Why can't I match True / False value in an IF statement?

Question

I'm trying to match the output of a re.search against a list and run code if the IF statement allows. However, even if I tell the IF statement to match True or False, the IF statement never proceeds.

Running the search in python:

>>> re.search(r'...$', 'abcde1234doc').group() in ['doc', 'rtf', 'txt']
>>> True

Returns True, so by asking the IF statement to match True, it should proceed with it's code. But this doesn't seem to work.

if re.search(r'...$', 'abcde1234doc').group() in ['doc', 'rtf', 'txt'] == True:
    print'regex search matched!'
else:
    print'regex search not matched'

I expected the IF statement proceed with it's code when the regex search returns True, however the IF statement doesn't match and goes straight to the ELSE.

Try putting parentheses around the whole expression before `== True` — CertainPerformance, Dec 24 '18 at 09:28
You just need `if re.search(r'...$', 'abcde1234doc').group() in ['doc', 'rtf', 'txt']:`. — Austin, Dec 24 '18 at 09:31
Basically, since the regex search in list already returns a True/False, I don't need to tell IF what to match since the condition already returns True/False? — MrU, Dec 24 '18 at 09:33
@MrU, exactly. Since the regex search returns a truthy value, this is expected to handle the case. — Austin, Dec 24 '18 at 09:34

score 1 · Answer 1 · answered Dec 24 '18 at 09:29

Boolean operators chain in Python.

x in y == z

means

x in y and y==z

You could write your condition as

if (re.search(r'...$', 'abcde1234doc').group() in ['doc', 'rtf', 'txt']) == True:

if you wanted, but you'd be better off just leaving off the ==True and sticking with

if re.search(r'...$', 'abcde1234doc').group() in ['doc', 'rtf', 'txt']:

Why can't I match True / False value in an IF statement?

1 Answers1