使用python对列表进行分组和排序

Question

我正在尝试使用一个列表对另一个列表进行排序，并使它们同时保持同步：

keys = [x,x,x,y,y,x,x,z,z,z,x,x]
data = [1,2,3,4,5,6,7,8,9,10,11,12]

我想使用键列表将数据列表组织到相同键的子组中。

result = [[1,2,3,6,7,11,12],[4,5,],[8,9,10]]

我还想确保列表在每个子组中排序。

到目前为止，我能够正确地进行所有排序：

group = []

data = sorted(zip(data, keys), key=lambda x: (x[1]))
for i, grp in groupby(data, lambda x: x[1]):
    sub_group = [], []
    for j in grp:
        sub_group.append(j[1])
    group.extend(sub_group)

我还想念什么？ 谢谢！

Answer 1

您快完成了。 试试这个代码

group = []
data = sorted(zip(data, keys), key=lambda x: (x[1]))
for i, grp in groupby(data, lambda x: x[1]):
    group.append([item[0] for item in grp])

grp具有(data, key)对，因此您需要从该对中选择data作为[item[0] for item in grp]

更新

我用来回答此代码。

from itertools import groupby

x, y, z = range(3)
keys = [x,x,x,y,y,x,x,z,z,z,x,x]
data = [1,2,3,4,5,6,7,8,9,10,11,12]

group = []
data = sorted(zip(data, keys), key=lambda x: (x[1]))
for i, grp in groupby(data, lambda x: x[1]):
    group.append([item[0] for item in grp])

print group

Answer 2

如果使用collections.OrderedDict及其setdefault方法，它将更加简单：

from collections import OrderedDict

# To demonstrate, I made the keys into strings
keys = ['x', 'x', 'x', 'y', 'y', 'x', 'x', 'z', 'z', 'z', 'x', 'x']
data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

dct = OrderedDict()
for key,val in zip(keys, data):
    dct.setdefault(key, []).append(val)

print(dct)
print(list(dct.values()))

输出：

OrderedDict([('x', [1, 2, 3, 6, 7, 11, 12]), ('y', [4, 5]), ('z', [8, 9, 10])])
[[1, 2, 3, 6, 7, 11, 12], [4, 5], [8, 9, 10]]

Answer 3

OrderedDict可能是更好的选择，但是..

import itertools as it
from operator import itemgetter
x = 1
y = 2
z = 3
keys = [x,x,x,y,y,x,x,z,z,z,x,x]
data = [1,2,3,4,5,6,7,8,9,10,11,12]
key = itemgetter(1)
value = itemgetter(0)

data = sorted(zip(data, keys), key=key)
print [map(value, grp) for k, grp in it.groupby(data, key)]

Answer 4

请注意，OrderedDict会按键的插入顺序对其进行排序，而不要像对事实进行排序一样。 如果“键”列表的顺序不符合要求，您将无法获得预期的结果。

我的解决方案：

from collections import defaultdict

keys = ['x', 'x', 'x', 'y', 'y', 'x', 'x', 'z', 'z', 'z', 'x', 'x']
data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

# group 'data' values by 'key'
grouped = defaultdict(list)
for key, data in zip(keys, data):
    grouped[key].append(data)

# construct the final list of subgroups
# contents of each subgroup must be sorted
# also sorting the keys so that the 'x' subgroup comes before the 'y' subgroup etc
grouped_and_ordered = [sorted(grouped[key]) for key in sorted(grouped.keys())]