Question

我有一个类似的列表：

A
B
C
D
E
F
G

如何解决此问题以查找3位数的所有组合。同一行不能使用相同的字母。

ABC
ABD
ABE
ABF
ABG
AGB

例如......：

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)

这不会给出所需的输入：

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

Answer 1

Python标准库itertools已经具有您尝试实现的功能。你也在你的代码中使用它（有趣）。

itertools.combinations(a,3)返回a的所有3种组合。要将其转换为“列表列表”，您应使用.extend()，如下所示;

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，我建议您搜索“Combination, Permutation Difference”。据我所知，你的问题;排列是你想要的。（如果你运行我分享的代码，你会理解差异很容易。）

Answer 2

要了解解决方案流程的工作原理，请尝试以下操作：

# get all combinations of n items from given list
def getCombinations(items, n):
    if len(items) < n: return [] # need more items than are remaining 
    if n == 0: return [''] # need no more items, return the combination of no items

    [fst, *rst] = items

    # all combinations including the first item in the list
    including = [fst + comb for comb in getCombinations(rst, n-1)]

    # all combinations excluding the first item in the list
    excluding = getCombinations(rst, n)

    both = including + excluding
    return both

x = ['a','b','c','d','e']
n = 3
print(getCombinations(x, n))
# ['abc', 'abd', 'abe', 'acd', 'ace', 'ade', 'bcd', 'bce', 'bde', 'cde']

Answer 3

组合适用于字符串而非列表，因此您应首先使用以下字符将其转换为字符串：''.join(x)

from itertools import combinations
x = ['a', 'b', 'c', 'd', 'e']
n = 3
aa = combinations(''.join(x), n)
for comb in aa:
    print(''.join(comb))

<强>输出

abc
abd
abe
acd
ace
ade
bcd
bce
bde
cde

或者作为一个单行：

[''.join(comb) for comb in combinations(''.join(x), n)]

列表的3位数的所有可能组合永远不会相同

3 个答案: