IF ELSE statement is not working properly in python

Question

I'm new to Python and trying to create a weight converter by using if, else statements.

This should ask the user to input weight and unit in pounds (lbs) or kilograms (kgs) and convert it, but the program is only executing the if condition.

Here is the snippet of my code:

weight = int(input("Weight = "))
unit = input('(L)bs or (K)')
if unit == "L" or "l":
  print(f'You are {weight*0.45} kilograms')
elif unit == "K" or "k":
  print(f'You are {weight*2.2} pounds')
else:
  print('The converter is only for Kgs and Lbs')

Here is a screenshot:

Welcome to SO. Please don't post images of code or links to images. Copy and paste as text. Why not upload images of code on SO when asking a question? — 001
– 001, Commented Jun 10, 2019 at 17:25
Possible duplicate of How to test multiple variables against a value? — Ari Cooper-Davis
– Ari Cooper-Davis, Commented Jun 10, 2019 at 17:29

Sunny Patel · Accepted Answer · 2019-06-10 17:32:03Z

1

You are lacking parenthesis. Please get familiar with Operator Precedence.

Currently your if-statements are being read by the interpreter as:

if (unit == "L") or "l":
    ...
elif (unit == "K") or "k":
    ...
else:
    ...

and so the first if-statement will always be true since bool("l") == True.

The correct way to do this is:

if unit in "Ll":
    ...
elif unit in "Kk":
    ...
else:
    ...

or even better:

unit = input('(L)bs or (K)g').lower()
if unit == "l":
    ...
elif unit == "k":
    ...
else:
    ...

answered Jun 10, 2019 at 17:32

Sunny Patel

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1 Comment

A_beginner Over a year ago

Thanks. I'll try to focus on operator precedence.

0xWish · Accepted Answer · 2019-06-10 17:31:24Z

1

You have to write the full condition before and after "or" For the line:

if unit == "L" or "l"

You have to write:

if unit == "L" or unit == "l"

also the line:

elif unit == "K" or "k"

to

elif unit == "K" or unit == "k"

answered Jun 10, 2019 at 17:31

0xWish

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Comments

Fatema Tuz Zuhora · Accepted Answer · 2019-06-10 18:00:04Z

1

You need to bring change in two lines:

weight  = int(input("Weight: "))
unit    = input("(L)bs or (K): ")

# if unit == "L" or "l":    # OLD_LINE
if unit == "L" or unit == "l":  # NEW_LINE
    print(f"You are {weight*0.45} kilograms")

# elif unit == "K" or "k":  # OLD_LINE
elif unit == "K" or unit == "k":    # NEW_LINE
    print(f"You are {weight*2.2} pounds")

else:
    print("The converter is only for Kgs and Lbs")

See the Reference

answered Jun 10, 2019 at 18:00

Fatema Tuz Zuhora

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Comments

RyRyCodz · Accepted Answer · 2023-09-10 07:50:10Z

1

I personally like to sort them out like this:

weight = int(input("Weight = "))
unit = input('(L)bs or (K)')
if unit in ["L", "l"]: #new code
print(f'You are {weight*0.45} kilograms')
elif unit in ["K", "k"]: # new code
  print(f'You are {weight*2.2} pounds')
else:
  print('The converter is only for Kgs and Lbs')

answered Sep 10, 2023 at 7:50

RyRyCodz

132 bronze badges

Comments

J-L · Accepted Answer · 2023-12-13 17:43:24Z

When you write:

if unit == 'L' or 'l':

it's getting parsed as:

if (unit == 'L') or ('l'):

which is equivalent to:

if (unit == 'L') or (bool('l') is True):

bool('l') is always true, which is why none of the elif/else blocks get executed.

This is what you intended:

if unit == 'L' or unit == 'l':

If you prefer to avoid the re-use of unit == , you can re-write it as:

if unit.upper() == 'L':

or even:

if unit in ('L', 'l'):

I hope this helps.

Addition: (2023-12-13)

An even better way of checking case-insensitive equivalence is to use the str.casefold method on both strings:

if unit.casefold() == 'L'.casefold():

The str.casefold method returns a version of its string suitable for caseless comparisons. It might seem a little overkill in the example I just gave, but consider a comparison against "Washington". You could write this:

if name.lower() == "washington":

or you could check against a tuple of possibilities, like this:

if name in ('WASHINGTON', 'Washington', 'washington'):

(Note that this last example is not a true case-less comparison, as it won't match "WashingTon".)

By using str.casefold you can simply write:

if name.casefold() == "Washington".casefold():

So how is this better than simply writing the slightly shorter:

if name.lower() == "washington":

Well, if you use str.lower, you'd have to hard-code the string in lower-case, which may not be the canonical form of the string. Which means that if you change the string, you'd have to be careful of keeping it lower-case; if you accidentally use one or more capital letters in your hard-coded string, your comparison will always fail.

If you want to do a case-less comparison against several hard-coded strings, you can map the possibilities with map(str.casefold, (...)) and use in to check for the choice among the possibilities:

if choice.casefold() in map(str.casefold, ('Y', 'yes', 'Affirmative')):

This check will evaluate to True if the choice is yes, Yes, YES, Y, y, YeS, yES, Affirmative, affirmative, AFFIRMative, etc. (It's nice that you do not have to worry about how you capitalize the possibilities, so you can focus on writing them out using the capitalizations that make the most sense to the program you're writing.)

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