Why is "if[ i %2==0]" always true?

Question

Need support/please clear my doubt:

def test(strValue):
    for i in range(len(strValue)):
        if[ i %2==0]:
            strValue.replace(strValue[i] ,strValue[i].upper())
            print('if loop '+strValue)
        else:
            strValue.replace(strValue[i] ,strValue[i].lower())
            print('else loop '+strValue)
**input:**
 test('apple')
**output:**
if loop apple
if loop apple
if loop apple
if loop apple
if loop apple

1.why if[ i %2==0] is always true 2.why strValue is not changed

`[ i %2==0]` is equal to either `[True]` or `[False]` depending on the value of `i`, so is a non empty list, which is truthy. — Sнаđошƒаӽ, Apr 18 '20 at 06:26
`str.replace` does not modify the string in-place, it returns the modified value. — bereal, Apr 18 '20 at 06:27
Does this answer your question? [In python, when to use a square or round brackets?](https://stackoverflow.com/q/46675477/188256) — snipsnipsnip, Apr 18 '20 at 06:28

A l w a y s S u n n y · Answer 1 · 2020-04-18T07:11:23.077

0

To answer your problems

The value of [i%2==0] is not always equal to True, it is either [True] or [False] depending on the i value but the non-empty list like [True] or [False] is always True so try to do this way

i%2==0 instead of [i%2==0]
You're replacing the strValue but not assigning it back. So do this way-

strValue = strValue.replace(strValue[i] ,strValue[i].upper())

DEMO & EXPLANATION: https://rextester.com/AHJHS98230

edited Apr 18 '20 at 07:11

answered Apr 18 '20 at 06:30

A l w a y s S u n n y

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The main issue is not that. The OP is using a list as the condition for if – Sнаđошƒаӽ Apr 18 '20 at 06:37
@Sнаđошƒаӽ Yes, sir I agreed with you. The main issue not only the list condition but also on string replace. I was editing my answer but in the meantime you downvoted. :) – A l w a y s S u n n y Apr 18 '20 at 07:08
Okay sir, now that your answer is complete, I take back my downvote ;-) – Sнаđошƒаӽ Apr 18 '20 at 09:07
@Sнаđошƒаӽ Thank you sir :) Happy coding! – A l w a y s S u n n y Apr 18 '20 at 09:08

score 0 · Accepted Answer · answered Apr 18 '20 at 06:31

1- [i%2==0] is a list of one element being the result of your test. In python, the brackets build a list. So you should just remove the bracket and your test will be fine.

2- strValue.replace(strValue[i] ,strValue[i].lower()) is not in-place modification. Strings are immutable in python so you need to store the result of this call in strValue and hence erase the former value.

So the working code in your case is:

def test(strValue):
    for i in range(len(strValue)):
        if i %2==0:
            strValue = strValue.replace(strValue[i] ,strValue[i].upper())
            print('if loop '+strValue)
        else:
            strValue = strValue.replace(strValue[i] ,strValue[i].lower())
            print('else loop '+strValue)

thanks .code works but **input:** test('apple') **output:** if loop Apple else loop Apple if loop APPle else loop APPle if loop APPlE -- In 3 rd line output is "if loop APPle" i am not clear why both p changed to P — Palaniappan Somu, Apr 18 '20 at 07:13

Why is "if[ i %2==0]" always true?

2 Answers2