优化速度条件

Question

example code:

e = [[1,2,3,4],[0,...,10000],[4,5,6,7]]
for n in range(0,1000000):
  if n in e[0] and n in e[1] and n in e[2]:
    yield n

I'm trying to optimize for speed to avoid e[1] lookup for every iteration if i can find out before that e[1] == range. Same goes for e[0] == range or e[2] == range

let me clarify this a bit more..im trying to write crontab http://unixhelp.ed.ac.uk/CGI/man-cgi?crontab+5 parser in python.

Format for crontab is * * * * *

where first field is minutes range(0,60), second hours range(0,23), third days of month range(1,32), 4th month range(1,13), 5th day of week range(0,8)

when every field is matched something happens, ie for every day at 5,30 am format is

30 5 * * *

now simplest solution is to iterate through lower,upper boundary with increment of 60 seconds, since smallest increment is 1 minute. and replace * with full range, matched fields with single number enclosed in list.

simplest algorithm is

for ts in range(startts,endts,60): 
  now = time.localtime(ts)
  if now.tm_min in [5,] and now.tm_hour in [30,] and now.tm_mon in range(0,13) ... etc:
    yield now

since some of those fields are sometimes * list lookup is not needed, and im trying to figure out way to dynamically generate conditions based on '*'

Answer 1

Looks like you want to find all elements that are common to all three lists. This is essentially the same as finding the intersection of the sets represented by the lists. In Python we can simply use set and find the intersection.

#! /usr/bin/env python

def get_values(lst):
    sets=map(set,lst)
    common=reduce(lambda x,y:x.intersection(y),sets,set(lst[0]))
    for value in common:
        yield value
if __name__=='__main__':
    e = [[1,2,3,4],range(10000),[4,5,6,7]]  
    for common_item in get_values(e):
        print common_item

Answer 2

if n==4:

will do it. At least in your example.

Perhaps you should tell us what you're actually trying to do? Are the ranges always contiguous? What exactly are the constraints you're working under?

Answer 3

I think this should work:

e = [[1,2,3,4],[0,...,10000],[4,5,6,7]]
for n in e[0]:
  if and n in e[2] and n in e[1]:
    yield n

It will only check the values which exist in e[0]. If the numbers of elements are different in each sub-list, you could add one further generic optimisation which checks them in order of the size of lists so you check against the smallest list first (I did that by hand in this case, but you get the idea).