在Python 3中将列表内的列表分组 [英] Grouping lists within lists in Python 3

问题描述

我有一个像这样的字符串列表列表:

I have a list of lists of strings like so:

List1 = [
          ['John', 'Doe'], 
          ['1','2','3'], 
          ['Henry', 'Doe'], 
          ['4','5','6']
        ]

我想变成这样的东西:

List1 = [
          [ ['John', 'Doe'], ['1','2','3'] ],
          [ ['Henry', 'Doe'], ['4','5','6'] ]
        ]

但是我似乎在这样做上有麻烦.

But I seem to be having trouble doing so.

推荐答案

List1 = [['John', 'Doe'], ['1','2','3'],
         ['Henry', 'Doe'], ['4','5','6'],
         ['Bob', 'Opoto'], ['10','11','12']]

def pairing(iterable):
    it = iter(iterable)
    itn = it.next
    for x in it :
        yield (x,itn())     

# The generator pairing(iterable) yields tuples:  

for tu in pairing(List1):
    print tu  

# produces:  

(['John', 'Doe'], ['1', '2', '3'])
(['Henry', 'Doe'], ['4', '5', '6'])
(['Bob', 'Opoto'], ['8', '9', '10'])    

# If you really want a yielding of lists:

from itertools import imap
# In Python 2. In Python 3, map is a generator
for li in imap(list,pairing(List1)):
    print li

# or defining pairing() precisely so:

def pairing(iterable):
    it = iter(iterable)
    itn = it.next
    for x in it :
        yield [x,itn()]

# produce   

[['John', 'Doe'], ['1', '2', '3']]
[['Henry', 'Doe'], ['4', '5', '6']]
[['Bob', 'Opoto'], ['8', '9', '10']]

不需要定义生成器功能，您可以动态配对列表:

Defining a generator function isn't required, you can do the pairing of a list on the fly:

List1 = [['John', 'Doe'], ['1','2','3'],
         ['Henry', 'Doe'], ['4','5','6'],
         ['Bob', 'Opoto'], ['8','9','10']]

it = iter(List1)
itn = it.next
List1 = [ [x,itn()] for x in it]

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