CODEWITHSUNDEEP
pythonraw-input
Kreatronik
Kreatronik

Reputation: 199

Python: Keyboard prompt with capital/noncapital letters

In this code:

    if raw_input("\n Enter 'y' or 'Y': ")==("y" or "Y"):
       print("\n Success!")

It doesn't take the "OR" properly, instead if the in this case noncapital 'y' is entered the condition is fulfilled. If I enter the capital 'Y' I don't get the Success!

What's wrong here?

Upvotes: 1

Views: 450

Answers (4)

das-g
das-g

Reputation: 9994

What's wrong here?

In many programming languages, OR is a boolean operator. You apply it to values that are TRUE or FALSE. The operation evaluates to TRUE if at least one operand is TRUE:

TRUE OR TRUE == TRUE
TRUE OR FALSE == TRUE
FALSE OR TRUE == TRUE
FALSE OR FALSE == FALSE

In Python, you can apply or on non-boolean operands:

x or y

returns x if x casted to boolean is True; else it returns y. For boolean operands, this leads to the same results as above, but for non-boolean operands this has interesting effects:

  • [] or {} is {} (because empty lists are False when casted to boolean)
  • [1] or {} is [1] (because non-empty lists are True when casted to boolean)
  • [1] or 1/0 is also [1] (the right operand doesn't even get evaluated when the left one is True, so we don't hit the ZeroDivisionError. This is known as (left-to-right) short-circuit evaluation.)

Thus, other than in natural language, the Python or cannot be interpreted as separating alternative values. (Only alternative conditions / boolean expressions.)

There are several possibilities on how to make your code behave as expected:

The naive approach:

answer = raw_input("\n Enter 'y' or 'Y': ")
if answer == "y" or answer == "Y":
   print("\n Success!")

Normalizing the input:

if raw_input("\n Enter 'y' or 'Y': ").lower() == 'y':
   print("\n Success!")

Comparing to set of alternative values with membership operator in:

if raw_input("\n Enter 'y' or 'Y': ") in {'y', 'Y'}:
   print("\n Success!")

Upvotes: 0

Quaker
Quaker

Reputation: 1553

The right-hand-side of your if-statement is wrong and I think you need to understand a little better how the or operator behaves between strings.

Keep in mind that the return value of or is the value that has been evaluated last, and that Python evaluates empty string as boolean False and non-empty strings as boolean True.

In your case, the interpreter reads ("y" or "Y"), it then evaluates the boolean value of "y" which is True, as it is a non-empty string. Therefore, the boolean value of the or statement it True and the return value of the statement becomes "y", the last evaluated value.

This is how I would write this code. I would keep the return value of raw_input in _input, which will make it easier for me and others to read and understand the if-statement:

_input = raw_input("\n Enter 'y' or 'Y': ")

if input in ["y", "Y"]:
    print("\n Success!")

Upvotes: 0

martianwars
martianwars

Reputation: 6500

Try to make a list of values and use the in keyword. Something like this will work,

if raw_input("\n Enter 'y' or 'Y': ") in ('y', 'Y'):
   print("\n Success!")

The in keyword tests the string against a tuple of strings and on a correct match it returns True.

Since here you have just one character, you can build a string "yY". Something like this will work,

if raw_input("\n Enter 'y' or 'Y': ") in "yY":
   print("\n Success!")

Here each character of the string acts like one element of the tuple above.

ERROR in your code: You used ("y" or "Y"). This does not work in Python. This will only return "y" as both "y" and "Y" are treated as True values. However, if you type (0 or "Y"), you will get "Y" as 0 is treated as a False value.

Upvotes: 1

ArunDhaJ
ArunDhaJ

Reputation: 631

Try this if raw_input("\n Enter 'y' or 'Y': ").lower() == "y":

Upvotes: 1

Related Questions