string goes away

[Andreas Beyer]

OK, you won. I read in an (regretably old) guidline for improving 
Python's performance that you should prefer map() compared to list 
comprehensions. Apparently the performance of list comprehensions has 
improved a lot, which is great. (Or the overhead of calling map() got 
too big, but I hope this is not the case.)  So, what is the purpose of 
map()? Should it too be deprecated?

Andreas

Skip Montanaro wrote:

>    Andreas> Yeeh, I was expecting something like that. The only reason to
>    Andreas> use map() at all is for improving the performance.  That is
>    Andreas> lost when using list comprehensions (as far as I know). So,
>    Andreas> this is *no* option for larger jobs.
>
>Did you test your hypothesis?
>
>    % python -m timeit -s 'lst = ("abc"*10).split() ; import string' 'map(string.upper, lst)'
>    100000 loops, best of 3: 9.24 usec per loop
>    % python -m timeit -s 'lst = ("abc"*10).split()' '[s.upper() for s in lst]'
>    100000 loops, best of 3: 4.18 usec per loop
>    % python -m timeit -s 'lst = ("abc"*100).split() ; import string' 'map(string.upper, lst)'
>    100000 loops, best of 3: 16.1 usec per loop
>    % python -m timeit -s 'lst = ("abc"*100).split()' '[s.upper() for s in lst]'
>    100000 loops, best of 3: 10.8 usec per loop
>    % python -m timeit -s 'lst = ("abc"*1000).split() ; import string' 'map(string.upper, lst)'
>    10000 loops, best of 3: 72.7 usec per loop
>    % python -m timeit -s 'lst = ("abc"*1000).split()' '[s.upper() for s in lst]'
>    10000 loops, best of 3: 67.7 usec per loop
>    % python -m timeit -s 'lst = ("abc"*10000).split() ; import string' 'map(string.upper, lst)'
>    1000 loops, best of 3: 844 usec per loop
>    % python -m timeit -s 'lst = ("abc"*10000).split()' '[s.upper() for s in lst]'
>    1000 loops, best of 3: 828 usec per loop
>
>    % python -m timeit -s 'lst = ["abc"*10]*10 ; import string' 'map(string.upper, lst)'
>    10000 loops, best of 3: 42.7 usec per loop
>    % python -m timeit -s 'lst = ["abc"*10]*10' '[s.upper() for s in lst]'
>    10000 loops, best of 3: 26.5 usec per loop
>    % python -m timeit -s 'lst = ["abc"*10]*100 ; import string' 'map(string.upper, lst)'
>    1000 loops, best of 3: 376 usec per loop
>    % python -m timeit -s 'lst = ["abc"*10]*100' '[s.upper() for s in lst]'
>    1000 loops, best of 3: 230 usec per loop
>    % python -m timeit -s 'lst = ["abc"*10]*1000 ; import string' 'map(string.upper, lst)'
>    100 loops, best of 3: 3.72 msec per loop
>    % python -m timeit -s 'lst = ["abc"*10]*1000' '[s.upper() for s in lst]'
>    100 loops, best of 3: 2.23 msec per loop
>
>The above results are using Python CVS (aka 2.5a0).
>
>Skip
>
>  
>