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Python- Searching codons using for/if

Hi everyone!

Can someone help me with this little problem? I am trying to look for the position of the first nucleotide of a metionine codon (or a start codon, whatever you like). For that reason, I am using for and if in this way (here is my code):

s_otro_dna = str(otro_dna)
donde_M = []
for i in range(0,len(s_otro_dna),3):
    codon = s_otro_dna[i:i+3]
    if (codon==['ATG']):
        donde_M.append(i)

Note: The type of the variable where I saved the sequence is string

This code doesn't give me any error message. However, when I try to look what is inside the list "donde_M" (where it is supposed to be the positions), Python "says" that is empty:

donde_M
Out[37]: []

What I am doing wrong? Thank you for your help!! :)

python
• 9,907 views • link
updated by Ram Ram • written by andreammo97 • andreammo97

I tried and the output that python gives me is the same

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written by andreammo97 • andreammo97

The approach seems like it works to me:

>>> dna = "ATGCGATCGATCGATCGATCGCGCGCGCAGCTA"
>>> for i in range(0, len(dna), 3):
...     codon = dna[i:i+3]
...     if codon == 'ATG':
...         print(str(i))

0

As alex points out, you don't need the [ ] or the ( ) for that matter, in your if

• 0 views • link
written by Joe Joe

Than you for your answer, but it also doesn't work. I've been investigating and the problem is within the if condition, it is something wrong.

• 0 views • link
written by andreammo97 • andreammo97 
> dna = "ATGCGATCGATCGATCGATCGCGCGCGCAGCTA" for i in range(0, len(dna),
> 3):
>     codon = dna[i:i+3]
> #    if codon == 'ATG':
> #        print(str(i))
>     print(codon, i)
> # why do you need str(i)?
>     print(codon, str(i)) # - nothing in this case
• 0 views • link
updated by Joe Joe • written by natasha.sernova natasha.sernova

Because i is an int and if you don't cast it to a str first, print will complain.

If you're printing i only, its not strictly necessary, but I usually print these things within other strings, so its just a habit I'm in.

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written by Joe Joe

Then something is wrong elsewhere in your code, because that works absolutely fine for me.

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written by Joe Joe

3 answers

Use re to find the start index positions of all instances of your string-of-interest (here, your start codon, or whatever). 

$ python
>>> import re
>>> [m.start() for m in re.finditer('ATG', 'ATGCGATCGATCGATCGATCGCGCGCGCAGCTA')]
[0]

Replace with ATGCGATCGATCGATCGATCGCGCGCGCAGCTA with s_otro_dna or whatever you want to search through.

Via: https://stackoverflow.com/questions/4664850/how-to-find-all-occurrences-of-a-substring

• 0 views • link
written by Alex Reynolds Alex Reynolds

I think I have found the problem! Ii is not in the if condition, it is in the way I have expressed the range. Instead of indicating the end of range with len(s_otro_dna) I introduce the length (in this case, 923) it works. Here is the code: 

for i in range(0,923, 3):
    codon = s_otro_dna[i:i+3]
    if codon == 'ATG':
        print(str(i))


231
348
606
792
864

Thank you for your help! :)

• 0 views • link
written by andreammo97 • andreammo97 
dna = "ATGCGATCGATCGATCGATCGCGCGCGCAGCTA"

for i in range(0, len(dna), 3): codon = dna[i:i+3]

if codon == 'ATG':

print(str(i))

print(codon, i)

why do you need str(i)?

print(codon, str(i)) # - nothing in this case
• 0 views • link
written by natasha.sernova natasha.sernova

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