连接列表的元素对

我知道一个列表可以连接成一个长字符串，如下所示:

x = ['a', 'b', 'c', 'd']
print ''.join(x)

显然，这将输出:

'abcd'

但是，我要做的只是简单地加入列表中的第一个和第二个字符串，然后加入第三个和第四个字符串，以此类推。简而言之，从上面的例子可以得到以下结果:

['ab', 'cd']

有什么简单的方法吗？我可能还应该提到，列表中字符串的长度是不可预测的，列表中字符串的数量也是不可预测的，尽管字符串的数量总是偶数。因此，最初的清单可能就是:

['abcd', 'e', 'fg', 'hijklmn', 'opq', 'r']
180965 次浏览
小开
最佳答案

You can use slice notation with steps:

>>> x = "abcdefghijklm"
>>> x[0::2] #0. 2. 4...
'acegikm'
>>> x[1::2] #1. 3. 5 ..
'bdfhjl'
>>> [i+j for i,j in zip(x[::2], x[1::2])] # zip makes (0,1),(2,3) ...
['ab', 'cd', 'ef', 'gh', 'ij', 'kl']

Same logic applies for lists too. String lenght doesn't matter, because you're simply adding two strings together.

小开 
>>> lst =  ['abcd', 'e', 'fg', 'hijklmn', 'opq', 'r']
>>> print [lst[2*i]+lst[2*i+1] for i in range(len(lst)/2)]
['abcde', 'fghijklmn', 'opqr']
小开

Without building temporary lists:

>>> import itertools
>>> s = 'abcdefgh'
>>> si = iter(s)
>>> [''.join(each) for each in itertools.izip(si, si)]
['ab', 'cd', 'ef', 'gh']

or:

>>> import itertools
>>> s = 'abcdefgh'
>>> si = iter(s)
>>> map(''.join, itertools.izip(si, si))
['ab', 'cd', 'ef', 'gh']
小开

Use an iterator.

List comprehension:

>>> si = iter(['abcd', 'e', 'fg', 'hijklmn', 'opq', 'r'])
>>> [c+next(si, '') for c in si]
['abcde', 'fghijklmn', 'opqr']
  • Very efficient for memory usage.
  • Exactly one traversal of s

Generator expression:

>>> si = iter(['abcd', 'e', 'fg', 'hijklmn', 'opq', 'r'])
>>> pair_iter = (c+next(si, '') for c in si)
>>> pair_iter # can be used in a for loop
<generator object at 0x4ccaa8>
>>> list(pair_iter)
['abcde', 'fghijklmn', 'opqr']
  • use as an iterator

Using map, str.__add__, iter

>>> si = iter(['abcd', 'e', 'fg', 'hijklmn', 'opq', 'r'])
>>> map(str.__add__, si, si)
['abcde', 'fghijklmn', 'opqr']

next(iterator[, default]) is available starting in Python 2.6

小开

Well I would do it this way as I am no good with Regs..

CODE

t = '1. eat, food\n\
7am\n\
2. brush, teeth\n\
8am\n\
3. crack, eggs\n\
1pm'.splitlines()


print [i+j for i,j in zip(t[::2],t[1::2])]

output: 

['1. eat, food   7am', '2. brush, teeth   8am', '3. crack, eggs   1pm']

Hope this helps :)

小开

just to be pythonic :-)

>>> x = ['a1sd','23df','aaa','ccc','rrrr', 'ssss', 'e', '']
>>> [x[i] + x[i+1] for i in range(0,len(x),2)]
['a1sd23df', 'aaaccc', 'rrrrssss', 'e']

in case the you want to be alarmed if the list length is odd you can try:

[x[i] + x[i+1] if not len(x) %2 else 'odd index' for i in range(0,len(x),2)]

Best of Luck

小开

I came across this page interesting yesterday while wanting to solve a similar issue. I wanted to join items first in pairs using one string in between and then together using another string. Based on the code above I came up with the following function:

def pairs(params,pair_str, join_str):
"""Complex string join where items are first joined in pairs
"""
terms = iter(params)
pairs = [pair_str.join(filter(len, [term, next(terms, '')])) for term in terms]
return join_str.join(pairs)

This results in the following:

a = ['1','2','3','4','5','6','7','8','9']
print(pairs(a, ' plus ', ' and '))
>>1 plus 2 and 3 plus 4 and 5 plus 6 and 7 plus 8 and 9

The filter step prevents the '' which is produced in case of an odd number of terms from putting a final pair_str at the end.


提交答案