时间戳

首页 > 解决方案 > Python中列表从2到N的唯一排列

python - Python中列表从2到N的唯一排列

问题描述

我正在尝试在循环中生成一个排列列表，并为每次迭代打印两个以上的输出（或写入文件的行）。

示例输入列表：

['one', 'two', 'three', 'four']

所需输出：

['one', 'two', 'three', 'four']
['two', 'three', 'four']
['one', 'three', 'four']
['one', 'two', 'four']
['one', 'two']
['one', 'three']
['one', 'four']
['two', 'three']
['two', 'four']
['three', 'four']

这是我迄今为止所管理的（在我的 Python 生命的早期，请原谅）：

from itertools import permutations

input = ['one', 'two', 'three', 'four']

def convertTuple(tup): 
    str =  ''.join(tup) 
    return str

while (len(input) > 1):    
    permlist = set(permutations(input))
    for i in permlist:
        print(i)
        i = convertTuple(i)
        outfile = open("out.txt", "w")
        outfile.write(i)
    input = input[:-1]
else:
    print("End of permutation cycle")

哪个输出：

('two', 'three', 'one', 'four')
('two', 'four', 'one', 'three')
('three', 'two', 'one', 'four')
('four', 'two', 'one', 'three')
('two', 'one', 'three', 'four')
('two', 'one', 'four', 'three')
('three', 'one', 'four', 'two')
('four', 'one', 'three', 'two')
('one', 'two', 'three', 'four')
('one', 'two', 'four', 'three')
('three', 'four', 'one', 'two')
('four', 'three', 'one', 'two')
('two', 'three', 'four', 'one')
('two', 'four', 'three', 'one')
('three', 'two', 'four', 'one')
('three', 'four', 'two', 'one')
('four', 'two', 'three', 'one')
('four', 'three', 'two', 'one')
('three', 'one', 'two', 'four')
('four', 'one', 'two', 'three')
('one', 'four', 'two', 'three')
('one', 'three', 'two', 'four')
('one', 'three', 'four', 'two')
('one', 'four', 'three', 'two')
('two', 'three', 'one')
('three', 'two', 'one')
('three', 'one', 'two')
('one', 'two', 'three')
('one', 'three', 'two')
('two', 'one', 'three')
('two', 'one')
('one', 'two')
End of permutation cycle

我明白我错了

input = input[:-1]

因为它只是删除了原始列表中的最后一个值，但我无法弄清楚如何仅获取唯一列表，每个列表中具有不同数量的值...

我是否使用了 itertools 的错误部分？我应该使用组合还是其他东西？

我被严重卡住了，所以非常感谢任何帮助！

谢谢！

标签： pythonloopscombinationspermutationitertools

解决方案

假设 ('one','two','three') 是您想要的列表中的意外遗漏，这应该可以解决问题：

from itertools import combinations

l = ['one', 'two', 'three', 'four']

for size in range(2,len(l)+1):
    for i in combinations(l,size):
        print(i)

推荐阅读