Compare strings and just keep those who have on same positions different characters [closed] - regex

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my question is the following: I have a file which contains around 70 strings, all of them have 6 characters (either a,c,g or t for every position -> these are short DNA-sequences).
For example:
accggt agctta gggatc gactta ccttgg
What I need are the strings which are completely unique. Which have on every position a different character (base) compared with the other strings.
In this case I would get two matches (I define them as lists but this is only an idea for the output format):
[accggt , gggatc]
[gggatc , ccttgg]
The elements of list one are on every position different and so are also the elements of list 2.
Is there a build-in function which can do it? I also thought of regular expression but I'm not that familar with this approach.
Thanks in advance!
Edit:
Ok, it seems it is not that easy to describe. So lets go into more detail:
Let's take the five strings mentioned above:
I would start to compare the first string with all the other strings and then continue with string 2 comparing with all other strings and so on.
The first character of the first string is an a.
The first character of the second string is also an a.
This means I would discard the second string.
The first character of the third string is an g.
Fine.
The second character of the first string is an c.
The second character of the third string is an g.
Fine.
The third character of the first string is an c.
The third character of the third string is an g.
Fine.
The fourth ... and so on.
And if all characters of a string are different from the characters of another string (on every position like described above) I would keep those two strings and would search for the next strings which are different on every position compared to the strings I already found. Because I only have four letters there should be only four possibilities fo different strings.
I should end up with, probably a list, which contains the groups of strings which are different in every position.
I hope this helps.

Answer 1

You can use the following algorithm: iterate through all possible word combinations in your string and check each pair for equality with if [x == y for (x, y) in zip(word, nextWord)].count(True) == 0:.
Here is a snippet:
s = "accggt agctta gggatc gactta ccttgg"
chks = s.split(" ");
for word in chks:
for nextWord in chks:
if word != nextWord:
if [x == y for (x, y) in zip(word, nextWord)].count(True) == 0:
print([word, nextWord])
Result of the IDEONE demo:
['accggt', 'gggatc']
['gggatc', 'accggt']
['gggatc', 'ccttgg']
['ccttgg', 'gggatc']
UPDATE
You can deduplicate the list with a custom function. Here is an updated snippet:
def dedup(lst):
seen = set()
result = []
for item in lst:
fs = frozenset(item)
if fs not in seen:
result.append(item)
seen.add(fs)
return result
res = []
s = "accggt agctta gggatc gactta ccttgg"
chks = s.split(" ");
for word in chks:
for nextWord in chks:
if word != nextWord:
if [x == y for (x, y) in zip(word, nextWord)].count(True) == 0:
res.append([word, nextWord])
print(dedup(res))
Result: [['accggt', 'gggatc'], ['gggatc', 'ccttgg']].
To check the words by 3, you need to create all possible permutations of the string into 3-word combinations and use something like:
from itertools import permutations
def dedup(lst):
seen = set()
result = []
for item in lst:
fs = frozenset(item)
if fs not in seen:
result.append(item)
seen.add(fs)
return result
res = []
s = "accggt agctta gggatc gactta ccttgg"
chks = s.split(" ");
perms = [p for p in permutations(chks, 3)]
for perm in perms:
if [(x == y or y == z or x == z) for (x, y, z) in zip(*perm)].count(True) == 0:
res.append(perm)
print(dedup(res))

Answer 2

To find the DNA strings which are completely different on every character you have to check every string against any other string if any character of the given string is the same character on the same position in the comparing string.
Here is an example code for that:
# read all dna strings into a list of strings
dna = ['accggt', 'agctta', 'gggatc', 'gactta', 'ccttgg', '123456']
def compare_two_dna(dna1, dna2):
i = 0
l = len(dna1)
while(i < l):
if dna1[i] == dna2[i]:
return True
i += 1
return False
def is_dna_unique(d, dna_strings):
return len(filter(lambda x: compare_two_dna(d, x), dna_strings)) == 1
# filter all items which only occure once in the list
unique_dna = filter(lambda d: is_dna_unique(d, dna), dna)
print(unique_dna)
The result here is: 123456

Answer 3

var dnaList = "accggt agctta gggatc gactta ccttgg".split( " " );
function getUniqueDnas( dna_list ){
var result = [];
for( var d1 in dna_list ){
var isRepeat = false;
var dna1 = dna_list[ d1 ];
for( var d2 in dna_list ){
var dna2 = dna_list[ d2 ];
if( dna1 == dna2 ){
isRepeat = true;
break;
}
}
if( !isRepeat )
result.push( dna1 );
}
return result;
}
var uniqueDnaList = getUniqueDnas( dnaList );