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Advent of Code 2015, Day 1: Not Quite Lisp

Description

Given a string s, scan it from left to right. Setting n to an initial value of 0, for each occurrence of an opening parenthesis (, increment n by one. For each occurrence of a closing parenthesis ), decrement n by one. Return n. (A return value of 0 indicates that parentheses are “balanced.”)

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(())    ⟹ 0
()()    ⟹ 0
(((     ⟹ 3
(()(()( ⟹ 3
))((((( ⟹ 3
())     ⟹ -1
))(     ⟹ -1
)))     ⟹ -3
)())()) ⟹ -3

My solution

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def parens(s):
    ps = {"(": 1, ")": -1}
    return sum(ps[c] for c in list(s))

Tests

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import pytest
from parens import *

a = ['(())', '()()'] # 0
b = ['(((', '(()(()(', '))((((('] # 3
c = ['())', '))('] # -1
d = [')))', ')())())'] # -3

def test_parens_balanced():
    """Should return 0."""
    for elt in a:
        assert parens(elt) == 0

def test_parens_pos():
    """Should return positive values."""
    for elt in b:
        assert parens(elt) == 3

def test_parens_neg():
    """Should return negative values."""
    for elt in c:
        assert parens(elt) == -1
    for elt in d:
        assert parens(elt) == -3