Question

我遇到Python问题，希望有人可以帮助我。我有一个列表，例如这个：

list = [['a','b','c'],['a','c1','d1'],['b','c1','c2']]

我希望以一种所有具有相同索引[0]的数组将在一起的方式组合列表，因此它将如下：

a, b, c, c1, d1
b, c1, c2

我试过这样的事情，但我没有让它发挥作用..

list = [['a','b','c'],['a','c1','d1'],['b','c1','c2']]
empty_list = []
for i in list:
    if i not in empty_list:
       empty_list.append(i)
    print empty_list

有人可以帮助我吗？

Answer 1

你可以试试这个：）

old_list = [['a','b','c'],['a','c1','d1'],['b','c1','c2']]
prev = None
empty_list = []
for l in old_list: # iterate through each sub list, sub list as l
    if l[0] == prev:
        # append your elements to existing sub list
        for i in l: # iterate through each element in sub list
            if i not in empty_list[-1]:
                empty_list[-1].append(i) 
    else:
        empty_list.append(l) # create new sub list
        prev = l[0] # update prev

print(empty_list)
# [['a', 'b', 'c', 'c1', 'd1'], ['b', 'c1', 'c2']]

Answer 2

使用itertools.groupby：

from itertools import groupby
from operator import itemgetter

listt = [['a','b','c'],['a','c1','d1'],['b','c1','c2']]

grouped = [list(g) for _,g in groupby(listt,itemgetter(0))]
result = [[item for sslist in slist for item in sslist] for slist in grouped]

Answer 3

OrderedDict可以完成大部分工作：

from collections import OrderedDict

l = [['a','b','c'], ['a','c1','d1'], ['b','c1','c2']]

d = OrderedDict()
for el in l:
  d.setdefault(el[0], el[0:1]).extend(el[1:])

print(d.values())

Answer 4

您也可以尝试使用defaultdict（list）

l = [['a','b','c'], ['a','c1','d1'], ['b','c1','c2']]
from collections import defaultdict
d_dict = defaultdict(list)
for i in l:
    d_dict[i[0]].extend(i[1:])

[ list(k) + v  for k, v in d_dict.items() ]

输出：

[['a', 'b', 'c', 'c1', 'd1'], ['b', 'c1', 'c2']]

结合列表的元素

4 个答案: