Finding patterns on strings on Python

Question

My function was suposed to receive a large string, go through it, and find the maximum number of times the pattern "AGATC" repeats consecutively. Regardless of what I feed this function, my return is always 1.

def agatc(s):
    maxrep = 0
    temp = 0
    for i in range(len(s) - 4):
        if s[i] == "A" and s[i + 1] == "G" and s[i + 2] == "A" and s[i + 3] == "T" and s[i + 4] == "C":
            temp += 1
            print(i)
            i += 3
        else:
            if temp > maxrep:
                maxrep = temp
            temp = 0
    return maxrep

Also tried initializing the for loop with (0, len(s) - 4, 1), got the same return.

I though the problem might be in adding 3 to the i variable (apparently it wasn't), so I added print(i) to see what was happening. I got the following:

Which implies that you probably wanna change range(len(s) - 4) to range(len(s) - 5). — goodvibration
– goodvibration, Commented Jun 14, 2020 at 12:58
i += 3 won't do anything, it's immediately replaced by the next value from the range by the loop itself. — jonrsharpe
– jonrsharpe, Commented Jun 14, 2020 at 12:58
It's because you don't let maxrep be any bigger than 1... Once you found a match, you do maxrep = temp but then initialize temp = 0. Now it will never hold that temp > maxrep so you will never change maxrep which is now 1... Why do you even need temp and the else clause? Why not just maxrep += 1 once the condition is true? — Tomerikoo
– Tomerikoo, Commented Jun 14, 2020 at 13:05

Riccardo Bucco · Accepted Answer · 2020-06-15 09:18:56Z

3

In this way you can find the number of overlapping matches:

def agatc(s):
    temp = 0
    for i in range(len(s) - len("AGATC") + 1):
        if s[i:i+len("AGATC")] == "AGATC":
            temp += 1
    return temp

If you want to find non-overlapping matches:

def agatc(s):
    temp = 0
    i = 0
    while i < len(s) - len("AGATC") + 1:
        if s[i:i+len("AGATC")] == "AGATC":
            temp += 1
            i += len("AGATC")
        else:
            i += 1
    return temp

edited Jun 15, 2020 at 9:18

answered Jun 14, 2020 at 12:59

Riccardo Bucco

15.5k4 gold badges29 silver badges57 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Henrique Baesa Over a year ago

Thank you! So I can't use for loops as I did in C, right? It's not possible to change the value of i in the middle of the iteration?

Riccardo Bucco Over a year ago

In C you perform an operation at the end of the for cycle body (for example, I++). In python, first all the values are generated (with range(n)), that is before the for cycle begins, and then i takes all those values. So even if you change i within the body, it has no effect on range values.

Tomerikoo Over a year ago

len(s) - len("AGATC") is wrong. Try s = "AGATCAGATCAAAAAGATCA"

Riccardo Bucco Over a year ago

The difference is 15. Therefore the last valid index is 14, because i is strictly less than 15

Tomerikoo Over a year ago

exactly, so you skip the last letter. You would expect the last index to be 15

|

Sơn Ninh · Accepted Answer · 2020-06-14 13:22:54Z

1

A simple solution with module re

import re

s = 'FGHAGATCATCFJSFAGATCAGATCFHGH'
match = re.finditer('(?P<name>AGATC)+', s)
max_len = 0
result = tuple()
for m in match:
    l = m.end() - m.start()
    if l > max_len:
        max_len = l
        result = (m.start(), m.end())

print(result)

answered Jun 14, 2020 at 13:22

Sơn Ninh

3311 silver badge10 bronze badges

Comments

Ronald · Accepted Answer · 2020-06-14 13:10:49Z

0

Personally I would use regular expressions. But if you do not want that, you could use the str.find() method. Here is my solution:

def agatc(s):
    cnt = 0
    findstr='aga'                             # pattern you are looking for
    for i in range(len(s)):
        index = s.find(findstr)
        if index != -1:
            cnt+=1
            s = s[index+1:]                   # overlapping matches
            # s = s[index+len(findstr):]      # non-overlapping matches only
            print(index, s)                   # just to see what happens
    return cnt

answered Jun 14, 2020 at 13:10

Ronald

3,3853 gold badges10 silver badges18 bronze badges

Comments

Red · Accepted Answer · 2020-06-14 13:28:48Z

0

This function counts the greatest amount of consecutive 'AGATC's in a string and returns the amount:

def agatc(s):
    w = "AGATC"
    maxrep = [m.start() for m in re.finditer(w,s)] # The beginning index fror each AGATC
    c = ''
    for i,v in enumerate(maxrep):
        if i < len(maxrep)-1:
            if v+5 == maxrep[i+1]:
                c+='y'
            else:
                c+='n'

    return len(max(c.split('n')))+1

print(agatc("oooooooooAGATCooooAGATCAGATCAGATCAGATCooooooAGATCAGATC"))

Output:

answered Jun 14, 2020 at 13:28

Red

27.7k8 gold badges44 silver badges63 bronze badges

Collectives™ on Stack Overflow

Finding patterns on strings on Python

4 Answers 4

6 Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

6 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related