[Python-ideas] Pass a function as the argument "step" of range()

[Ryan Gonzalez]

I like the idea, but this would be nicer in a language with implicit currying so that your example could be:

for i in Range(1, N, (*2)):
    ...

Or in Haskell:

range' v a b f
  | a == b = []
  | otherwise = f a:range v a+v b f

range a b
  | a > b = range' -1 a b
  | a < b = range' 1 a b
  | a == b = []

Since Python doesn't have this, the generator forms others showed would likely end up more concise.


On July 2, 2015 1:30:53 AM CDT, Pierre Quentel <pierre.quentel at gmail.com> wrote:
>In languages such as Javascript, the incrementation of a for loop
>counter
>can be done by an operation, for instance :
>
>for(i=1; i<N; i*=2)
>
>would iterate on the powers of 2 lesser than N.
>
>To achieve the same thing in Python we currently can't use range()
>because
>it increments by an integer (the argument "step"). An option is to
>build a
>generator like :
>
>def gen(N):
>    i = 1
>    while i<=N:
>        yield i
>        i *= 2
>
>then we can iterate on gen(N).
>
>My proposal is that besides an integer, range() would accept a function
>as
>the "step" argument, taking the current counter as its argument and
>returning the new counter value. Here is a basic pure-Python
>implementation
>:
>
>import operator
>
>class Range:
>
>    def __init__(self, start, stop, incrementor):
>        self.start, self.stop = start, stop
>        self.incrementor = incrementor
>        # Function to compare current counter and stop value : <= or >=
>       self.comp = operator.ge if self.stop>self.start else operator.le
>        self.counter = None
>
>    def __iter__(self):
>        return self
>
>    def __next__(self):
>        if self.counter is None:
>            self.counter = self.start
>        else:
>            self.counter = self.incrementor(self.counter)
>        if self.comp(self.counter, self.stop):
>            raise StopIteration
>        return self.counter
>
>Iterating on the powers of 2 below N would be done by :
>
>for i in Range(1, N, lambda x:x*2)
>
>I haven't seen this discussed before, but I may not have searched
>enough.
>
>Any opinions ?
>
>
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