列表中的平均时间增量

Question

我想计算列表中日期之间的平均时间差。尽管以下方法效果很好，但我想知道是否有更明智的方法？

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

顺便说一句，如果您有时间增量或日期时间列表，为什么还要自己做任何数学运算？

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)

Answer 2

尝试一下：

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)

我认为这样做的做法更具可读性。对数学上不太偏爱代码的读者的注释可能有助于解释您如何计算每个增量。就其价值而言，一个生成器表达式具有我所看过的任何事物中最少（而且我认为最慢）的操作码指令。

  # 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，放下lambda可以避免创建闭包，构建元组并加载两个闭包。两种方法都调用了五个函数。当然，这种对性能的关注有点荒谬，但是很高兴知道引擎盖下发生了什么。最重要的是可读性，我认为以此方式也能获得很高的评分。