Calculating the square numbers within a range python

Question

Have a look at the following code:

for i in Squares(5, 50):

      print(i)

Presently this is exceptionally simple to execute utilizing a loop, anyway I need to utilize an iterator. 

So I have characterized the accompanying class:

import math

class Squares(object):

    def __init__(self, start, stop):

       self.start = start

       self.stop = stop

    def __iter__(self): 

        return self

    def __next__(self):

        start = self.start

        stop = self.stop

        squareroot = math.sqrt(start)

        if self.start > self.stop:

            raise StopIteration

        if squareroot == math.ceil(squareroot):

            start += 1

Be that as it may, right now this is returning None a boundless measure of times. This implies the none should be on the grounds that the StopIteration is being attempted in any event, when it shouldn't. I think me if squareroot == math.ceil(squareroot): the condition is right since I tried it independently, however, I can't sort out what to change to get the yield I need. Any assistance is valued.

For the below code:

for i in Squares(4,16):

print(i)

What I expect the output to me:

4

9

16

Answer 1

Just try to create a generator function:

from math import sqrt, ceil

def Squares(start, stop):

    for i in range(start, stop+1):

        sqrti = sqrt(i)

        if sqrti == ceil(sqrti):

            yield i

Then, loop it:

for i in Squares(4, 20):

    print i,

which gives:

4 9 16

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