给定一个日期时间列表:
A= [datetime.datetime(2019, 12, 14, 0, 0), datetime.datetime(2019, 12, 16, 0, 0),
datetime.datetime(2019, 12, 24, 0, 0), datetime.datetime(2019, 12, 25, 0, 0),
datetime.datetime(2019, 12, 26, 0, 0), datetime.datetime(2019, 12, 27, 0, 0),
]
我想对相差 1 天的日期时间进行分组，以便输出为:
[datetime.datetime(2019, 12, 14, 0, 0),
datetime.datetime(2019, 12, 16, 0, 0),
[datetime.datetime(2019, 12, 24, 0, 0), datetime.datetime(2019, 12, 25, 0, 0),
datetime.datetime(2019, 12, 26, 0, 0), datetime.datetime(2019, 12, 27, 0, 0)]]
我已经试过了，但是失败了:
res = []
temp = []
l = len(A)-1
for i,v in enumerate(A):
if i+1 > l:
break
if abs(v.days() - A[i+1].days()) == 1:
temp.append(v)
else:
if temp:
temp.append(v)
res.append(temp)
temp = []
print(res)
请您参考如下方法:
问题的第一个版本是关于整数的，所以这里首先回答“将差为 1 的整数分组”(请参阅下面的日期时间对象)。
您可以使用 itertools.groupby和 itertools.count像这样对它们进行分组:
from itertools import groupby, count
counter = count()
A= [1,14,24,25,26,27,28,29,30,31]
out = [list(group) for _, group in groupby(A, key=lambda v:v-next(counter))]
print(out)
# [[1], [14], [24, 25, 26, 27, 28, 29, 30, 31]]
如果您真的希望孤立的值出现在任何列表之外，只需添加:
out = [sublist[0] if len(sublist)==1 else sublist for sublist in out]
print(out)
[1, 14, [24, 25, 26, 27, 28, 29, 30, 31]]
next(counter) 每次被调用都会返回 0, 1, 2...等等。如果值之间的差异为 1，则 v-next(counter) 将保持不变，并且 groupby 对保持不变的项目进行分组。
对于问题的第二个版本，我们可以对具有一天差异的 datetime 对象保持相同的原则:我们减去一个 datetime.timedelta(days=next(counter)) 而不仅仅是 next(counter)。
from itertools import groupby, count
import datetime
A= [datetime.datetime(2019, 12, 14, 0, 0), datetime.datetime(2019, 12, 16, 0, 0),
datetime.datetime(2019, 12, 24, 0, 0), datetime.datetime(2019, 12, 25, 0, 0),
datetime.datetime(2019, 12, 26, 0, 0), datetime.datetime(2019, 12, 27, 0, 0),
]
counter = count()
out = [list(group) for _, group in groupby(A, key=lambda v:v-datetime.timedelta(days=next(counter)))]
# print(out)
# [1], [14], [24, 25, 26, 27, 28, 29, 30, 31]]
out = [sublist[0] if len(sublist)==1 else sublist for sublist in out]
print(out)
# [datetime.datetime(2019, 12, 14, 0, 0), datetime.datetime(2019, 12, 16, 0, 0),
# [datetime.datetime(2019, 12, 24, 0, 0), datetime.datetime(2019, 12, 25, 0, 0),
# datetime.datetime(2019, 12, 26, 0, 0), datetime.datetime(2019, 12, 27, 0, 0)]]
