['bill, 's'], ['a', 's']
。 What I'm not expected to obtain is output like this ['s', 'bill, 'smith']
as the first element (s) is already taken into account by the third element ('smith').我不希望得到的是 output 这样['s', 'bill, 'smith']
作为第一个元素(s)已经被第三个元素('smith')考虑在内。 Can someone help me?有人能帮我吗?
This is what I've done so far:这是我到目前为止所做的:
mapping = dict(enumerate(['a', 'bill', 'smith']))
for i in mapping.items():
if len(i[1])>1:
mapping[i[0]] = [i[1], i[1][0]]
else:
mapping[i[0]] = [i[1]]
print(mapping)
{0: ['a'], 1: ['bill', 'b'], 2: ['william', 'w'], 3: ['stein', 's']}
I'm now stucked.我现在被困住了。 I would like to use itertools library to iterate over the dict values to create all possible combinations.我想使用 itertools 库来迭代 dict 值以创建所有可能的组合。
Thanks in advance:)提前致谢:)
You can use some itertools
:您可以使用一些itertools
:
from itertools import product, permutations
lst = [list({s, s[:1]}) for s in ['a', 'bill', 'smith']]
# [['a'], ['bill', 'b'], ['s', 'smith']]
for perms in map(permutations, product(*lst)):
for p in perms:
print(p)
('a', 'bill', 's')
('a', 's', 'bill')
('bill', 'a', 's')
('bill', 's', 'a')
('s', 'a', 'bill')
('s', 'bill', 'a')
('a', 'bill', 'smith')
('a', 'smith', 'bill')
('bill', 'a', 'smith')
('bill', 'smith', 'a')
('smith', 'a', 'bill')
('smith', 'bill', 'a')
('a', 'b', 's')
('a', 's', 'b')
('b', 'a', 's')
('b', 's', 'a')
('s', 'a', 'b')
('s', 'b', 'a')
('a', 'b', 'smith')
('a', 'smith', 'b')
('b', 'a', 'smith')
('b', 'smith', 'a')
('smith', 'a', 'b')
('smith', 'b', 'a')
The first step creates the list of equivalent lists:第一步创建等效列表的列表:
[['a'], ['bill', 'b'], ['s', 'smith']]
then, product
produces the cartesian product of the lists in said lists:然后, product
生成所述列表中列表的笛卡尔积:
('a', 'bill', 's')
('a', 'bill', 'smith')
('a', 'b', 's')
...
and for each of those, permutations
gives you, well, all permutations:对于其中的每一个, permutations
都会为您提供所有排列:
('a', 'bill', 's')
('a', 's', 'bill')
('bill', 'a', 's')
...
You could do something like this with combinations
from itertools
:您可以使用来自itertools
的combinations
来做这样的事情:
Here I am assuming you want the first letter of each word in the list only if it has a length greater than 1. If not you can change the if condition.在这里,我假设您只希望列表中每个单词的第一个字母的长度大于 1。如果不是,您可以更改 if 条件。
from itertools import combinations
lst = ['a', 'bill', 'smith']
lst_n=[]
for words in lst:
lst_n.append(words)
if len(words)>1:
lst_n.append(words[0])
for t in range(1,len(lst_n)+1):
for comb in combinations(lst_n,r=t):
print(list(comb))
OUTPUT: OUTPUT:
['a']
['bill']
['b']
['smith']
['s']
['a', 'bill']
['a', 'b']
['a', 'smith']
['a', 's']
['bill', 'b']
['bill', 'smith']
['bill', 's']
['b', 'smith']
['b', 's']
['smith', 's']
['a', 'bill', 'b']
['a', 'bill', 'smith']
['a', 'bill', 's']
['a', 'b', 'smith']
['a', 'b', 's']
['a', 'smith', 's']
['bill', 'b', 'smith']
['bill', 'b', 's']
['bill', 'smith', 's']
['b', 'smith', 's']
['a', 'bill', 'b', 'smith']
['a', 'bill', 'b', 's']
['a', 'bill', 'smith', 's']
['a', 'b', 'smith', 's']
['bill', 'b', 'smith', 's']
['a', 'bill', 'b', 'smith', 's']
Here if you want combinations to be of length 3
only remove the for loop
with range
and set r=3
.在这里,如果您希望组合的长度为3
,则只需删除带有range
的for loop
并设置r=3
。
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