Python's any() function on a list of lists gives weird outputs

Question

Given a list of list s named a , I got the following outputs on exploration of the any() function. It would be great if I could get some help in understanding the logic behind these outputs. Why are some outputs True and some False , even though all should be True ?

a = [[1,2,3], [0,3,6], 5, [4,5,7], [0,1,2]]

print(any(a) in [0,3,4,[5],100])
print(any(a) in [1,2,3,4,5])     # Gives True
print(any(a) in [4,5,7,6])       # Gives False
print(any(a) in [0,1,2,4])       # Gives True
print(any(a) in [5])             # Gives False

Answer 1

Your syntax is wrong. You are first computing any(a) (which is True ) then check if True is in your target list. The reason it sometimes returns True is the fact that int(True) is equal to 1 , so it only prints True if your target list contains a 1 .

Try this, for example:

print(any(x in [4,5,7,6] for x in a))     # Should print True

Answer 2

Actually it is not the functioning of any but instead the functioning of in operator in python.

Basically in works in the following way:-

for i in iterable:
    if i == value:
        return True
else:
    return False

Now if you apply this logic manually over one of your conditions that return True , you can find the element which causes this peculiar behaviour:-

for i in [1,2,3,4,5]:
    if i == True: # Since any(a) is True
        print(i)

and the output this produces is:-

1

Thus, one thing which you can infer from this is that the integer 1 represents the boolean True also, which you can confirm by checking in the interactive shell:-

Thus, you can observe that in python the integers 0 and 1 represent the booleans False and True respectively, too.

Therefore, now you can understand why your code executes in a peculiar way, like this...

a = [[1,2,3], [0,3,6], 5, [4,5,7], [0,1,2]]

print(any(a) in [0,3,4,[5],100]) # Will give False because 1 is NOT present
print(any(a) in [1,2,3,4,5])     # Gives True because 1 is present
print(any(a) in [4,5,7,6])       # Gives False because 1 is NOT present
print(any(a) in [0,1,2,4])       # Gives True because 1 is present
print(any(a) in [5])             # Gives False because 1 is NOT present