将字符串拆分为元组列表？

Question

我有一串键值对,不幸的是用相同的符号分隔.有没有办法将它"拆分"成元组列表,而不使用lambda？

这是我有的:

Moscow|city|London|city|Royston Vasey|vilage

我想要的是:

[("Moscow","city"), ("London", "city")....] 

Answer 1

这真的很容易......

第一,拆分对串'|'然后zip每隔一个元素一起:

data = s.split('|')
print zip(data[::2],data[1::2])

在python3中,您将需要: print(list(zip(data[::2],data[1::2]))

Answer 2

s = 'Moscow|city|London|city|Royston Vasey|vilage'

it = iter(s.split('|'))
print [(x,next(it)) for x in it]

Answer 3

def group(lst, n):
    for i in range(0, len(lst), n):
        val = lst[i:i+n]
        if len(val) == n:
            yield tuple(val)

a = 'Moscow|city|London|city|Royston Vasey|vilage'
list(group(a.split('|'), 2))

输出是 [('Moscow', 'city'), ('London', 'city'), ('Royston Vasey', 'vilage')]

Answer 4

对于Python2

>>> s = "Moscow|city|London|city|Royston Vasey|vilage"
>>> zip(*[iter(s.split('|'))]*2)
[('Moscow', 'city'), ('London', 'city'), ('Royston Vasey', 'vilage')]

list(zip(...))当然需要Python3