Do python "in" statements automatically return as true

Question

I wrote this bit of unnecessarily complicated code on my way to learning about how to use the in statement to make if statements work better. I have two questions following the code snippet.

answer = ['Yes', 'yes', 'YES']
answer2 = ['No', 'no', 'NO']
ans = raw_input()
for i in range(0, 3):
    if ans in answer[i]:
        print "Yes!"
    elif ans in answer2[i]:
        print "No!"
    else:
        print "Don't know what that means"

First question: I think if n in listname: returns as True or False automatically. Does anyone know if that's the case?

Second question: the above code returns 3 lines that vary depending on if ans is actually in answer or answer2. I tried to eliminate that by replacing the relevant portions like so:

if ans in answer[i] == True:

This had the strange effect of making the code output only the else: statement. So can anyone explain to me the difference between how python interprets if ans in answer[i]: and if ans in answer[i] == True:, please?

Answer 1

To answer your questions in reverse order, the reason why explicitly comparing with True did not work for you is that Python did not interpret the expression they way you expected. The Python parser has special handling of compare expressions so that you can chain them together and get a sensible result, like this:

>>> "a" == "a" == "a"
True

Notice that Python has to treat this whole thing as one operation, because if you split it into two operations either way you don't get the same result:

>>> ("a" == "a") == "a"
False
>>> "a" == ("a" == "a")
False

These behave differently because the part in the parentheses is evaluated first and returns True, but True != "a" so the whole expression returns false.

By rights the above shouldn't actually have any impact on your program at all. Unfortunately, Python handles in via the same mechanism as == so when you chain these together they are interpreted as a sequence like the above, so Python actually evaluates it as follows:

>> "a" in ["a"] == True
False
>>> ("a" in ["a"]) and ("a" == True)
False

It's wacky and arguably counter-intuitive, but that's unfortunately just how it works. To get the behavior you wanted you need to use parentheses to force Python to evaluate the first part separately:

>>> ("a" in ["a"]) == True
True

With all of that said, the == True is redundant because, as you suspected, the expression already returns a boolean and the if statement can just evaluate it as-is.

To return now to your other problem, I believe what you're trying to do is take one line of input and produce one corresponding line of output depending on what the user entered. You can apply the in operator to a string and a list to see if the string is in the list, which allows you to eliminate your for loop altogether:

answer = ['Yes', 'yes', 'YES']
answer2 = ['No', 'no', 'NO']
ans = raw_input()
if ans in answer:
    print "Yes!"
elif ans in answer2:
    print "No!"
else:
    print "Don't know what that means"

This first tests if the input matches any of the strings in answer, then the same for answer2. Of course, you could achieve a similar effect but also support other forms like YeS by simply converting the input to lowercase and comparing it to the lowercase form:

if ans.lower() == "yes":
    print "Yes!"
# (and so forth)