One line if-condition-assignment
Question:
I have the following code
num1 = 10
someBoolValue = True
I need to set the value of num1 to 20 if someBoolValue is True; and do nothing otherwise. So, here is my code for that
num1 = 20 if someBoolValue else num1
Is there someway I could avoid the ...else num1 part to make it look cleaner? An equivalent to
if someBoolValue:
num1 = 20
I tried replacing it with ...else pass like this: num1=20 if someBoolValue else pass. All I got was syntax error. Nor I could just omit the ...else num1 part.
Answers:
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.
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
Use this:
num1 = 20 if someBoolValue else num1
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.
num1 = 20 * someBoolValue or num1
you can use one of the following:
(falseVal, trueVal)[TEST]
TEST and trueVal or falseVal
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
You can definitely use num1 = (20 if someBoolValue else num1) if you want.
Here is what i can suggest.
Use another variable to derive the if clause and assign it to num1.
Code:
num2 =20 if someBoolValue else num1
num1=num2
Another way
num1 = (20*boolVar)+(num1*(not boolVar))
If you wish to invoke a method if some boolean is true, you can put else None to terminate the trinary.
>>> a=1
>>> print(a) if a==1 else None
1
>>> print(a) if a==2 else None
>>> a=2
>>> print(a) if a==2 else None
2
>>> print(a) if a==1 else None
>>>
You can do it this way.
try:
a = [i for i in [20] if False][0]
except IndexError:
print("Do what ever you want here")
You can solve your problem this way but, using ‘try/except block’ is not the best practice for python.
For the future time traveler from Google, here is a new way (available from Python 3.8 onward):
b = 1
if a := b:
# this section is only reached if b is not 0 or false.
# Also, a is set to b
print(a, b)
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
I have the following code
num1 = 10
someBoolValue = True
I need to set the value of num1 to 20 if someBoolValue is True; and do nothing otherwise. So, here is my code for that
num1 = 20 if someBoolValue else num1
Is there someway I could avoid the ...else num1 part to make it look cleaner? An equivalent to
if someBoolValue:
num1 = 20
I tried replacing it with ...else pass like this: num1=20 if someBoolValue else pass. All I got was syntax error. Nor I could just omit the ...else num1 part.
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.
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
Use this:
num1 = 20 if someBoolValue else num1
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.
num1 = 20 * someBoolValue or num1
you can use one of the following:
(falseVal, trueVal)[TEST]
TEST and trueVal or falseVal
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
You can definitely use num1 = (20 if someBoolValue else num1) if you want.
Here is what i can suggest.
Use another variable to derive the if clause and assign it to num1.
Code:
num2 =20 if someBoolValue else num1
num1=num2
Another way
num1 = (20*boolVar)+(num1*(not boolVar))
If you wish to invoke a method if some boolean is true, you can put else None to terminate the trinary.
>>> a=1
>>> print(a) if a==1 else None
1
>>> print(a) if a==2 else None
>>> a=2
>>> print(a) if a==2 else None
2
>>> print(a) if a==1 else None
>>>
You can do it this way.
try:
a = [i for i in [20] if False][0]
except IndexError:
print("Do what ever you want here")
You can solve your problem this way but, using ‘try/except block’ is not the best practice for python.
For the future time traveler from Google, here is a new way (available from Python 3.8 onward):
b = 1
if a := b:
# this section is only reached if b is not 0 or false.
# Also, a is set to b
print(a, b)
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