Itertools.count Python

Question

I am taking a datacamp tutorial and am wondering if someone could please explain why this code prints out the ODD numbers up to 11? I was thinking each next() call steps through the sequence, but if I follow this thought, the first print should be the number 2 (count(0) goes to count(1) when it first hits the while and then when it prints, it should go to count(2).

TIA.

from itertools import count
sequence = count(start=0, step=1)
while(next(sequence) <= 10):
    print(next(sequence))

1 3 5 7 9 11

No, the very first call to next yields 0, not 1.

chepner
– chepner

2018-01-31 16:44:04 +00:00
Commented Jan 31, 2018 at 16:44 — chepner
– chepner, Commented Jan 31, 2018 at 16:44

chepner · Accepted Answer · 2018-01-31 17:16:06Z

next(sequence), as the name says, will get the next element in the provided sequence.

Once an element is obtained, a further call to next(sequence) will simply yield the next number after that one (providing we are reusing the same sequence).

>>> sequence = count(start=0, step=1)
>>> next(sequence)
0
>>> next(sequence)
1
>>> next(sequence)
2

Note that, if you re-generate the sequence, you'll start over:

>>> sequence = count(start=0, step=1)
>>> next(sequence)
0
>>> next(sequence)
1
>>> sequence = count(start=0, step=1)
>>> next(sequence)
0

In your code you are calling next(sequence) on the while condition, and then next(sequence) for the print function, so both values will be different. Basically, as the next calls are alternated between the while and the print you'll see that the print will print out the odd values.

You can see a similar behavior if you add one more next:

from itertools import count
sequence = count(start=0, step=1)
while(next(sequence) <= 10):
    noop = next(sequence)
    print(next(sequence))

This will print out 2 5 8 11

If you'd want to print out the even numbers instead, you can:

Start counting from 1:

from itertools import count
sequence = count(start=1, step=1)
while(next(sequence) <= 10):
    print(next(sequence))

This will print out 2 4 6 8 10

Or, if you'd like 0 to be there too, just add a single print statement before the while loop:

from itertools import count
sequence = count(start=0, step=1)
print(next(sequence))
while(next(sequence) <= 10):
    print(next(sequence))

This will print out 0 2 4 6 8 10

If you'd want to print each of the numbers, simply persist the output of next(sequence):

from itertools import count
sequence = count(start=0, step=1)
element = next(sequence)
while(element <= 10):
    print(element)
    element = next(sequence)

Great explanation, thanks. My mistake was not understanding the initial next() call takes the sequence start value, not stepping through it.

Albin Paul · Accepted Answer · 2018-01-31 16:54:49Z

0

this is just a modified version of your code that produce same output i have added comments as to why the code works in the particular way

from itertools import count
sequence = count(start=0, step=1)
# a has value count(0)
a=next(sequence)
while(a <= 10):
    # b has value count(1) 3 5 and so on
    b=next(sequence)
    print(b)
     # a has value count(2) 4 6 and so on
    a=next(sequence)

answered Jan 31, 2018 at 16:54

Albin Paul

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Comments

Mike Müller · Accepted Answer · 2018-01-31 17:04:25Z

0

You need to start at 1 to get your desired output:

sequence = count(start=1, step=1)
while(next(sequence) <= 10):
    print(next(sequence))

2
4
6
8
10

edited Jan 31, 2018 at 17:04

answered Jan 31, 2018 at 16:54

Mike Müller

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