Avoid conditional counter increment in Python loop

Question

How can I avoid having to increment the count_odd_values variable manually in the following Python code:

count_odd_values = 0
for value in random.sample(range(1000), 250):
  if value % 2 == 1:
    count_odd_values += 1

@Josh Thanks, but I don't see how. I am new to Python.

necromancer
– necromancer

2013-08-23 03:23:18 +00:00
Commented Aug 23, 2013 at 3:23 — necromancer
– necromancer, Commented Aug 23, 2013 at 3:23

mishik · Accepted Answer · 2013-08-23 03:16:20Z

3

You can do:

count_odd_values = sum(value % 2 for value in random.sample(range(1000), 250))

All the even numbers will give value % 2 == 0 and will not change the sum. All the odd numbers will give value % 2 == 1 and sum will be incremented by 1.

answered Aug 23, 2013 at 3:16

mishik

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3 Comments

necromancer Over a year ago

thanks, i assume this would generalize to a condition that checks a database object in django?

mishik Over a year ago

Sorry, can't say much for Django, not experienced. All I know is that this code will count odd numbers in "whatever's provided by random.sample(...)"

martineau Over a year ago

Less useless addition would be required with sum(1 for value in random.sample(range(1000), 250) if value % 2).

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