按项目分组的元组列表

问题描述

我以以下列表为例:

[(148, Decimal('3.0')), (325, Decimal('3.0')), (148, Decimal('2.0')), (183, Decimal('1.0')), (308, Decimal('1.0')), (530, Decimal('1.0')), (594, Decimal('1.0')), (686, Decimal('1.0')), (756, Decimal('1.0')), (806, Decimal('1.0'))]

现在我想按ID分组，所以我将使用itemgetter(0):

Now i want to group by the id, so I will use itemgetter(0):

import operator, itertools
from decimal import *
test=[(148, Decimal('3.0')), (325, Decimal('3.0')), (148, Decimal('2.0')), (183, Decimal('1.0')), (308, Decimal('1.0')), (530, Decimal('1.0')), (594, Decimal('1.0')), (686, Decimal('1.0')), (756, Decimal('1.0')), (806, Decimal('1.0'))]

for _k, data in itertools.groupby(test, operator.itemgetter(0)):
    print list(data)

我不知道为什么，但是我得到了错误的输出:

I don't know why but I am getting this wrong output:

[(148, Decimal('3.0'))]
[(325, Decimal('3.0'))]
[(148, Decimal('2.0'))]
[(183, Decimal('1.0'))]
[(308, Decimal('1.0'))]
[(530, Decimal('1.0'))]
[(594, Decimal('1.0'))]
[(686, Decimal('1.0'))]
[(756, Decimal('1.0'))]
[(806, Decimal('1.0'))]

如您所见，输出未按ID分组.但是，如果我使用itemgetter(1)，则上面的代码可以正常工作.输出按十进制值分组.

As you can see the output is not grouped by id. However the code above works fine if I use itemgetter(1). The output is grouped by decimal val.

[(148, Decimal('3.0')), (325, Decimal('3.0'))]
[(148, Decimal('2.0'))]
[(183, Decimal('1.0')), (308, Decimal('1.0')), (530, Decimal('1.0')), (594, Decimal('1.0')), (686, Decimal('1.0')), (756, Decimal('1.0')), (806, Decimal('1.0'))]

我在这里想念什么?

推荐答案

您首先需要对 groupby 的数据进行排序，然后将分组连续元素(根据您提供的密钥):

You would first need to sort the data for the groupby to work, it groups consecutive elements based on the key you provide:

import operator, itertools
from decimal import *
test=[(148, Decimal('3.0')), (325, Decimal('3.0')), (148, Decimal('2.0')), (183, Decimal('1.0')), (308, Decimal('1.0')), (530, Decimal('1.0')), (594, Decimal('1.0')), (686, Decimal('1.0')), (756, Decimal('1.0')), (806, Decimal('1.0'))]

for _k, data in itertools.groupby(sorted(test), operator.itemgetter(0)):
    print list(data)

但是您最好使用字典进行分组以避免不必要的O(n log n)排序:

But you would be better using a dict to group to avoid an unnecessary O(n log n) sort:

from collections import defaultdict

d = defaultdict(list)

for t in test:
    d[t[0]].append(t)

for v in d.values():
    print(v)

两者都会给您相同的分组，只是不一定要以相同的顺序.

Both would give you the same groupings, just not necessarily in the same order.

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