Average timedelta in list

Question

I want to calculate the avarage timedelta between dates in a list. Although the following works well, I\'m wondering if there\'s a smarter way?

delta = lambd         


        

Answer 1

Try this:

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)

In my opinion doing it like this is more readable. A comment for less mathematically inclined readers of your code might help to explain how your are calculating each delta. For what it's worth, one generator expression has the least (and I think least slow) opcode instructions of anything I looked at.

  # 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

In particular, dropping the lambda allows us to avoid making a closure, building a tuple and loading two closures. Five functions get called either way. Of course this sort of concern with performance is sort of ridiculous but it is nice to know what's going on under the hood. The most important thing is readability and I think that doing it this way scores high on that as well.

Answer 2

sum(timedelta_list ,datetime.timedelta())

Answer 3

Btw, if you have a list of timedeltas or datetimes, why do you even do any math yourself?

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)