Search a sequence for its minimum and stop as soon as the lowest possible value is found
__peter__ at web.de
Sat Jan 7 06:42:58 EST 2017
Jussi Piitulainen wrote:
> Peter Otten writes:
>> Example: you are looking for the minimum absolute value in a series of
>> integers. As soon as you encounter the first 0 it's unnecessary extra
>> work to check the remaining values, but the builtin min() will continue.
>> The solution is a minimum function that allows the user to specify a stop
>>>>> from itertools import count, chain
>>>>> stopmin(chain(reversed(range(10)), count()), key=abs, stop=0)
>> How would you implement stopmin()?
> Only let min see the data up to, but including, the stop value:
I should have mentioned that my actual use case has a costly key() function.
> from itertools import groupby
> def takeuntil(data, pred):
> '''Take values from data until and including the first that
> satisfies pred (until data is exhausted if none does).'''
> for kind, group in groupby(data, pred):
> if kind:
> yield next(group)
> yield from group
A clever takeuntil() implementation ;) I may steal it for my current
def stopmin_du(items, *, key, stop):
# decorate, process, undecorate
pairs = ((key(item), item) for item in items)
pairs = takeuntil(pairs, lambda pair: pair <= stop)
return min(pairs, key=firstitem)
> def stopmin(data, key, stop):
> return min(takeuntil(data, lambda o : key(o) == stop),
> key = key)
> data = '31415926'
> for stop in range(5):
> '=>', repr(''.join(takeuntil(data, lambda o : int(o) == stop))),
> '=>', repr(stopmin(data, int, stop)))
> # 0 => '31415926' => '1'
> # 1 => '31' => '1'
> # 2 => '3141592' => '1'
> # 3 => '3' => '3'
> # 4 => '314' => '1'
> from itertools import count, chain
> print(stopmin(chain(reversed(range(10)), count()), key=abs, stop=0))
> print(stopmin(chain(reversed(range(10)), count()), key=abs, stop=3))
> # 0
> # 3
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