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Finding matching motifs on sequence and their positions

I am trying to find some matching motifs on a sequence, as well as the position that the motif is located in and then output that into a fasta file. The code below shows that the motif [L**L*L] is present in the sequence, when I run it returns as "YES" but I do not know where it is positioned
The ** inside the square bracket is to show that any amino acid there is permited.
`
This is the code I used to check whether the motif is present in the sequence, and it worked because it returned "YES".
peptide1= "MKFSNEVVHKSMNITEDCSALTGALLKYSTDKSNMNFETLYRDAAVESPQHEVSNESGSTLKEHDYFGLSEVSSSNSSSGKQPEKCCREELNLNESATTLQLGPPAAVKPSGHADGADAHDEGAGPENPAKRPAHHMQQESLADGRKAAAEMGSFKIQRKNILEEFRAMKAQAHMTKSPKPVHTMQHNMHASFSGAQMAFGGAKNNGVKRVFSEAVGGNHIAASGVGVGVREGNDDVSRCEEMNGTEQLDLKVHLPKGMGMARMAPVSGGQNGSAWRNLSFDNMQGPLNPFFRKSLVSKMPVPDGGDSSANASNDCANRKGMVASPSVQPPPAQNQTVGWPPVKNFNKMNTPAPPASTPARACPSVQRKGASTSSSGNLVKIYMDGVPFGRKVDLKTNDSYDKLYSMLEDMFQQYISGQYCGGRSSSSGESHWVASSRKLNFLEGSEYVLIYEDHEGDSMLVGDVPWELFVNAVKRLRIMKGSEQVNLAPKNADPTKVQVAVG"
if re.search(r"L*L*L", peptide1):
print("YES")
else:
print("NO")
The code that I wrote to find the position is below, but when I run it says invalid syntax. Could you please assist as I have no clue whether in the right track or not, as I am still new in the field and python.
for position in range(len(s)):
if peptide[position:].startswith(r"L*L*L"):
print(position+1)
I am expecting to see the position of these motifs has been identified, for example the output should state whether the motif is found in position [2, 10] or any other number. This is just random posiitions that I chose since I dont know where this is positioned

You can use re.finditer() to search for multiple regex pattern matches within a string. Your peptide1 example does not contain an "L*L*L" motif, so I designated a random simple string as a demo.
simple_demo_string = "ABCLXLYLZLABC" # use a simple string to demonstrate code
The demo string contains two overlapping motifs. Normally, regex matches do not account for overlap
Example 1
simple_regex = "L.L.L" # in regex, periods are match-any wildcards
for x in re.finditer(simple_regex, simple_demo_string):
print( x.start(), x.end(), x.group() )
# Output: 3 8 LXLYL
However, if you use a capturing group inside a lookahead, you'll be able to get everything even if there's overlap.
Example 2
lookahead_regex = "(?=(L.L.L))"
for x in re.finditer(lookahead_regex, simple_demo_string):
# note - x.end() becomes same as x.start() due to lookahead
# but can be corrected by simply adding length of match
print( x.start(), x.start()+len(x.group(1)), x.group(1) )
# Output: 3 8 LXLYL
#. 5 10 LYLZL

New to regex -- unexpected results in for loop

I'm not sure if this is a problem in my understanding of regex modules, or a silly mistake I'm making in my for loop.
I have a list of numbers that look like this:
4; 94
3; 92
1; 53
etc.
I made a regex pattern to match just the last two digits of the string:
'^.*\s([0-9]+)$'
This works when I take each element of the list 1 at a time.
However when I try and make a for loop
for i in xData:
if re.findall('^.*\s([0-9]+)$', i)
print i
The output is simply the entire string instead of just the last two digits.
I'm sure I'm missing something very simple here but if someone could point me in the right direction that would be great. Thanks.

You are printing the whole string, i. If you wanted to print the output of re.findall(), then store the result and print that result:
for i in xData:
results = re.findall('^.*\s([0-9]+)$', i)
if results:
print results
I don't think that re.findall() is the right method here, since your lines contain just the one set of digits. Use re.search() to get a match object, and if the match object is not None, take the first group data:
for i in xData:
match = re.search('^.*\s([0-9]+)$', i)
if match:
print match.group(1)

I might be missing something here, but if all you're looking to do is get the last 2 characters, could you use the below?
for i in xData:
print(i[-2:])

Can't convert 'list'object to str implicitly Python

I am trying to import the alphabet but split it so that each character is in one array but not one string. splitting it works but when I try to use it to find how many characters are in an inputted word I get the error 'TypeError: Can't convert 'list' object to str implicitly'. Does anyone know how I would go around solving this? Any help appreciated. The code is below.
import string
alphabet = string.ascii_letters
print (alphabet)
splitalphabet = list(alphabet)
print (splitalphabet)
x = 1
j = year3wordlist[x].find(splitalphabet)
k = year3studentwordlist[x].find(splitalphabet)
print (j)
EDIT: Sorry, my explanation is kinda bad, I was in a rush. What I am wanting to do is count each individual letter of a word because I am coding a spelling bee program. For example, if the correct word is 'because', and the user who is taking part in the spelling bee has entered 'becuase', I want the program to count the characters and location of the characters of the correct word AND the user's inputted word and compare them to give the student a mark - possibly by using some kind of point system. The problem I have is that I can't simply say if it is right or wrong, I have to award 1 mark if the word is close to being right, which is what I am trying to do. What I have tried to do in the code above is split the alphabet and then use this to try and find which characters have been used in the inputted word (the one in year3studentwordlist) versus the correct word (year3wordlist).

There is a much simpler solution if you use the in keyword. You don't even need to split the alphabet in order to check if a given character is in it:
year3wordlist = ['asdf123', 'dsfgsdfg435']
total_sum = 0
for word in year3wordlist:
word_sum = 0
for char in word:
if char in string.ascii_letters:
word_sum += 1
total_sum += word_sum
# Length of characters in the ascii letters alphabet:
# total_sum == 12
# Length of all characters in all words:
# sum([len(w) for w in year3wordlist]) == 18
EDIT:
Since the OP comments he is trying to create a spelling bee contest, let me try to answer more specifically. The distance between a correctly spelled word and a similar string can be measured in many different ways. One of the most common ways is called 'edit distance' or 'Levenshtein distance'. This represents the number of insertions, deletions or substitutions that would be needed to rewrite the input string into the 'correct' one.
You can find that distance implemented in the Python-Levenshtein package. You can install it via pip:
$ sudo pip install python-Levenshtein
And then use it like this:
from __future__ import division
import Levenshtein
correct = 'because'
student = 'becuase'
distance = Levenshtein.distance(correct, student) # distance == 2
mark = ( 1 - distance / len(correct)) * 10 # mark == 7.14
The last line is just a suggestion on how you could derive a grade from the distance between the student's input and the correct answer.

I think what you need is join:
>>> "".join(splitalphabet)
'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'

join is a class method of str, you can do
''.join(splitalphabet)
or
str.join('', splitalphabet)

To convert the list splitalphabet to a string, so you can use it with the find() function you can use separator.join(iterable):
"".join(splitalphabet)
Using it in your code:
j = year3wordlist[x].find("".join(splitalphabet))

I don't know why half the answers are telling you how to put the split alphabet back together...
To count the number of characters in a word that appear in the splitalphabet, do it the functional way:
count = len([c for c in word if c in splitalphabet])

import string
# making letters a set makes "ch in letters" very fast
letters = set(string.ascii_letters)
def letters_in_word(word):
return sum(ch in letters for ch in word)
Edit: it sounds like you should look at Levenshtein edit distance:
from Levenshtein import distance
distance("because", "becuase") # => 2

While join creates the string from the split, you would not have to do that as you can issue the find on the original string (alphabet). However, I do not think is what you are trying to do. Note that the find that you are trying attempts to find the splitalphabet (actually alphabet) within year3wordlist[x] which will always fail (-1 result)
If what you are trying to do is to get the indices of all the letters of the word list within the alphabet, then you would need to handle it as
for each letter in the word of the word list, determine the index within alphabet.
j = []
for c in word:
j.append(alphabet.find(c))
print j
On the other hand if you are attempting to find the index of each character within the alphabet within the word, then you need to loop over splitalphabet to get an individual character to find within the word. That is
l = []
for c within splitalphabet:
j = word.find(c)
if j != -1:
l.append((c, j))
print l
This gives the list of tuples showing those characters found and the index.
I just saw that you talk about counting the number of letters. I am not sure what you mean by this as len(word) gives the number of characters in each word while len(set(word)) gives the number of unique characters. On the other hand, are you saying that your word might have non-ascii characters in it and you want to count the number of ascii characters in that word? I think that you need to be more specific in what you want to determine.
If what you are doing is attempting to determine if the characters are all alphabetic, then all you need to do is use the isalpha() method on the word. You can either say word.isalpha() and get True or False or check each character of word to be isalpha()

Python: how to count overlapping occurrences of a substring [duplicate]

This question already has answers here:
How can I find the number of overlapping sequences in a String with Python? [duplicate]
(4 answers)
Closed 9 years ago.
I wanted to count the number of times that a string like 'aa' appears in 'aaa' (or 'aaaa').
The most obvious code gives the wrong (or at least, not the intuitive) answer:
'aaa'.count('aa')
1 # should be 2
'aaaa'.count('aa')
2 # should be 3
Does anyone have a simple way to fix this?

From str.count() documentation:
Return the number of non-overlapping occurrences of substring sub in the range [start, end]. Optional arguments start and end are interpreted as in slice notation.
So, no. You are getting the expected result.
If you want to count number of overlapping matches, use regex:
>>> import re
>>>
>>> len(re.findall(r'(a)(?=\1)', 'aaa'))
2
This finds all the occurrence of a, which is followed by a. The 2nd a wouldn't be captured, as we've used look-ahead, which is zero-width assertion.

haystack = "aaaa"
needle = "aa"
matches = sum(haystack[i:i+len(needle)] == needle
for i in xrange(len(haystack)-len(needle)+1))
# for Python 3 use range instead of xrange

The solution is not taking overlap into consideration.
Try this:
big_string = "aaaa"
substring = "aaa"
count = 0
for char in range(len(big_string)):
count += big_string[char: char + len(subtring)] == substring
print count

You have to be careful, because you seem to looking for non-overlapping substrings. To fix this I'd do:
len([s.start() for s in re.finditer('(?=aa)', 'aaa')])
And if you don't care about the position where the substring starts you can do:
len([_ for s in re.finditer('(?=aa)', 'aaa')])
Although someone smarter than myself might be able to show that there are performances differences :)

regular expressions to extract phone numbers

I am new to regular expressions and I am trying to write a pattern of phone numbers, in order to identify them and be able to extract them. My doubt can be summarized to the following simple example:
I try first to identify whether in the string is there something like (+34) which should be optional:
prefixsrch = re.compile(r'(\(?\+34\)?)?')
that I test in the following string in the following way:
line0 = "(+34)"
print prefixsrch.findall(line0)
which yields the result:
['(+34)','']
My first question is: why does it find two occurrences of the pattern? I guess that this is related to the fact that the prefix thing is optional but I do not completely understand it. Anyway, now for my big doubt
If we do a similar thing searching for a pattern of 9 digits we get the same:
numsrch = re.compile(r'\d{9}')
line1 = "971756754"
print numsrch.findall(line1)
yields something like:
['971756754']
which is fine. Now what I want to do is identify a 9 digits number, preceded or not, by (+34). So to my understanding I should do something like:
phonesrch = re.compile(r'(\(?\+34\)?)?\d{9}')
If I test it in the following strings...
line0 = "(+34)971756754"
line1 = "971756754"
print phonesrch.findall(line0)
print phonesrch.findall(line1)
this is, to my surprise, what I get:
['(+34)']
['']
What I was expecting to get is ['(+34)971756754'] and ['971756754']. Does anybody has the insight of this? thank you very much in advance.

Your capturing group is wrong. Make the country code within a non-capturing group and the entire expression in the capturing group
>>> line0 = "(+34)971756754"
>>> line1 = "971756754"
>>> re.findall(r'((?:\(?\+34\)?)?\d{9})',line0)
['(+34)971756754']
>>> re.findall(r'((?:\(?\+34\)?)?\d{9})',line1)
['971756754']
My first question is: why does it find two occurrences of the pattern?
This is because, ? which means it match 0 or 1 repetitions, so an empty string is also a valid match