遍历python中的嵌套列表

Question

I'm trying to iterate through a list and depending on several conditions to rearrange the items in the list in sublists, all inside the original list that is. so with the code below in Python, while the list1 prints correctly by grouping 0s, 1s and 2s :

new list 1 = [['A0', 'B0', 'C0'], ['A1', 'B1', 'C1'], ['A2', 'B2', 'C2']]

everything gets mixed up on the 2nd list trying to re-group in further sublists As Bs Cs by getting :

new list 2 = [[['A0', 'A1', 'A2'], ['B0', 'B1', 'B2'], ['C0', 'C1', 'C2']]]

while expecting to get:

new list 1 = [[['A0'], ['B0'], ['C0']], [['A1'], ['B1'], ['C1']], [['A2'], ['B2'], ['C2']]]

below is the code I used (sorry for the length, i'm sure there is a very reduced version).

def Main():

    array = ['A0', 'A1', 'A2', 'B0', 'B1', 'B2', 'C1', 'C0', 'C2']

    list1 = NewList1(array)
    print 'new list 1 = ' + str(list1)

    list2 = NewList2(list1)
    print 'new list 2 = ' + str(list2)

def NewList1(thelist):
    sublist1=[]
    sublist2=[]
    sublist3=[]
    sublist4=[]
    newlist=[]
    for item in thelist:
        if hasattr(item,"__iter__"):
            for subitem in item:
                sublist1.append(subitem)
        elif "0" in item:
                sublist2.append(item)
        elif "1" in item:
                sublist3.append(item)
        elif "2" in item:
                sublist4.append(item)
        else:
            newlist.append(item)
    if sublist2: newlist.append(sublist2)
    if sublist3: newlist.append(sublist3)
    if sublist4: newlist.append(sublist4)

    if sublist1: newlist.append(NewList1(sublist1))

    return newlist


def NewList2(thelist):
    sublist1=[]
    sublist2=[]
    sublist3=[]
    sublist4=[]
    newlist=[]
    for item in thelist:
        if hasattr(item,"__iter__"):
            for subitem in item:
                sublist1.append(subitem)
        elif "A" in item:
                sublist2.append(item)
        elif "B" in item:
                sublist3.append(item)
        elif "C" in item:
                sublist4.append(item)
        else:
            newlist.append(item)
    if sublist2: newlist.append(sublist2)
    if sublist3: newlist.append(sublist3)
    if sublist4: newlist.append(sublist4)

    if sublist1: newlist.append(NewList2(sublist1))

    return newlist

Main();

Answer 1

Sort on your grouping criteria, then use itertools.groupby() :

from itertools import groupby

key = lambda el: el[1]
list2 = [[[el] for el in grouped] for group, grouped in groupby(sorted(input, key=key), key)]

Demonstration:

>>> from itertools import groupby
>>> input = ['A0', 'A1', 'A2', 'B0', 'B1', 'B2', 'C1', 'C0', 'C2']
>>> key = lambda el: el[1]
>>> [[[el] for el in grouped] for group, grouped in groupby(sorted(input, key=key), key)]
[[['A0'], ['B0'], ['C0']], [['A1'], ['B1'], ['C1']], [['A2'], ['B2'], ['C2']]]

Your output requirement is a little convoluted; if returning lists of the grouped elements is enough, calling list() on grouped is enough:

>>> [list(grouped) for group, grouped in groupby(sorted(input, key=key), key)]
[['A0', 'B0', 'C0'], ['A1', 'B1', 'C1'], ['A2', 'B2', 'C2']]

Answer 2

     python 3.2

     [[[i] for i in v] for v in list1]

Answer 3

Consider more_itertools.chunked , which accepts an iterable and a chunk size n :

import more_itertools as mit

data = ['A0', 'A1', 'A2', 'B0', 'B1', 'B2', 'C1', 'C0', 'C2']

list(mit.chunked(data, 3))
# [['A0', 'A1', 'A2'], ['B0', 'B1', 'B2'], ['C1', 'C0', 'C2']]

list(mit.chunked(list(mit.chunked(data, 1)), 3))
# [[['A0'], ['A1'], ['A2']], [['B0'], ['B1'], ['B2']], [['C1'], ['C0'], ['C2']]]