Question

我有一个可以包含None和datetime对象的列表。我需要在连续的datetime个对象的子列表中拆分它，并且需要在原始列表中记录该子列表的第一个datetime对象的索引。

例如，我需要能够转向

original = [None, datetime(2013, 6, 4), datetime(2014, 5, 12), None, None, datetime(2012, 5, 18), None]

成：

(1, [datetime.datetime(2013, 6, 4, 0, 0), datetime.datetime(2014, 5, 12, 0, 0)])
(5, [datetime.datetime(2012, 5, 18, 0, 0)])

我尝试了两种方法。一个使用find：

binary = ''.join('1' if d else '0' for d in original)
end = 0
start = binary.find('1', end)
while start > -1:
    end = binary.find('0', start)
    if end < 0:
        end = len(binary)
    dates = original[start:end]
    print (start, dates)
    start = binary.find('1', end)

和一个使用groupby：

from itertools import groupby
for key, group in groupby(enumerate(original), lambda x: x[1] is not None):
    if key:
        group = list(group)
        start = group[0][0]
        dates = [t[1] for t in group]
        print (start, dates)

但两者对我来说似乎并不过分Pythonic。还有更好的方法吗？

Answer 1

我使用生成器生成元素，封装分组：

from itertools import takewhile

def indexed_date_groups(it):
    indexed = enumerate(it)
    for i, elem in indexed:
        if elem is not None:
           yield (
             i, [elem] + [v for i, v in takewhile(
                 lambda v: v[1] is not None, indexed)])

我们在找到初始非 - None对象后，使用itertools.takewhile()生成子列表。

当然，您仍然可以使用itertools.groupby()执行相同操作：

from itertools import groupby

def indexed_date_groups(it):
    for key, group in groupby(enumerate(it), lambda v: v[1] is not None):
        if key:
           indices, elems = zip(*group)
           yield indices[0], elems

演示：

>>> list(indexed_date_groups(original))
[(1, [datetime.datetime(2013, 6, 4, 0, 0), datetime.datetime(2014, 5, 12, 0, 0)]), (5, [datetime.datetime(2012, 5, 18, 0, 0)])]
>>> original = [None, datetime(2013, 6, 4), datetime(2014, 5, 12), None, None, datetime(2012, 5, 18), None]
>>> for index, group in indexed_date_groups(original):
...     print index, group
... 
1 [datetime.datetime(2013, 6, 4, 0, 0), datetime.datetime(2014, 5, 12, 0, 0)]
5 [datetime.datetime(2012, 5, 18, 0, 0)]

Answer 2

from itertools import groupby, count
idx = count()
for key, group in groupby(original, lambda x: x is not None):
    indices, group = zip(*((next(idx), i) for i in group))
    if key:
        print (indices[0], group)

Answer 3

基于gnibbler's answer。它省去count和zip并直接计算索引：

from itertools import groupby
idx = 0
for key, group in groupby(original, lambda x: x is not None):
    group = list(group)
    if key:
        print idx, group
    idx += len(group)

拆分无列表和记录索引

3 个答案: