Why does my code produce this output (basic Python)?

Asked Jan 14 '22 at 02:40

Active Jan 14 '22 at 02:40

Viewed 16 times

This is a program to count the number of times the letter "o" appears in a sentence.

sentence = "We hold these truths to be self-evident..."
number_of_Os = 0
for i in sentence:
    if i == "o":
        number_of_Os += 1
print(number_of_Os)

This program works, but it only counts lowercase. I want to count BOTH uppercase AND lowercase.

So I changed this line:

if i == "o":

to this:

if i == "o" or "O":

sentence = "We hold these truths to be self-evident..."
number_of_Os = 0
for i in sentence:
    if i == "o" or "O":
        number_of_Os += 1
print(number_of_Os)

The second program counts EVERY character in the string, not just the o's.

Why does it do this?

asked Jan 14 '22 at 02:40

Indy Duran

`if i.lower() == 'o':` – Barmar Jan 14 '22 at 02:42
In Python, `”O”` is truthy (it’s not the empty string) so `i == "o" or "O"` is always `True` because it’s parsed as `(i == "o") or "O"`. You want `i in “oO”`, I think. – Ben Jan 14 '22 at 02:42
i in “oO” works for what I was trying to do, thanks! – Indy Duran Jan 14 '22 at 02:46
if i.lower() == 'o' also works! – Indy Duran Jan 14 '22 at 02:55
I was a little nervous bc this was my first question, so thanks Barmar and Ben for helping! – Indy Duran Jan 14 '22 at 02:56

Why does my code produce this output (basic Python)?

0 Answers0