列表的3位数字的所有可能组合永远都不相同 [英] All possible combinations for 3 digits of list never the same

问题描述

我有一个看起来像这样的列表:

I have a list that looks like:

A
B
C
D
E
F
G

如何解决此问题以找到3位数字的所有组合.同一字母不能在同一行中使用.

How do I solve this to find all combinations for 3 digits. The same letter cannot be used in same row.

ABC
ABD
ABE
ABF
ABG
AGB

例如类似...:

x = ['a','b','c','d','e']
n = 3
import itertools
aa = [list(comb) for i in range(1, n+2) for comb in itertools.combinations(x, i)]
print(aa)

这没有提供所需的输入:

This does not give desired input:

[['a'], ['b'], ['c'], ['d'], ['e'], ['a', 'b'], ['a', 'c'], ['a', 'd'], ['a', 'e'], ['b', 'c'], ['b', 'd'], ['b', 'e'], ['c'

推荐答案

Python标准库

The Python Standard Library itertools already has the functionality you are trying to implement. Also you are using it in your code (funnily).

itertools.combinations(a,3)返回a的所有3个组合.要将其转换为列表列表"，您应按以下方式使用.extend()；

itertools.combinations(a,3) returns all 3-combinations of the a. To convert that to "list of list" you should use .extend() as follows;

x = ['a','b','c','d','e']
n = 3
import itertools
permutations = []
combinations = []
combinations.extend(itertools.combinations(x,n))
permutations.extend(itertools.permutations(x,n))

print("Permutations;", permutations)
print("\n")
print("Combinations;", combinations)

此外，建议您搜索"组合，排列差异 ".从您的问题中我了解到；排列就是您想要的. (如果运行我共享的代码，您将容易理解它们之间的区别.)

Additionally, I suggest you to search on "Combination, Permutation Difference". As I understood from your question; permutation is what you want. (If you run the code I shared, you will understand the difference easliy.)

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