One line if-condition-assignment

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挽巷
挽巷 2020-12-02 08:46

I have the following code

num1 = 10
someBoolValue = True

I need to set the value of num1 to 20 if someBoolV

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  • 2020-12-02 09:31

    I don't think this is possible in Python, since what you're actually trying to do probably gets expanded to something like this:

    num1 = 20 if someBoolValue else num1
    

    If you exclude else num1, you'll receive a syntax error since I'm quite sure that the assignment must actually return something.

    As others have already mentioned, you could do this, but it's bad because you'll probably just end up confusing yourself when reading that piece of code the next time:

    if someBoolValue: num1=20
    

    I'm not a big fan of the num1 = someBoolValue and 20 or num1 for the exact same reason. I have to actually think twice on what that line is doing.

    The best way to actually achieve what you want to do is the original version:

    if someBoolValue:
        num1 = 20
    

    The reason that's the best verison is because it's very obvious what you want to do, and you won't confuse yourself, or whoever else is going to come in contact with that code later.

    Also, as a side note, num1 = 20 if someBoolValue is valid Ruby code, because Ruby works a bit differently.

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  • 2020-12-02 09:36

    If one line code is definitely going to happen for you, Python 3.8 introduces assignment expressions affectionately known as “the walrus operator”.

    :=

    someBoolValue and (num := 20)
    

    The 20 will be assigned to num if the first boolean expression is True. The assignment must be inside parentheses here otherwise you will get a syntax error.

    num = 10
    someBoolValue = True
    
    someBoolValue and (num := 20)
    print(num) # 20
    
    num = 10
    someBoolValue = False
    
    someBoolValue and (num := 20)
    print(num) # 10
    
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  • 2020-12-02 09:41

    you can use one of the following:

    (falseVal, trueVal)[TEST]
    
    TEST and trueVal or falseVal
    
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  • 2020-12-02 09:42
    num1 = 10 + 10*(someBoolValue is True)
    

    That's my new final answer. Prior answer was as follows and was overkill for the stated problem. Getting_too_clever == not Good. Here's the prior answer... still good if you want add one thing for True cond and another for False:

    num1 = 10 + (0,10)[someBoolValue is True]
    

    You mentioned num1 would already have a value that should be left alone. I assumed num1 = 10 since that's the first statement of the post, so the operation to get to 20 is to add 10.

    num1 = 10
    someBoolValue = True
    
    num1 = 10 + (0,10)[someBoolValue is True]
    
    print(f'num1 = {num1}\nsomeBoolValue = {someBoolValue}')
    

    produced this output

    num1 = 20
    someBoolValue = True
    
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  • 2020-12-02 09:45

    In one line:

    if someBoolValue: num1 = 20
    

    But don’t do that. This style is normally not expected. People prefer the longer form for clarity and consistency.

    if someBoolValue:
        num1 = 20
    

    (Equally, camel caps should be avoided. So rather use some_bool_value.)

    Note that an in-line expression some_value if predicate without an else part does not exist because there would not be a return value if the predicate were false. However, expressions must have a clearly defined return value in all cases. This is different from usage as in, say, Ruby or Perl.

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  • 2020-12-02 09:45

    No. I guess you were hoping that something like num1 = 20 if someBoolValue would work, but it doesn't. I think the best way is with the if statement as you have written it:

    if someBoolValue:
        num1 = 20
    
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