Why index is showing wrong in Python list

Question

I was writing a program to find out the index number of a given value of 'o' in a string="Hello World". I have done it using three ways, first is using the index(), second is using the for loop with a counter variable, third is using the for loop with enumerate() function.

I have a problem with the counter variable. Hence, there are two 'o' in "Hello World" first is at the 4th index and the second is at the 7th index. but my counter variable is showing me 4 and 6 as the output. But the enumerate() is showing me 4 and 7 as the output. I want to know what is wrong with my counter variable.

Thanks in advance.

Here is my code

val = 'Hello World'
counter = 0
print("Index of 'o' Using Index Function: ", val.index('o', 5))
print('---------------------------------------')

for i in val:
    if i == 'o':
        print(i, counter)
              #I don't want to use break here because I want to print all the index of 'o'
    else:
        counter += 1
print("------------------")

for i, var in enumerate(val):
    print(i, var)

Here Is the Output

Index of 'o' Using Index Function:  7
---------------------------------------
o 4
o 6
------------------
0 H
1 e
2 l
3 l
4 o
5  
6 W
7 o
8 r
9 l
10 d

Would you please Explain it a bit more, actually I am beginner. — Kapil Yadav
– Kapil Yadav, Commented Jul 4, 2021 at 11:20
You are doing counter += 1 only when i != "o" i.e., in the else part; but counter must go up whether it hits "o" or not, right? — Mustafa Aydın
– Mustafa Aydın, Commented Jul 4, 2021 at 11:21
@KapilYadav remove the else block and let counter stay in the loop — user14887424
– user14887424, Commented Jul 4, 2021 at 11:21
Try changing val = "ooooooooooo" and run your code and see what counter has. — Mustafa Aydın
– Mustafa Aydın, Commented Jul 4, 2021 at 11:22

Suravam · Accepted Answer · 2021-07-04 11:25:59Z

2

Seems like error in the counting logic. first 'o' is not counted as counter increment was placed on else block. Try this,

for i in val:
    if i == 'o':
        print(i, counter)
    counter += 1
print("------------------")

answered Jul 4, 2021 at 11:25

Suravam

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Comments

Ninja · Accepted Answer · 2021-07-04 11:21:55Z

2

You need to update the counter variable even when o is found.

if i == 'o':
    print(i, counter)
counter += 1

answered Jul 4, 2021 at 11:21

Ninja

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1 Comment

user14887424 Over a year ago

I think you deleted your previous post as it was shown to me on my screen. Please use the edit button instead of delete. Welcome to StackOverflow

khadim husen · Accepted Answer · 2021-07-04 11:32:09Z

0

val = 'Hello World' 
counter = 0
print("Index of 'o' Using Index Function: ", val.index('o', 5))

print('---------------------------------------')

for i in val: 
    if i == 'o': 
        print(i, counter)
    counter += 1     

print("------------------")

for i, var in enumerate(val):
    print(i, var)

answered Jul 4, 2021 at 11:32

khadim husen

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Comments

Jonathan1609 · Accepted Answer · 2021-07-04 11:27:58Z

-1

I suggest you using this code i have written

def count(string, letter):
    assert letter in string, f"The letter {letter!r} is not in the string {string!r}"
    return [i for i, j in enumerate(string) if j == letter]
    

print(count('Hello World', "o"))

Which it prints out [4, 7] as you want.

it is called list-comprehension, it goes character by character and if the character equals to letter, then it adds it to the list. equals to:

counter = []
for i, j in enumerate(string):
    if j == letter:
        counter.append(i)

edited Jul 4, 2021 at 11:27

answered Jul 4, 2021 at 11:21

Jonathan1609

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5 Comments

DeepSpace Over a year ago

OP isn't sure what's going on with their own code, I don't think throwing another, unexplained piece of code at them (with constructs they probably didn't see before) is going to be of much help

user15801675 Over a year ago

[i for i,j in enumerate(string) if j==letter]. Use this

Jonathan1609 Over a year ago

@Sujay ?, That's exactly what i did

user15801675 Over a year ago

I am telling to shorten it down to 1 line

Jonathan1609 Over a year ago

Ah ok, i was actually about to doing that

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