Iterate string to compare an i element to i 1 element python-CodePudding

I have a DNA sequence:

seq='AACGTTCAA'

I want to count how many letters are equal to the next one. In this example I should get 3 (because of AA-TT-AA).

In my first try I found out that this doesn't work, because i is a string and 1 an integer.

seq='AACGTTCAA'
count=[]
for i in seq:
    if i == i 1: #neither i =1
        count.append(True)
    else: count.append(False)
print(sum(count))

So I tried this:

seq='AACGTTCAA'
count=[]
for i in seq:
    if i == seq[seq.index(i) 1]:
        count.append(True)
    else: count.append(False)
print(sum(count))

Then I receive this output which I cannot understand. 3 of these True should be False (1,5,8) Especially 8 as it is the last element of the string.

6
[True, True, False, False, True, True, False, True, True]

If thought about doing this with arrays but I think there might be a easy way to do this just in strings. Thanks

CodePudding user response：

To answer your question, the statement for i in seq yields a series of string variables like 'A', 'A', 'C' etc. so when in your first case when you are attempt to compare i == i 1: you are adding 1 to a string variable which throw a TypeError. In your second example, where you execute if i == seq[seq.index(i) 1] gives a false result, since the seq.index(i) always returns the first occurrence of the value. To do what you want on a basic level you can do the following:

def countPairedLetters(seq):
    count = 0
    for i in range(1, len(seq)):
        # i starts with 1 and ends with len(seq)-1
        if seq[i-1] == seq[i]:
            count  = 1
    return count

Note: by starting with the index 1 and going to last, you avoid the issue with overrunning the sequence.

CodePudding user response：

You can do this:

for i in range(0, len(seq)):
   if seq[i] == seq[i 1]: # <- this causes an error
      count.append(True)

Though you have to check if seq[i 1] does not cause an error.

Update

count = 0
for i in range(0, len(seq)-1): # this prevents an error
   if seq[i] == seq[i 1]:
      count  = 1

CodePudding user response：

Using itertools is one way:

from itertools import groupby
seq = 'AACGTTCAA'
print(sum(len(list(g))-1 for k,g in groupby(seq)))

This splits the sequence into groups of consecutive letters, then counts each group's lenght-1 into the total.

Edit: Updated with mozway's comments.

CodePudding user response：

The reason for unwanted True's is because of seq.index()

index(), would always return the first occurrence of the character you are searching for. When you have 2 consecutive characters, its actually returning the index of the first occurrence of that character and they always match.

here is a quick solution:

seq='AACGTTCAA'
count=[]
for i in range(0,len(seq)-1):
    print(i)
    if seq[i]==seq[i 1]:
        count.append(True)
    else: count.append(False)
print(count)