计算连续两次之间出现的次数 python

Question

我有一些日期时间信息如下：

DateTime                          
2018/01/01 01:00:00             
2018/01/01 02:30:00
2018/01/01 03:10:00
2018/01/01 04:00:00
2018/01/01 05:25:00

我还有一些其他的事件发生时间：

occurrence
2018/01/01 01:01:00             
2018/01/01 01:02:00
2018/01/01 02:31:00
2018/01/01 04:05:00

我想计算任意两个连续时间间隔之间事件发生的次数，以生成以下内容：

Occurrence      Start Time               End Time
2               2018/01/01 01:00:00      2018/01/01 02:30:00
1               2018/01/01 02:30:00      2018/01/01 03:10:00
0               2018/01/01 03:10:00      2018/01/01 04:00:00
1               2018/01/01 04:00:00      2018/01/01 05:25:00

我正在考虑使用类似

的东西

sum(1 if meets_condition(x) else 0 for x in my_list)

但我不知道如何实现。有人可以帮忙吗？

Answer 1

您可以使用：

from datetime import datetime
from collections import Counter

t = """2018/01/01 01:00:00         
2018/01/01 02:30:00
2018/01/01 03:10:00
2018/01/01 04:00:00
2018/01/01 05:25:00"""

occurrence = """2018/01/01 01:01:00             
2018/01/01 01:02:00
2018/01/01 02:31:00
2018/01/01 04:05:00"""


fmt ='%Y/%m/%d %H:%M:%S'
dates = [datetime.strptime(d.strip(), fmt) for d in t.split('\n')]
intervals = [(d1, d2) for d1, d2 in zip(dates, dates[1:])] # already sorted
occ = [datetime.strptime(d.strip(), fmt) for d in occurrence.split('\n')]

count = Counter()
for o in occ:
    for d1, d2 in intervals:
        if o < d2:
            count[(d1, d2)] += 1
            break


print('Occurrence      Start Time               End Time')
for d1, d2 in intervals:
    print(str(count[(d1, d2)]).ljust(15), d1.strftime(fmt).ljust(24), d2.strftime(fmt))

输出：

Occurrence      Start Time               End Time
2               2018/01/01 01:00:00      2018/01/01 02:30:00
1               2018/01/01 02:30:00      2018/01/01 03:10:00
0               2018/01/01 03:10:00      2018/01/01 04:00:00
1               2018/01/01 04:00:00      2018/01/01 05:25:00

Answer 2

试试这个：

# pip install dateutil-python
from collections import Counter
import dateutil.parser

bases = [
    "2018/01/01 01:00:00",
    "2018/01/01 02:30:00",
    "2018/01/01 03:10:00",
    "2018/01/01 04:00:00",
    "2018/01/01 05:25:00"
]

occurances = [
    "2018/01/01 01:01:00",
    "2018/01/01 01:02:00",
    "2018/01/01 02:31:00",
    "2018/01/01 04:05:00"
]

class DateInterval:
    def __init__(self, start, end):
        self.start = dateutil.parser.parse(start)
        self.end = dateutil.parser.parse(end)
    def __contains__(self, other) -> bool:
        return self.start <= dateutil.parser.parse(other) <= self.end
    def __repr__(self):
        return f'<Interval:{self.start} ~ {self.end}>'

my_intervals = Counter({DateInterval(i, bases[n+1]): 0 for n, i in enumerate(bases) if n < len(bases)-1})

for occ in occurances:
    for intr in my_intervals:
        my_intervals[intr] += int(occ in intr)

print(my_intervals)

Answer 3

您的 sum/for 组合是正确的，但我们必须对 python 的日期时间 class 做一些额外的魔法。在这里，我们只是将字符串解析为日期时间对象，然后使用比较运算符来检查它是否在范围内。

from datetime import datetime as dt

dates = ["2018/01/01 01:00:00", "2018/01/01 02:30:00", "2018/01/01 04:00:00"]
events = ["2018/01/01 01:01:00", "2018/01/01 01:02:00", "2018/01/01 02:31:00"]

dates = [dt.fromisoformat(date.replace("/", "-")) for date in dates]
events = [dt.fromisoformat(event.replace("/", "-")) for event in events] 
buckets = [(x,y) for x,y in zip(dates, dates[1:])]

result = dict()

for start, end in buckets:  
  result[str(start) + " to " + str(end)] = sum([1 if start >= date < end else 0 for date in dates])

print(result)
# {'2018-01-01 01:00:00 to 2018-01-01 02:30:00': 1, '2018-01-01 02:30:00 to 2018-01-01 04:00:00': 2}

计算连续两次之间出现的次数 python

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