Question

我正在搜索连接列表的元素。

例如： l=[0,1,2,3,4,5,6,7,8,9]

我希望：

l_concatenated=45

我的代码是：

  from datetime import *
  data=[[datetime.datetime(2012, 8, 19, 15, 28, 2, 810000), 159],
  [datetime.datetime(2012, 8, 19, 15, 28, 2, 811000), 159],
  [datetime.datetime(2012, 8, 19, 15, 28, 3, 108000), 159],
  [datetime.datetime(2012, 8, 19, 15, 28, 3, 108000), 159],
  [datetime.datetime(2012, 8, 19, 15, 28, 3, 182000), 159],
  [datetime.datetime(2012, 8, 19, 15, 28, 3, 182000), 159],
  [datetime.datetime(2012, 8, 19, 15, 28, 3, 326000), 159],
  [datetime.datetime(2012, 8, 19, 15, 28, 3, 326000), 159]]

  date=[i[0] for i in data]
  interval=[b-a for a,b in zip(date[:-1],date[1:])]
  sorted_interval=[item for item in interval if item!=timedelta(0)]
  # It doesn't function and just give wrong information
  interval_concatenated=[ko[i]+ko[i+1] for i in range(len(sorted_interval)-1)]

编辑：我需要恢复两个日期之间的平均间隔。与interval=[b-a for a,b in zip(date[:-1],date[1:])]我设置2个连续值之间的间隔。例如：它可能看起来interval=[timedelta(0, 0, 297000),timedelta(0, 0, 297000),timedelta(0, 0, 123000),timedelta(0, 1, 300000)]，我想要间隔的平均值。

Answer 1

首先，您可以使用zip计算连续日期之间的间隔：

intervals = [d2[0] - d1[0] for (d1, d2) in zip(data, data[1:])]

然后，得到总timedelta：

total_interval = sum(intervals, timedelta(0, 0, 0))

最后，计算平均值：

average_interval = total_interval / len(intervals)

Answer 2

我认为你在寻找这个：

import datetime
import time

data=[[datetime.datetime(2012, 8, 19, 15, 28, 2, 810000), 159],
      [datetime.datetime(2012, 8, 19, 15, 28, 2, 811000), 159],
      [datetime.datetime(2012, 8, 19, 15, 28, 3, 108000), 159],
      [datetime.datetime(2012, 8, 19, 15, 28, 3, 108000), 159],
      [datetime.datetime(2012, 8, 19, 15, 28, 3, 182000), 159],
      [datetime.datetime(2012, 8, 19, 15, 28, 3, 182000), 159],
      [datetime.datetime(2012, 8, 19, 15, 28, 3, 326000), 159],
      [datetime.datetime(2012, 8, 19, 15, 28, 3, 326000), 159]]

date = [time.mktime(i[0].timetuple()) for i in data]

average = datetime.datetime.fromtimestamp(sum(date)/len(date))

首先将datetime对象转换为Posix time，然后计算平均值并最终将其转换回来。

在一个值中对列表（日期时间）的元素求和

2 个答案: