Question

我正在处理日志。需要计算流程运行时持续时间的总和，而不会造成长时间的中断。将最大可能中断时间设置为30秒。每3秒发出一次日志。

因此，例如，如果它从10:20:00（小时）到10:30:00（小时）开始运行，并且从10:24:10到10:27:10被中断，则期望的结果是10:24:10-10:20:00和10:30:00-10:27:10 = 420的总和（以秒为单位）。但是，使用datetime类型计算时差不能提供有效的解决方案-我想它可以计算时差而不包含开始/结束秒数。

这是我想出的解决方案（['timestamps']是通常每3秒发出的datetime个时间戳的列表）

for k, v in proc_activity.items():
        proc_activity[k]['duration'] = 0

        start, next = v['timestamps'][0], ''
        for time in v['timestamps']:
            next = time
            diff = next - start

            if diff.seconds < 30:
                proc_activity[k]['duration'] += diff.seconds
            else:
                print("diff: %s" % diff.seconds)

            start = next

        print(f"added: {proc_activity[k]['duration']}")
        diff = v['timestamps'][-1] - v['timestamps'][0]
        print(f"real: {diff.seconds}")

输出：

added: 39
real: 45
added: 39
real: 45
diff: 36
added: 155
real: 218

任何建议如何解决？

更新，示例输入数据：

{'service_0': {'timestamps': [datetime.datetime(2018, 7, 1, 22, 33, 39, 86170),
                                     datetime.datetime(2018, 7, 1, 22, 33, 42, 33213),
                                     datetime.datetime(2018, 7, 1, 22, 33, 44, 898234),
                                     datetime.datetime(2018, 7, 1, 22, 33, 47, 893731),
                                     datetime.datetime(2018, 7, 1, 22, 33, 50, 928946),
                                     datetime.datetime(2018, 7, 1, 22, 33, 53, 895617),
                                     datetime.datetime(2018, 7, 1, 22, 35, 7, 116182),
                                     datetime.datetime(2018, 7, 1, 22, 35, 10, 105035),
                                     datetime.datetime(2018, 7, 1, 22, 35, 13, 193428),
                                     datetime.datetime(2018, 7, 1, 22, 35, 16, 210135),
                                     datetime.datetime(2018, 7, 1, 22, 35, 19, 168881),
                                     datetime.datetime(2018, 7, 1, 22, 35, 22, 114653),
                                     datetime.datetime(2018, 7, 1, 22, 35, 25, 102365),
                                     datetime.datetime(2018, 7, 1, 22, 35, 43, 46950),
                                     datetime.datetime(2018, 7, 1, 22, 35, 46, 15435),
                                     datetime.datetime(2018, 7, 1, 22, 35, 49, 23333),
                                     datetime.datetime(2018, 7, 1, 22, 35, 52, 22164),
                                     datetime.datetime(2018, 7, 1, 22, 35, 55, 78615),
                                     datetime.datetime(2018, 7, 1, 22, 35, 58, 78573)]}}

Answer 1

简而言之，我认为您缺少的关键是使用timedelta.total_seconds()而不是timedelta.seconds

这对我来说似乎很好：

import datetime
from pprint import pprint

def get_duration(timestamps):
    max_interruption = 30
    starts = timestamps[:-1]
    ends = timestamps[1:]
    durations = zip(starts, ends)
    accumulated = 0
    for start, end in durations:
        delta = (end - start).total_seconds()
        if delta < max_interruption:
            accumulated += delta
    return accumulated

proc_activity = {
    'service_0': {
        'timestamps': [
            datetime.datetime(2018, 7, 1, 22, 33, 39, 86170),
            datetime.datetime(2018, 7, 1, 22, 33, 42, 33213),
            datetime.datetime(2018, 7, 1, 22, 33, 44, 898234),
            datetime.datetime(2018, 7, 1, 22, 33, 47, 893731),
            datetime.datetime(2018, 7, 1, 22, 33, 50, 928946),
            datetime.datetime(2018, 7, 1, 22, 33, 53, 895617),
            datetime.datetime(2018, 7, 1, 22, 35, 7, 116182),
            datetime.datetime(2018, 7, 1, 22, 35, 10, 105035),
            datetime.datetime(2018, 7, 1, 22, 35, 13, 193428),
            datetime.datetime(2018, 7, 1, 22, 35, 16, 210135),
            datetime.datetime(2018, 7, 1, 22, 35, 19, 168881),
            datetime.datetime(2018, 7, 1, 22, 35, 22, 114653),
            datetime.datetime(2018, 7, 1, 22, 35, 25, 102365),
            datetime.datetime(2018, 7, 1, 22, 35, 43, 46950),
            datetime.datetime(2018, 7, 1, 22, 35, 46, 15435),
            datetime.datetime(2018, 7, 1, 22, 35, 49, 23333),
            datetime.datetime(2018, 7, 1, 22, 35, 52, 22164),
            datetime.datetime(2018, 7, 1, 22, 35, 55, 78615),
            datetime.datetime(2018, 7, 1, 22, 35, 58, 78573)
        ],
    }
}

for k,v in proc_activity.items():
    proc_activity[k]['duration'] = get_duration(v['timestamps'])

pprint(proc_activity)

持续时间65.77183800000002秒

Answer 2

My try to the problem using generators:

import datetime

def calculate(timestamps, largest_interrupt = 30):
    begin_t, last_good_t = timestamps[0], timestamps[0]
    for current_t, previous_t in zip(timestamps[1:], timestamps):
        if (current_t - last_good_t).total_seconds() < largest_interrupt:
            last_good_t = current_t
            continue
        yield (previous_t - begin_t).total_seconds()
        last_good_t, begin_t = current_t, current_t
    yield (current_t - begin_t).total_seconds()

sample_data = {'service_0': {'timestamps': [datetime.datetime(2018, 7, 1, 22, 33, 39, 86170),
                                     datetime.datetime(2018, 7, 1, 22, 33, 42, 33213),
                                     datetime.datetime(2018, 7, 1, 22, 33, 44, 898234),
                                     datetime.datetime(2018, 7, 1, 22, 33, 47, 893731),
                                     datetime.datetime(2018, 7, 1, 22, 33, 50, 928946),
                                     datetime.datetime(2018, 7, 1, 22, 33, 53, 895617),
                                     datetime.datetime(2018, 7, 1, 22, 35, 7, 116182),
                                     datetime.datetime(2018, 7, 1, 22, 35, 10, 105035),
                                     datetime.datetime(2018, 7, 1, 22, 35, 13, 193428),
                                     datetime.datetime(2018, 7, 1, 22, 35, 16, 210135),
                                     datetime.datetime(2018, 7, 1, 22, 35, 19, 168881),
                                     datetime.datetime(2018, 7, 1, 22, 35, 22, 114653),
                                     datetime.datetime(2018, 7, 1, 22, 35, 25, 102365),
                                     datetime.datetime(2018, 7, 1, 22, 35, 43, 46950),
                                     datetime.datetime(2018, 7, 1, 22, 35, 46, 15435),
                                     datetime.datetime(2018, 7, 1, 22, 35, 49, 23333),
                                     datetime.datetime(2018, 7, 1, 22, 35, 52, 22164),
                                     datetime.datetime(2018, 7, 1, 22, 35, 55, 78615),
                                     datetime.datetime(2018, 7, 1, 22, 35, 58, 78573)]}}


for k, v in sample_data.items():
    s = sum(calculate(v['timestamps']))
    print(f"Service '{k}' has duration of '{s}' seconds")

The program prints Service 'service_0' has duration of '65.771838' seconds

python 3.6时间戳之间的短时间之和

2 个答案: