根据python中的条件将子列表分别分组在列表中。

Question

我有个子目录

list=[['RD-2','a',120],
      ['RD-2','b',125],
      ['RD-2','c',127],
      ['RD-3','a',120],
      ['RD-3','b',140],
      ['RD-5','a',120]]

如果每个子列表相同的话，我想根据每个子列表的第一个元素在列表中做单独的子列表。输出如下所示

finallist=[
           [['RD-2','a',120],['RD-2','b',125],['RD-2','c',127]],   
           [['RD-3','a',120],['RD-3','b',140]],
           ['RD-5','a',120]]

我试过使用if条件，但它不起作用。帮帮我

Answer 1

使用itertools.groupby

Ex:

from itertools import groupby

lst=[['RD-2','a',120],['RD-2','b',125],['RD-2','c',127],['RD-3','a',120],['RD-3','b',140],['RD-5','a',120]]

res = [list(v) for i, v in groupby(lst, lambda x: x[0])]   #GroupBy first element. 
print(res)

输出：

[[['RD-2', 'a', 120], ['RD-2', 'b', 125], ['RD-2', 'c', 127]], [['RD-3', 'a', 120], ['RD-3', 'b', 140]], [['RD-5', 'a', 120]]]

Answer 2

由于不需要使用collections.defaultdict，所以可以使用

from collections import defaultdict

L = [['RD-2','a',120],['RD-2','b',125],['RD-2','c',127],
     ['RD-3','a',120],['RD-3','b',140],['RD-5','a',120]]

dd = defaultdict(list)

for key, value1, value2 in L:
    dd[key].append([value1, value2])

## Python 3.x, use * unpacking
# for key, *values in L:
#     dd[key].append(values)

res = [[[key, value] for value in values] for key, values in dd.items()]

print(res)

# [[['RD-2', ['a', 120]], ['RD-2', ['b', 125]], ['RD-2', ['c', 127]]],
#  [['RD-3', ['a', 120]], ['RD-3', ['b', 140]]],
#  [['RD-5', ['a', 120]]]]

Answer 3

如果组始终是连续的，则可以使用群比，否则可以根据第一个元素对子列表进行排序，然后按组排序：

from itertools import groupby
from operator import itemgetter

lst=[['RD-2','a',120],['RD-2','b',125],['RD-2','c',127],['RD-3','a',120],['RD-3','b',140],['RD-5','a',120]]
result = [list(v) for _, v in groupby(lst, key=itemgetter(0))]

print(result)

输出

[[['RD-2', 'a', 120], ['RD-2', 'b', 125], ['RD-2', 'c', 127]], [['RD-3', 'a', 120], ['RD-3', 'b', 140]], [['RD-5', 'a', 120]]]