Incorrect lloc count

[shabi9605]

I'm getting incorrect lloc count, when a dictionary is in the code.

def sample_dict():
    a = {'user1': 'password1', 'user2': 'password2'}
    print(a)
    return a

I'm calculating lloc of this function with this below command :-
'radon raw 'path' -s'
Result:-
LOC: 4
LLOC: 5
SLOC: 4
Comments: 0
Single comments: 0
Multi: 0
Blank: 0
- Comment Stats
(C % L): 0%
(C % S): 0%
(C + M % L): 0%

In this result lloc is incorrect , there is only 4 lloc. If I remove the dictionary, then the lloc will be correct. I think, here the issue is colon is present in the dictionary.

[devdanzin]

Confirmed, and you have identified the right reason.

A simple (but not perfect) fix would be to, besides checking for a colon, also check for a compound statement keyword on the line. That would fix the provided example.

So instead of:

radon/radon/raw.py

Line 163 in 3b75f79

return 2 - (token_pos == len(processed) - 2)

We'd have something like(untested):

            # Only consider colons when a compound statement is found in the line
            if any(token.string in COMPOUND_KEYWORDS for token in processed): 
                return 2 - (token_pos == len(processed) - 2)
            return 1

Where COMPOUND_KEYWORDS is:

COMPOUND_KEYWORDS = {"if", "while", "for", "try", "with", "match", "def", "class", "else", "elif", "except", "finally", "case"}

That still leaves at least one issue: both match and case are soft keywords, so we might get wrong lloc counts for something like:

case = {1: 2}
match = {3: 4}

It seems that a line with a soft keyword match always ends with a colon, so it would be possible to check for that. But a line with a case can have further statements, so it might be trickier to get right.

Edited to add: It would also be necessary to decide on how to treat lambda: currently it counts as two logical lines because it has a colon. Should it? And if it's two logical lines, how do we count the following:

def a(b=lambda x: 1):
    pass

[devdanzin]

I've added a PR to fix this, #263. It doesn't consider lambda x: y as two logical lines, as it would make a mess of counting lambda x: lambda y: lambda z: 1.