+ 2

In operator confusing example

Hi, In the following code snippet: arr = ["a", True] print("a" in arr in arr) #False I feel like this should either err (trying to check if ‚Äúa‚ÄĚ is in True) or give True (True is in arr). However it gives false, how do you explain it?

4th Dec 2023, 7:11 PM
Edward Finkelstein
Edward Finkelstein - avatar
3 Respuestas
+ 11
Edward Finkelstein , python evaluates this expression like: (read more in python reference/ expressions) see also this thread in stackoverflow.com: https://stackoverflow.com/questions/60400708/why-a-in-arr-in-arr-a-in-arr-in-arr#60400956 first evaluation: ("a" in arr) => True second evaluation: (arr in arr) => False, because arr is not in arr True and False => False if we use parenthesis for grouping, the result is different: print(("a" in arr) in arr) => True
4th Dec 2023, 8:08 PM
Lothar - avatar
+ 4
Here is a similar question: https://www.sololearn.com/Discuss/2166930/?ref=app If I understand it correct it first checks if "a" in arr (True), then it checks if arr in arr (False). True and False leads to False.
4th Dec 2023, 7:47 PM
Denise Roßberg
Denise Roßberg - avatar
+ 2
EDIT: I'm wrong. Always learning! Check answers below ūüĎć
4th Dec 2023, 7:27 PM
StuartH - avatar