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| From | Terry Reedy <tjreedy@udel.edu> |
| Subject | Re: Fast recursive generators? |
| Date | Fri, 28 Oct 2011 22:27:40 -0400 |
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On 10/28/2011 8:49 PM, Michael McGlothlin wrote: >> Better to think of a sequence of values, whether materialized as a 'list' or >> not. > > The final value will actually be a string but it seems it is usually > faster to join a list of strings than to concat them one by one. .join() takes an iterable of strings as its argument >> Comprehensions combine map and filter, both of which conceptually work on >> each item of a pre-existing list independently. (I am aware that the >> function passed can stash away values to create dependence. > > The problem is I don't really have a pre-existing list. So, as I said, map, filter, and comprehensions are not applicable to your problem. >> def do(func, N, value): >> yield value >> for i in range(1,N): >> value = func(value) >> yield value >> >> For more generality, make func a function of both value and i. >> If you need a list, "l = list(do(f,N,x))", but if you do not, you can do >> "for item in do(f,N,x):" and skip making the list. > > Generators seem considerably slower than using comprehension or > map/filter So what? A saw may cut faster than a hammer builds, but I hope you don't grab a saw when you need a hammer. > /reduce. Do you actually have an example where ''.join(reduce(update, N, start)) is faster than ''.join(update_gen(N, start))? Resuming a generator n times should be *faster* than n calls to the update function of reduce (which will actually have to build a list). --- Terry Jan Reedy
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Re: Fast recursive generators? Terry Reedy <tjreedy@udel.edu> - 2011-10-28 22:27 -0400
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