Detect if there are 7 or more of the same symbol

Question

So I need to know if in a string which contains only '1's and '0's has 7 of one of those in a row. My first thought was to iterate through the string like so:

for i in inp:
    if count == 6:
        answ = True
    this = i
    if this == other:
        count+=1
        other = this
    else:
        other = this
        count = 0

But it randomly gives out a wrong answer. My next idea was to use regex (.)\1{7,} like this:

pattern = re.compile(r'(.)\1{7,}')
found = re.search(pattern, inp)
if found != None:
    answ = True

But it still sometimes makes mistakes. Anyone has any ideas on why that would happen? Regex was taken from here

Your regular expression checks for *8* or more occurrence of a value: `(.)` is the first one, `\1{7,}` checks for 7 more. — chepner, Dec 12 '21 at 17:27
A counter-based approach is natural and, when properly debugged, won't fail randomly. You just need to think about the logic a bit more. — John Coleman, Dec 12 '21 at 17:27
Until you can prove it's a bottleneck, I wouldn't do anything more complicated than `'1111111' in inp or '0000000' in inp`. — chepner, Dec 12 '21 at 17:28

score 3 · Answer 1 · answered Dec 12 '21 at 17:50

3

cheper's comment actually worked flawlessly. I don't know why I didn't think of it sooner

Until you can prove it's a bottleneck, I wouldn't do anything more complicated than '1111111' in inp or '0000000' in inp

answered Dec 12 '21 at 17:50

Ive

51
1
7

score 1 · Answer 2 · answered Dec 12 '21 at 17:31

I'm sure that there is a regex pattern that could work for this, but like many other programmers, I've never quite figured out how regex works. However, the following script should be able to determine if there is a substring that matches your description.

current = None
current_len = 0
answer = False

for i in inp:
    if i == current:
        current_len += 1
        if current_len >= 7:
            answer = True
            break
    else:
        current = i
        current_len = 1

print(answer)

Obviously, the answer variable in this case would be True if there are 7 consecutive characters and False if there aren't. Hopefully this helps :)

nikeros · Answer 3 · 2021-12-12T17:56:18.893

0

Can't you follow this approach?

def check_line(l, target = 7):
     return l if any(map(lambda y: any(map(lambda x:len(x) == target, l.split(y))), ["1","0"])) else ""
    
print(check_line("00000001111"))
print(check_line("000000001111"))

OUTPUT

00000001111

This solution works if you want sequences of exactly 7 elements: if you need 7 or more, chepner approach is the cleanest.

edited Dec 12 '21 at 17:56

answered Dec 12 '21 at 17:39

nikeros

3,302
2
10
26

1

@JonSG - oh, he needs consecutive - well spotted – nikeros Dec 12 '21 at 17:42

score 0 · Answer 4 · answered Dec 12 '21 at 17:53

I like @chepner's comment and I might start with that as echoed by @ive. Since you started a regex based answer, you might also try:

import re

def check_line(line):
    return re.search(r"0{7,}|1{7,}", line) is not None

tests = [
    "00001111",
    "00000001",
    "01111111",
    "100000001111",
    "10000100001111"
]

for test in tests:
    print(test, check_line(test))

That should give you:

00001111 False
00000001 True
01111111 True
100000001111 True
10000100001111 False

Detect if there are 7 or more of the same symbol

4 Answers4