Question

说我有以下数据：

data = [['John', 1], ['Ada', 2], ['Ada', 3], ['Paul', 4],
        ['Paul', 5], ['Paul', 6], ['Kat', 7], ['Kat', 8]]

我可以按groupby：

按人分组

In [37]:

from itertools import groupby, izip_longest
from operator import itemgetter

for name, g in groupby(data, key=itemgetter(0)):
    print name, list(g)

John [['John', 1]]
Ada [['Ada', 2], ['Ada', 3]]
Paul [['Paul', 4], ['Paul', 5], ['Paul', 6]]
Kat [['Kat', 7], ['Kat', 8]]

我还可以使用recipe tools' grouper对每两个条目进行分组。我将复制/粘贴它以供参考：

In [38]:

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

for g in grouper(data, 2):
    print g
(['John', 1], ['Ada', 2])
(['Ada', 3], ['Paul', 4])
(['Paul', 5], ['Paul', 6])
(['Kat', 7], ['Kat', 8])

但是现在，我想迭代数据，使第一个元素包含John和Ada的数据，第二个元素包含Paul和Kat的数据。换句话说，我想像这样结合groupby和grouper：

In [39]:

person_iterator = groupby(data, key=itemgetter(0))
for group_iterator in grouper(person_iterator, 2):
    print [(keyvalue[0], list(keyvalue[1])) for keyvalue in group_iterator]

但输出并非我的预期：

[('John', []), ('Ada', [['Ada', 2], ['Ada', 3]])]
[('Paul', []), ('Kat', [['Kat', 7], ['Kat', 8]])]

为什么John和Paul有空列表？如何解决？

Answer 1

iterator产生的group_iterator[1] itertools.groupby在iterator屈服时会耗尽。

您需要将迭代器转换为序列，然后再将其传递给grouper以防止：

person_iterator = ((key, list(grp)) for key, grp in groupby(data, key=itemgetter(0)))
for group_iterator in grouper(person_iterator, 2):
    print [(key, value) for key, value in group_iterator]

输出：

[('John', [['John', 1]]), ('Ada', [['Ada', 2], ['Ada', 3]])]
[('Paul', [['Paul', 4], ['Paul', 5], ['Paul', 6]]), ('Kat', [['Kat', 7], ['Kat', 8]])]

Python itertools - 结合groupby和配方工具'石斑鱼

1 个答案: