Python 3 - Counting up with two different values

Question

I am trying to figure out how to count up to a certain integer (as a range) alternating between two different numbers like 2 and 3. So that the output would be 2 5 7 10 12 15 etc.

I started off trying to alter a simple while loop like the following to take two values:

a = 0 
while a < 100: 
    a = a + 2 + 3
    print(a, end=' ')

But it just ends up counting up to 100 by 5.

I have tried number ranges and for loops and modules like itertools to try and figure out a way to do this and I am completely stumped.

I have performed search after search on this and all I can find is counting up by a single number with loops and ranges.

Answer 1

You can use itertools.cycle for this

import itertools
a = 0
it = itertools.cycle((2,3))
while a < 100:
    a += next(it)
    print(a)

Output

2
5
7
10
...
95
97
100

The itertools.cycle generator will just continuously loop back over the tuple as many times as you call it.

Answer 2

You need to print the content of a after adding two, and after adding three:

a = 0
while a < 100: 
    a = a + 2
    print(a, end=' ')
    if a < 100:
        a = a + 3
        print(a, end=' ')

That being said, you can better construct a generator, that iteratively adds two and three interleaved:

def add_two_three(a):
    while True:
        a += 2
        yield a
        a += 3
        yield a

You can then print the content of the generator until it hits 100 or more:

from itertools import takewhile

print(' '.join(takewhile(lambda x: x < 100,add_two_three(0))))

Answer 3

You have to increment a with different value each time (2 and 3). You can just swap the two values being used to increment a on each iteration to achieve this.

a = 0
inc1 = 2   # values being used to increment a
inc2 = 3
while a < 100:
    a = a + inc1
    inc1, inc2 = inc2, inc1    # swap the values
    print(a, end=' ')