python-如何连接列表中的两个索引

Question

我有一个列表：

listvalue = ['charity','hospital','carrefour']

我试图从列表中合并两个索引：

twoconcat = [listvalue[i:i + 2] for i in range(len(listvalue))]

我得到的输出是：

[['charity', 'hospital'], ['hospital', 'carrefour'], ['carrefour']]`

我希望输出为

[['charity','hospital'],['charity','carrefour'],['hospital','charity'],['hospital','carrefour'],['carrefour','charity'],['carrefour','hospital']]

有什么建议吗？

Answer 1

您可以使用itertools.permutations进行此操作。

>>> places = ['charity','hospital','carrefour']
>>> list(itertools.permutations(places, 2))
[('charity', 'hospital'), ('charity', 'carrefour'), ('hospital', 'charity'), 
('hospital', 'carrefour'), ('carrefour', 'charity'), ('carrefour', 'hospital')]

Answer 2

如果您不关心列表理解。这会有所帮助， 在python 2.x上运行

listvalue = ['charity','hospital','carrefour']

arr_len= len(listvalue)

result=[]
for i in range(arr_len):
    for j in range(arr_len):
        if i !=j:
            result.append([listvalue[i],listvalue[j]])
print result    

结果

[['charity', 'hospital'], ['charity', 'carrefour'], ['hospital', 'charity'], ['hospital', 'carrefour'], ['carrefour', 'charity'], ['carrefour', 'hospital']]

Answer 3

l = ['a', 'b', 'c']
[[x, y] for x in l for y in l if y != x]

输出：

[['a', 'b'], ['a', 'c'], ['b', 'a'], ['b', 'c'], ['c', 'a'], ['c', 'b']]

Answer 4

@Alasdair是正确的...使用itertools。

代码：

 from itertools import permutations

 places = ['charity','hospital','carrefour']
 result = list(permutations(places, 2)

输出：

[('charity', 'hospital'), ('charity', 'carrefour'), ('hospital', 'charity'), ('hospital', 'carrefour'), ('carrefour', 'charity'), ('carrefour', 'hospital')]

代码：

from itertools import permutations
places = ['charity','hospital','carrefour']
result = [list(place) for place in list(permutations(places, 2))]

输出：

[['charity','hospital'],['charity','carrefour'],['hospital','charity'],['hospital','carrefour'],['carrefour','charity'],['carrefour','hospital']]