Question

我正在尝试为使用情况统计信息生成一个报告。下面是我在数组中的示例数据，这是从Mysql表中获取的。如何实现一个逻辑，说明如果用户空闲时间超过30分钟他没有使用系统，否则计算使用时间的平均时间。

timestamp=[]
for i in timestamp:
   print i 

        2010-04-20 10:07:30
        2010-04-20 10:07:38
        2010-04-20 10:07:52
        2010-04-20 10:08:22
        2010-04-20 10:08:22
        2010-04-20 10:09:46
        2010-04-20 10:10:37
        2010-04-20 10:10:58
        2010-04-20 10:11:50
        2010-04-20 10:12:13
        2010-04-20 10:12:13
        2010-04-20 10:25:38
        2010-04-20 10:26:01
        2010-04-20 10:26:01
        2010-04-20 10:26:06
        2010-04-20 10:26:29
        2010-04-20 10:26:29
        2010-04-20 10:26:35
        2010-04-20 10:27:21
        2010-04-20 01:32:46
        2010-04-20 01:32:47
        2010-04-20 01:32:57
        2010-04-20 01:32:59
        2010-04-20 01:33:03
        2010-04-20 01:33:03
        2010-04-20 01:33:05
        2010-04-20 01:33:11
        2010-04-20 01:33:15
        2010-04-20 01:34:49
        2010-04-20 01:34:55
        2010-04-20 01:35:02
        2010-04-20 01:35:17
        2010-04-20 01:35:20
        2010-04-20 01:36:49
        2010-04-20 01:36:52
        2010-04-20 01:36:52
        2010-04-20 01:37:11
        2010-04-20 01:37:15
        2010-04-20 01:37:17
        2010-04-20 01:50:11
        2010-04-20 01:50:15
        2010-04-20 01:50:18
        2010-04-20 01:50:20
        2010-04-20 01:50:33
        2010-04-20 01:50:36
        2010-04-20 01:51:56

Answer 1

我认为这就是你想要的。它通过列表计算每个条目和前一个条目之间的差异。如果差异大于或等于30分钟，则忽略它。如果小于30分钟，则将其添加到该用户的总使用量中。（我假设所有时间戳都是针对同一个用户的。）

from datetime import datetime,timedelta

# Convert the timestamps to datetime objects
usetimes = sorted(datetime.strptime(d, '%Y-%m-%d %H:%M:%S') for d in timestamp)

# Set the idle time to compare with later
idletime = timedelta(minutes = 30)

# Start the running total with a timedelta of 0
usage = timedelta()
last = usetimes[0]

for d in usetimes[1:]:
    delta = d - last
    if delta < idletime:
        usage += delta
    last = d

print "total usage:",usage

如果你想使用sum()和zip()，你可以减少代码行，但我不确定它是否可读：

from datetime import datetime,timedelta
usetimes = sorted(datetime.strptime(d, '%Y-%m-%d %H:%M:%S') for d in timestamp)
idletime = timedelta(minutes = 30)
usage = sum((x - y for x,y in zip(usetimes[1:],usetimes[:-1]) if x - y < idletime),timedelta())
print "total usage:", usage

在这种情况下，如果时间戳列表很长，您可以考虑使用izip中的itertools而不是zip。

Python - 生成时差

1 个答案: