列表中的平均时间差

Question

我想计算一个日期列表中日期之间的平均时间差。虽然下面的方法效果不错，但我在想有没有更聪明的办法？

delta = lambda last, next: (next - last).seconds + (next - last).days * 86400   
total = sum(delta(items[i-1], items[i]) for i in range(1, len(items)))
average = total / (len(items) - 1)

Answer 1

试试这个：

from itertools import izip

def average(items):   
    total = sum((next - last).seconds + (next - last).days * 86400
                for next, last in izip(items[1:], items))
     return total / (len(items) - 1)

我觉得这样写更容易读懂。对于那些数学基础不太好的读者，添加一个注释可以帮助解释你是如何计算每个增量的。顺便说一下，我看到的所有方法中，使用一个生成器表达式的指令是最少的（而且我觉得执行速度也最慢）。

  # The way in your question compiles to....
  3           0 LOAD_CONST               1 (<code object <lambda> at 0xb7760ec0, file 

"scratch.py", line 3>)
              3 MAKE_FUNCTION            0
              6 STORE_DEREF              1 (delta)

  4           9 LOAD_GLOBAL              0 (sum)
             12 LOAD_CLOSURE             0 (items)
             15 LOAD_CLOSURE             1 (delta)
             18 BUILD_TUPLE              2
             21 LOAD_CONST               2 (<code object <genexpr> at 0xb77c0a40, file "scratch.py", line 4>)
             24 MAKE_CLOSURE             0
             27 LOAD_GLOBAL              1 (range)
             30 LOAD_CONST               3 (1)
             33 LOAD_GLOBAL              2 (len)
             36 LOAD_DEREF               0 (items)
             39 CALL_FUNCTION            1
             42 CALL_FUNCTION            2
             45 GET_ITER            
             46 CALL_FUNCTION            1
             49 CALL_FUNCTION            1
             52 STORE_FAST               1 (total)

  5          55 LOAD_FAST                1 (total)
             58 LOAD_GLOBAL              2 (len)
             61 LOAD_DEREF               0 (items)
             64 CALL_FUNCTION            1
             67 LOAD_CONST               3 (1)
             70 BINARY_SUBTRACT     
             71 BINARY_DIVIDE       
             72 STORE_FAST               2 (average)
             75 LOAD_CONST               0 (None)
             78 RETURN_VALUE        
None
#
#doing it with just one generator expression and itertools...

  4           0 LOAD_GLOBAL              0 (sum)
              3 LOAD_CONST               1 (<code object <genexpr> at 0xb777eec0, file "scratch.py", line 4>)
              6 MAKE_FUNCTION            0

  5           9 LOAD_GLOBAL              1 (izip)
             12 LOAD_FAST                0 (items)
             15 LOAD_CONST               2 (1)
             18 SLICE+1             
             19 LOAD_FAST                0 (items)
             22 CALL_FUNCTION            2
             25 GET_ITER            
             26 CALL_FUNCTION            1
             29 CALL_FUNCTION            1
             32 STORE_FAST               1 (total)

  6          35 LOAD_FAST                1 (total)
             38 LOAD_GLOBAL              2 (len)
             41 LOAD_FAST                0 (items)
             44 CALL_FUNCTION            1
             47 LOAD_CONST               2 (1)
             50 BINARY_SUBTRACT     
             51 BINARY_DIVIDE       
             52 RETURN_VALUE        
None

特别是，去掉lambda函数可以让我们避免创建一个闭包、构建一个元组和加载两个闭包。无论如何，五个函数都会被调用。当然，过于关注性能有点过头，但知道底层是怎么运作的也不错。最重要的是可读性，我觉得这样做在可读性上也得分很高。

Answer 2

如果你有一个时间差的列表：

import pandas as pd
avg=pd.to_timedelta(pd.Series(yourtimedeltalist)).mean()

Answer 3

顺便问一下，如果你手里有一堆时间差（timedelta）或者日期时间（datetime），那你为什么还要自己去算呢？

import datetime

datetimes = [ ... ]

# subtracting datetimes gives timedeltas
timedeltas = [datetimes[i-1]-datetimes[i] for i in range(1, len(datetimes))]

# giving datetime.timedelta(0) as the start value makes sum work on tds 
average_timedelta = sum(timedeltas, datetime.timedelta(0)) / len(timedeltas)