unexpected result when using a list comprehension on a list of booleans Python

Question

r=[True, False,True,False, False]
print([i for i in r if str(r[i])=="True"])

this code gives the following unexpected result: [False, False, False]

Why is this the behavior? I would expect: [True, True]

Answer 1

i is a boolean with a value of True or False. If you use it as an index in r[i], you will get either r[0] or r[1] because bool is a subclass of int.

You can picture it as r[int(i)] and int(True) == 1, int(False) == 0.

What you probably mean is:

print([i for i in range(len(r)) if str(r[i])=="True"])

where i is an integer or:

print([i for i in r if str(i)=="True"])

where i is a boolean.

Note that if i is a boolean, there is little point in if str(i) == "True". A more concise way of writing it would be:

print([i for i in r if i])

or even using the filter built-in function:

it = filter(None, r)
print(list(it))

Answer 2

You meant str(i).

With your original code,

• on the first iteration, i is True, and as True has the integer value 1, r[i] will be r[1], which is False. The if condition fails.

• on the second iteration, i is False, and as False has the integer value 0, r[i] will be r[0], which is True. The if condition succeeds and i (which is False) gets added to the result.