Code Golf - bisulfite conversion
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3.1 years ago

Been a while since we had a code golf challenge (shortest script wins, but expect to receive upvotes for cleverness)!

Simplifying slightly, bisulfite conversion means that Cs get converted to Ts, but some bases may be protected by methylgroups and thus remain unconverted. (This happens on the opposite strand too, so some complimentary Gs could get converted to As)

So, given an input sequence read (assume a [ACTGN] string of no more than 150 bp), output all reads that might result from partial or complete bisulfite conversion of that sequence.

EDIT: To be clear, C>T happens on one read, G>A happens on the other. Reads will not exist with both conversions! See this screenshot: Small example: Input:

AACGCGAA


Output:

AATGTGAA
AATGCGAA
AACGTGAA
AACACAAA
AACACGAA
AACGCAAA
AACGCGAA


(the original string could be in there as well, since all bases could be protected!)

Happy Golfing! golf codegolf bsseq bisulfite Forum • 1.1k views
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So that we can test our outputs, what is the output for ACTGN?

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after I saw the other's answers, i'm not sure if we only need to change the 'C' in the sequence or the 'C' AND the 'G'.

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Both need to change since the methylation and read you are looking at could be from either strand.

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If I am understanding the problem correctly, you can't change C->T and G->A for the same read.

5'- ATGCCCATG -3'
3'- TACGGGTAC -5'


The only conversion chemically occurring is C->T, so there are 3 conversions possible for the upper strand, and two for the lower strand. If my interpretation is correct, Pierre Lindenbaum and jrj.healey solutions are incomplete, and zx8754 solution is incorrect.

edit: my interpretation was right, but all answers have been corrected and updated.

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This is correct, it's either C->T or G->A, but not both!

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OK, R code corrected, gives same output as expected.

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Updated this post with more details and example input/output. Sorry for the confusion!

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3.1 years ago
Joe 19k

## A python solution

### Edit II: gained 8 bytes (now 108):

NB This doesn’t solve the problem completely correctly as present (was written before OP clarified).

from itertools import product as p
print(["".join(s) for s in p(*[i+{'C':'t','G':'a'}.get(i,'') for i in ""])])


### Edit: a little enhancement at the cost of readability:

As a pseudo-oneliner, without counting the input string as bytes:

from itertools import product
print(["".join(s) for s in product(*[i + {'C':'t'}.get(i, '') for i in ""])])


112 bytes, or 116, if the dictionary should read {'C':'t','G':'a'}.

Not 100% sure if this is correct, but I get the same output as Pierre:

from itertools import product
s = [i + {'C':'t'}.get(i, '') for i in "ATGCCCATG"]
print(["".join(sub) for sub in product(*s)])


### 127 bytes.

['ATGCCCATG', 'ATGCCtATG', 'ATGCtCATG', 'ATGCttATG', 'ATGtCCATG', 'ATGtCtATG', 'ATGttCATG', 'ATGtttATG']


If the G substitution is also needed (OP was unclear):

from itertools import product
s = [i + {'C':'t','G':'a'}.get(i, '') for i in "ATGCCCATG"]
print(["".join(sub) for sub in product(*s)])


I think I get the same as zx8754:

['ATGCCCATG', 'ATGCCCATa', 'ATGCCtATG', 'ATGCCtATa', 'ATGCtCATG', 'ATGCtCATa', 'ATGCttATG', 'ATGCttATa', 'ATGtCCATG', 'ATGtCCATa', 'ATGtCtATG', 'ATGtCtATa', 'ATGttCATG', 'ATGttCATa', 'ATGtttATG', 'ATGtttATa', 'ATaCCCATG', 'ATaCCCATa', 'ATaCCtATG', 'ATaCCtATa', 'ATaCtCATG', 'ATaCtCATa', 'ATaCttATG', 'ATaCttATa', 'ATatCCATG', 'ATatCCATa', 'ATatCtATG', 'ATatCtATa', 'ATattCATG', 'ATattCATa', 'ATatttATG', 'ATatttATa']


@Chris, it would be useful if you had some 100% correct example output?

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Very interesting ! Could you detailed please :)

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Each iterable variation on the input string is stored, with a dictionary guiding the substitutions for each character in the string (for i in "ATG...").

According to the docs, the imported itertools module is functionally equivalent to nested for-loops (https://docs.python.org/3/library/itertools.html#itertools.product), allowing you to yield permutations of the input iterables, as 'cartesian products'. Its kind of similar to itertools.permutations, but that would give all the shuffled possibilities (though its following that train of thought that got me to this).

I confess I'm black-boxing this a little as it's heavily based upon https://stackoverflow.com/a/29184387/3691040 and I can't take all the credit. You'll be much better guided reading the docs than listening to me :P

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Yeah the product(*s) is quite murky to me. Anyhow it's really good coding !

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The (*s) syntax is just telling the product function that it will receive a variable number of args, as the number of iterables returned will depend on the sequence and its length. The inner workings of product I don't have a very strong feeling for, as I suspect it required better linear algebra than I have haha..

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3.1 years ago

In C :

compilation:

gcc -O3 -o biostar341051 biostar341051.c


usage:

$./biostar341051 ATGCCCATG AACGCGAA AACGCAAA AACACGAA AACACAAA AACGCGAA AACGTGAA AATGCGAA AATGTGAA  Edit: fixed after I saw your comments about 'G'->A Edit2 : fixed after Chris's comment. ADD COMMENT 4 Entering edit mode 3.1 years ago zx8754 10k Using R, 212 bytes: Note: I am converting C > t and G > a so it is easier to see the changes in the output The function with whitespaces: foo <- function(s){ x = strsplit(s, "")[] c(s, unlist( lapply(c("C", "G"), function(i){ y = which(x == i) sapply(seq(y), function(j) combn(y, j, function(k){ z = x z[ k ] = setNames(c("t", "a"), c("C", "G"))[ z[ k ] ] paste(z, collapse = "")})) }))) }  Usage: foo("AACGCGAA") #  "AACGCGAA" "AAtGCGAA" "AACGtGAA" "AAtGtGAA" "AACaCGAA" "AACGCaAA" "AACaCaAA"  ADD COMMENT 2 Entering edit mode 3.1 years ago EDIT I may have a correct (Python-based) answer. Please see edited code below. In addition to being shorter, jrj.healey's answer is also a bit more than twice as fast as building a powerset of substitution indices and reducing: from functools import reduce from itertools import product as p import timeit e = 'AACGCGAA' m = {'C':'t','G':'a'} def f(k,s,t): for i in s: t[i] = ord(m[k]) return t.decode('ascii') def a1(): return [i for s in [[f(k,x,bytearray(e,'ascii')) for x in [reduce(lambda z,x: z + [y + [x] for y in z], s, [[]]) for s in [[i for i,v in enumerate(e) if v == k] for k in m.keys()]][i] if x != []] for i,k in enumerate(m.keys())] + [[e]] for i in s] def a2(): return ["".join(s) for s in p(*[i+{'C':'t','G':'a'}.get(i,'') for i in e])] print("a1 ->", timeit.timeit(stmt=a1, number=1000)) print(a1) print("a2 ->", timeit.timeit(stmt=a2, number=1000)) print(a2) """$ python test341051.py
a1 -> 0.015046401007566601
['AAtGCGAA', 'AACGtGAA', 'AAtGtGAA', 'AACaCGAA', 'AACGCaAA', 'AACaCaAA', 'AACGCGAA']
a2 -> 0.006538485002238303
['AACGCGAA', 'AACGCaAA', 'AACGtGAA', 'AACGtaAA', 'AACaCGAA', 'AACaCaAA', 'AACatGAA', 'AACataAA', 'AAtGCGAA', 'AAtGCaAA', 'AAtGtGAA', 'AAtGtaAA', 'AAtaCGAA', 'AAtaCaAA', 'AAtatGAA', 'AAtataAA']
"""


For code golf purposes, the minified version uses 421 characters 310 bytes. Might be worth comparing with R solution on time or memory, to see how it scales.

from functools import reduce
e='AACGCGAA'
m={'C':'t','G':'a'}
def f(k,s,t):
for i in s:
t[i]=ord(m[k])
return t.decode('ascii')
print([i for s in[[f(k,x,bytearray(e,'ascii'))for x in[reduce(lambda z,x:z+[y+[x]for y in z],s,[[]])for s in[[i for i,v in enumerate(e)if v==k]for k in m.keys()]][i]if x!=[]]for i,k in enumerate(m.keys())]+[[e]]for i in s])


More positive controls may be useful.

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Forget efficiency :)

shortest script wins

Use http://bytesizematters.com/ and tick "Ignore whitespace".

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haha okay 310 bytes, then!

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mine still needs some work now Chris has clarified, it actually substitutes too many things :/

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Hmm, mine will work on each key of the substitution map m on its own, one at a time — so either C or G is substituted, but not both — so I think mine should give correct output. It would be useful to have some positive examples though, just to test this out.

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When I edited the post, I added a small example!

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Okay, I made some edits and re-ran things with output. I think mine matches your example. Some other input or edge cases may be useful to test.