根据python中的条件,在列表中单独分组子列表

使用 itertools.groupby

前任:

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]]]

无需订购,您可以使用 collections.defaultdict :

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]]]]

如果组总是连续的,则可以使用 groupby ,否则,可以根据第一个元素对子列表进行排序,然后分组:

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)

产量

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