Returning the product of a list

Question

Is there a more concise  efficient or simply pythonic way to do the following  def product lst       p   1     for i in lst          p    i     return p  EDIT  I actually find that this is marginally faster than using operator mul  from operator import mul   from functools import reduce   python3 compatibility  def with lambda lst       reduce lambda x  y  x   y  lst   def without lambda lst       reduce mul  lst   def forloop lst       r   1     for x in lst          r    x     return r  import timeit  a   range 50  b   range 1 50  no zero t   timeit Timer  quot with lambda a  quot    quot from   main   import with lambda a quot   print  quot with lambda  quot   t timeit    t   timeit Timer  quot without lambda a  quot    quot from   main   import without lambda a quot   print  quot without lambda  quot   t timeit    t   timeit Timer  quot forloop a  quot    quot from   main   import forloop a quot   print  quot for loop  quot   t timeit     t   timeit Timer  quot with lambda b  quot    quot from   main   import with lambda b quot   print  quot with lambda  no 0   quot   t timeit    t   timeit Timer  quot without lambda b  quot    quot from   main   import without lambda b quot   print  quot without lambda  no 0   quot   t timeit    t   timeit Timer  quot forloop b  quot    quot from   main   import forloop b quot   print  quot for loop  no 0   quot   t timeit     gives me   with lambda    17 755449056625366    without lambda    8 2084708213806152    for loop    7 4836349487304688    with lambda  no 0     22 570688009262085    without lambda  no 0     12 472226858139038    for loop  no 0     11 04065990447998

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Python 3 result for the OP s tests   best of 3 for each   with lambda  18 978000981995137 without lambda  8 110567473006085 for loop  10 795806062000338 with lambda  no 0   26 612515013999655 without lambda  no 0   14 704098362999503 for loop  no 0   14 93075215499266

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This also works though its cheating  def factorial n       x        if n  lt   1          return 1     else          for i in range 1 n 1                p  i             x append p          print x n-1

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The intent of this answer is to provide a calculation that is useful in certain circumstances -- namely when a  there are a large number of values being multiplied such that the final product may be extremely large or extremely small  and b  you don t really care about the exact answer  but instead have a number of sequences  and want to be able to order them based on each one s product   If you want to multiply the elements of a list  where l is the list  you can do   import math math exp sum map math log  l      Now  that approach is not as readable as  from operator import mul reduce mul  list    If you re a mathematician who isn t familiar with reduce   the opposite might be true  but I wouldn t advise using it under normal circumstances  It s also less readable than the product   function mentioned in the question  at least to non-mathematicians     However  if you re ever in a situation where you risk underflow or overflow  such as in   gt  gt  gt  reduce mul   10   309  inf   and your purpose is to compare the products of different sequences rather than to know what the products are  then   gt  gt  gt  sum map math log   10   309   711 49879373515785   is the way to go because it s virtually impossible to have a real-world problem in which you would overflow or underflow with this approach   The larger the result of that calculation is  the larger the product would be if you could calculate it

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I remember some long discussions on comp lang python  sorry  too lazy to produce pointers now  which concluded that your original product   definition is the most Pythonic   Note that the proposal is not to write a for loop every time you want to do it  but to write a function once  per type of reduction  and call it as needed   Calling reduction functions is very Pythonic - it works sweetly with generator expressions  and since the sucessful introduction of sum    Python keeps growing more and more builtin reduction functions - any   and all   are the latest additions     This conclusion is kinda official - reduce   was removed from builtins in Python 3 0  saying       Use functools reduce   if you really need it  however  99 percent of the time an explicit for loop is more readable     See also The fate of reduce   in Python 3000 for a supporting quote from Guido  and some less supporting comments by Lispers that read that blog    P S  if by chance you need product   for combinatorics  see math factorial    new 2 6

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if you just have numbers in your list   from numpy import prod prod list    EDIT  as pointed out by  off99555 this does not work for large integer results in which case it returns a result of type numpy int64 while Ian Clelland s solution based on operator mul and reduce works for large integer results because it returns long

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import operator reduce operator mul  list  1

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One option is to use numba and the  jit or  njit decorator  I also made one or two little tweaks to your code  at least in Python 3   list  is a keyword that shouldn t be used for a variable name     njit def njit product lst       p   lst 0     first element     for i in lst 1       loop over remaining elements         p    i     return p   For timing purposes  you need to run once to compile the function first using numba  In general  the function will be compiled the first time it is called  and then called from memory after that  faster    njit product  1  2      execute once to compile   Now when you execute your code  it will run with the compiled version of the function  I timed them using a Jupyter notebook and the  timeit magic function   product b     yours   32 7   s    510 ns per loop  mean    std  dev  of 7 runs  10000 loops each   njit product b    92 9   s    392 ns per loop  mean    std  dev  of 7 runs  10000 loops each    Note that on my machine  running Python 3 5  the native Python for loop was actually the fastest  There may be a trick here when it comes to measuring numba-decorated performance with Jupyter notebooks and the  timeit magic function  I am not sure that the timings above are correct  so I recommend trying it out on your system and seeing if numba gives you a performance boost

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Without using lambda   from operator import mul reduce mul  list  1    it is better and faster  With python 2 7 5  from operator import mul import numpy as np import numexpr as ne   from functools import reduce   python3 compatibility  a   range 1  101   timeit reduce lambda x  y  x   y  a       1   timeit reduce mul  a                      2   timeit np prod a                          3   timeit ne evaluate  prod a                4    In the following configuration   a   range 1  101     A a   np array a       B a   np arange 1  1e4  dtype int   C a   np arange 1  1e5  dtype float   D   Results with python 2 7 5                 1           2           3           4       ------- ----------- ----------- ----------- -----------   A       20 8   s     13 3   s     22 6   s     39 6   s       B        106   s     95 3   s     5 92   s     26 1   s  C       4 34 ms     3 51 ms     16 7   s     38 9   s  D       46 6 ms     38 5 ms      180   s      216   s   Result  np prod is the fastest one  if you use np array as data structure  18x for small array  250x for large array   with python 3 3 2                  1           2           3           4       ------- ----------- ----------- ----------- -----------   A       23 6   s     12 3   s     68 6   s     84 9   s       B        133   s      107   s     7 42   s     27 5   s  C       4 79 ms     3 74 ms     18 6   s     40 9   s  D       48 4 ms     36 8 ms      187   s      214   s   Is python 3 slower

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I ve tested various solutions with perfplot  a small project of mine  and found that numpy prod lst   is by far the fastest solution  if the list isn t very short     Code to reproduce the plot  import perfplot import numpy  import math from operator import mul from functools import reduce  from itertools import accumulate   def reduce lambda lst       return reduce lambda x  y  x   y  lst    def reduce mul lst       return reduce mul  lst    def forloop lst       r   1     for x in lst          r    x     return r   def numpy prod lst       return numpy prod lst    def math prod lst       return math prod lst    def itertools accumulate lst       for value in accumulate lst  mul           pass     return value   perfplot show      setup numpy random rand      kernels  reduce lambda  reduce mul  forloop  numpy prod  itertools accumulate  math prod       n range  2    k for k in range 15        xlabel  quot len a  quot       logx True      logy True

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Starting Python 3 8  a prod function has been included to the math module in the standard library      math prod iterable     start 1    which returns the product of a start value  default  1  times an iterable of numbers   import math  math prod  2  3  4     24   Note that if the iterable is empty  this will produce 1  or the start value if provided

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Well if you really wanted to make it one line without importing anything you could do   eval     join str item  for item in list     But don t

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The fastest way I found was  using while   mysetup       import numpy as np from find intervals import return intersections         code snippet whose execution time is to be measured mycode        x    4 5 6 7 8 9 10  prod   1 i   0 while True      prod   prod   x i      i   i   1     if i    len x           break        timeit statement for while  print  using while      timeit timeit setup mysetup                stmt mycode      timeit statement for mul  print  using mul          timeit timeit  from functools import reduce      from operator import mul      c   reduce mul   4 5 6 7 8 9 10         timeit statement for mul  print  using lambda                timeit timeit  from functools import reduce      from operator import mul      c   reduce lambda x  y  x   y   4 5 6 7 8 9 10        and the timings are    gt  gt  gt  using while   0 8887967770060641   gt  gt  gt  using mul   2 0838719510065857   gt  gt  gt  using lambda   2 4227715369997895

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reduce lambda x  y  x   y  list  1

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I am surprised no-one has suggested using itertools accumulate with operator mul  This avoids using reduce  which is different for Python 2 and 3  due to the functools import required for Python 3   and moreover is considered un-pythonic by Guido van Rossum himself   from itertools import accumulate from operator import mul  def prod lst       for value in accumulate lst  mul           pass     return value   Example   prod  1 5 4 3 5 6     1800

[python] Returning the product of a list

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