CODEWITHSUNDEEP
pythonrangeyieldxrange
user2239318
user2239318

Reputation: 2784

Python best practices xrange yield

def list_step(max = 268):    
    """ split in steps of 120 elements at time
    """
    if max > 120:
        for i in xrange(0,max,120):
            if i <= max:
                rest =  max % xrange(0,max,120)[-1]
                if rest != 120:
                    yield(i, rest)
                else:
                    yield(i, 0)
    else:
        yield(max)


a = list_step()
a.next()      return > (0,28) (120,28),ecc

Would it be possible to return the rest on the execution of last next(), instead of the tuple with the rest?

so that :

    a.next()      return > (0) (120),ecc.. (28)

Upvotes: 0

Views: 682

Answers (2)

syntonym
syntonym

Reputation: 7384

You can use itertools.chain to chain iterators together (documentation). If you simply want a single value "appended" to your generator you can use it (note that you need to somehow turn a single item into an iterable).

Also your max % xrange(0, max, 120)[-1] will always be max % 120, because xrange(0, max, 120) is the biggest value that is a multiple of 120 that is smaller as max, so dividing by it will yield the same result as dividing by 120 (modulo).

import itertools

itertools.chain(xrange(0,max,120), [max % 120])

Upvotes: 1

SolaWing
SolaWing

Reputation: 1722

something like this ?

def list_step(max = 268):    
    """ split in steps of 120 elements at time
    """
    if max > 120:
        rest_list = [];
        for i in xrange(0,max,120):
            if i <= max:
                rest =  max % xrange(0,max,120)[-1]
                if rest != 120:
                    yield(i)
                    rest_list.append(rest)
                else:
                    yield(i)
                    rest_list.append(0)
        for i in rest_list:
            yield(i)
    else:
        yield(max)


a = list_step()
for i in a: print i;

Upvotes: 0

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