How to add or subtract 1 to a sum based on a certain value in a text?-CodePudding

After reading a text, I need to add 1 to a sum if I find a ( character, and subtract 1 if I find a ) character in the text. I can't figure out what I'm doing wrong.

This is what I tried at first:

file = open("day12015.txt")

sum = 0
up = "("

for item in file:
    if item is up:
        sum  = 1
    else:
        sum -= 1
print(sum)

I have this long text like the following example (((())))((((( .... If I find a ), I need to subtract 1, if I find a (, I need to add 1. How can I solve it? I'm always getting 0 as output even if I change my file manually.

CodePudding user response：

your for loop only gets all the string in the file so you have to loop through the string to get your desired output.

Example .txt

(((())))(((((

Full Code

file = open("Data.txt")

sum = 0
up = "("

for string in file:
    for item in string:
        if item is up:
            sum  = 1
        else:
            sum -= 1
print(sum)

Output

Hope this helps.Happy Coding :)

CodePudding user response：

So you need to sum 1 for "(" character and -1 for ")".

Do it directly specifying what to occur when you encounter this character. Also you need to read the lines from a file as you're opening it. In your code, you are substracting one for every case that is not "(".

file = open("day12015.txt")
total = 0

for line in file:
    for character in line:
        if character == "(":
            total  = 1
        elif character == ")":
            total -= 1

print(sum)

CodePudding user response：

That's simply a matter of counting each character in the text. The sum is the difference between those counts. Look:

from pathlib import Path

file = Path('day12015.txt')

text = file.read_text()

char_sum = text.count('(') - text.count(')')

For the string you posted, for example, we have this:

>>> p = '(((())))((((('
>>> p.count('(') - p.count(')')
5
>>>

Just for comparison and out of curiosity, I timed the str.count() and a loop approach, 1,000 times, using a string composed of 1,000,000 randoms ( and ). Here is what I found:

import random
from timeit import timeit

random.seed(0)

p = ''.join(random.choice('()') for _ in range(1_000_000))

def f():
    return p.count('(') - p.count(')')


def g():
    a, b = 0, 0
    for c in p:
        if c == '(':
            a = a   1
        else:
            b = b   1
    return a - b


print('f', timeit(f, number=1_000))
print('g', timeit(g, number=1_000))

f 8.194599456997821
g 49.339132178

It means the loop approach is 6 times slower, even though the str.count() one is iterating over p two times to compute the result.