I've been trying to understand if it is possible to use an if statement similar to the likes of what I have demonstrated here below. It is my understand that it is not?
for i in range(10):
if i == (3 or 5) or math.sqrt(i) == (3 or 5):
numbers.append(i)
With this block of code I only get the numbers 3 & 9, while I should be getting 3, 5, 9. Is there another way of doing so without listing the code below?
for i in range(10):
if i == 3 or i == 5 or math.sqrt(i) == 3 or math.sqrt(i) == 5:
numbers.append(i)
You can use in operator:
for i in range(10):
if i in (3, 5) or math.sqrt(i) in (3, 5):
numbers.append(i)
or in case you expect each of the calculations to be in the same group of results, you can use any()
results = [1, 2, ..., too long list for single line]
expected = (3, 5)
if any([result in expected for result in results]):
print("Found!")
Just a minor nitpick, sqrt will most likely return a float sooner or later and this approach will be silly in the future, therefore math.isclose() or others will help you not to encounter float "bugs" such as:
2.99999999999999 in (3, 5) # False
which will cause your condition to fail.
if i == (3 or 5) or math.sqrt(i) == (3 or 5):
is equivalent to
if i == 3 or math.sqrt(i) == 3:
leading to 3 and 9 as results.
This because 3 or 5 is evaluated as 3 by being 3 the first non-zero number (nonzero numbers are considered True). For instance, 0 or 5 would be evaluated as 5.
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