find frequency of substring in a set of strings - r

Question

I have as input a gene list where each genes has a header like >SomeText.
For each gene I would like to find the frequency of the string GTG. (number of occurences divided by length of gene). The string should only be counted if it starts at position 1,4,7,10 etc (every thids position).
>ENST00000619537.4 cds:known chromosome:GRCh38:21:6560714:6564489:1 gene:ENSG00000276076.4 gene_biotype:protein_coding transcript_biotype:protein_coding gene_symbol:CH507-152C13.3 description:alpha-crystallin A chain [Source:RefSeq peptide;Acc:NP_001300979]
ATGGATGTGACCATCCAGCACCCCTGGTTCAAGCGCACCCTGGGGCCCTTCTACCCCAGC
CGGCTGTTCGACCAGTTTTTCGGCGAGGGCCTTTTTGAGTATGACCTGCTGCCCTTCCTG
TCGTCCACCATCAGCCCCTACTACCGCCAGTCCCTCTTCCGCACCGTGCTGGACTCCGGC
ATCTCTGAGGTTCGATCCGACCGGGACAAGTTCGTCATCTTCCTCGATGTGAAGCACTTC
TCCCCGGAGGACCTCACCGTGAAGGTGCAGGACGACTTTGTGGAGATCCACGGAAAGCAC
AACGAGCGCCAGGACGACCACGGCTACATTTCCCGTGAGTTCCACCGCCGCTACCGCCTG
CCGTCCAACGTGGACCAGTCGGCCCTCTCTTGCTCCCTGTCTGCCGATGGCATGCTGACC
TTCTGTGGCCCCAAGATCCAGACTGGCCTGGATGCCACCCACGCCGAGCGAGCCATCCCC
GTGTCGCGGGAGGAGAAGCCCACCTCGGCTCCCTCGTCCTAA
>ENST00000624019.3 cds:known chromosome:GRCh38:21:6561284:6563978:1 gene:ENSG00000276076.4 gene_biotype:protein_coding transcript_biotype:protein_coding gene_symbol:CH507-152C13.3 description:alpha-crystallin A chain [Source:RefSeq peptide;Acc:NP_001300979]
ATGGACGCCCCCCCCCCCCACCCAACCACAGGCCTCCTCTCTGAGCCACGGGTTCGATCC
GACCGGGACAAGTTCGTCATCTTCCTCGATGTGAAGCACTTCTCCCCGGAGGACCTCACC
GTGAAGGTGCAGGACGACTTTGTGGAGATCCACGGAAAGCACAACGAGCGCCAGGACGAC
CACGGCTACATTTCCCGTGAGTTCCACCGCCGCTACCGCCTGCCGTCCAACGTGGACCAG
TCGGCCCTCTCTTGCTCCCTGTCTGCCGATGGCATGCTGACCTTCTGTGGCCCCAAGATC
CAGACTGGCCTGGATGCCACCCACGCCGAGCGAGCCATCCCCGTGTCGCGGGAGGAGAAG
CCCACCTCGGCTCCCTCGTCCTAA
>ENST00000624932.1 cds:known chromosome:GRCh38:21:6561954:6564203:1 gene:ENSG00000276076.4 gene_biotype:protein_coding transcript_biotype:protein_coding gene_symbol:CH507-152C13.3 description:alpha-crystallin A chain [Source:RefSeq peptide;Acc:NP_001300979]
ATGCCTGTCTGTCCAGGAGACAGTCACAGGCCCCCGAAAGCTCTGCCCCACTTGGTGTGT
GGGAGAAGAGGCCGGCAGGTTCGATCCGACCGGGACAAGTTCGTCATCTTCCTCGATGTG
AAGCACTTCTCCCCGGAGGACCTCACCGTGAAGGTGCAGGACGACTTTGTGGAGATCCAC
GGAAAGCACAACGAGCGCCAGGACGACCACGGCTACATTTCCCGTGAGTTCCACCGCCGC
TACCGCCTGCCGTCCAACGTGGACCAGTCGGCCCTCTCTTGCTCCCTGTCTGCCGATGGC
ATGCTGACCTTCTGTGGCCCCAAGATCCAGACTGGCCTGGATGCCACCCACGCCGAGCGA
GCCATCCCCGTGTCGCGGGAGGAGAAGCCCACCTCGGCTCCCTCGTCCTAA
Output:
Gene Frequency
Gene1: 3
Gene2 6.3
....
I was thinging of something like this, but I dont now how to define the positions requirements:
freq <- sapply(gregexpr("GTG",x),function(x)if(x[[1]]!=-1) length(x) else 0)

Answer 1

Here is an idea in R using stringi.
We use stri_locate_all_fixed() to find the start and end position of each GTG occurence. Then we create a column condition to test if start position is in 1,4,7,10,13,16,19,22 ....
library(stringi)
library(dplyr)
data.frame(stri_locate_all_fixed(gene1, "GTG")) %>%
mutate(condition = start %in% seq(1, nchar(gene), 3))
Which gives:
# start end condition
#1 4 6 TRUE
If you want to generalize this to a list of genes, you could do:
lst <- list(gene1, gene2, gene3)
res <- lapply(lst, function(x) {
data.frame(stri_locate_all_fixed(x, "GTG")) %>%
mutate(condition = start %in% seq(1, nchar(x), 3))
})
Which would give:
#[[1]]
# start end condition
#1 4 6 TRUE
#
#[[2]]
# start end condition
#1 NA NA FALSE
#
#[[3]]
# start end condition
#1 3 5 FALSE
#2 9 11 FALSE
#3 21 23 FALSE
#4 70 72 TRUE
#5 75 77 FALSE
Following #Sobrique's comment, if divided by length means number of occurences respecting condition divided by total number of char in each gene, you could do:
lapply(1:length(res), function(x) sum(res[[x]][["condition"]]) / nchar(lst[[x]]))
Which would give:
#[[1]]
#[1] 0.004830918
#
#[[2]]
#[1] 0
#
#[[3]]
#[1] 0.003021148

Answer 2

Here's a Perl solution that does as you ask
But I don't understand how your example output is derived: the first and last sequences have only one occurrence of GTG in the positions you require, and the second sequence has none at all. That means the outputs are 1 / 207, 0 / 74, and 1 / 331 respectively. None of those are anything like 3 and 6.3 that you say you're expecting
This program expects the path to the input file as a parameter on the command line
use strict;
use warnings 'all';
print "Gene Frequency\n";
my $name;
local $/ = '>';
while ( <> ) {
chomp;
next unless /\S/;
my ($name, $seq) = split /\n/, $_, 2;
$seq =~ tr/A-Z//cd;
my $n = 0;
while ( $seq =~ /(?=GTG)/g ) {
++$n if $-[0] % 3 == 0;
}
printf "%-7s%.6f\n", $name, $n / length($seq);
}
output
Gene Frequency
Gene1 0.004831
Gene2 0.000000
Gene3 0.003021

Answer 3

Here is an alternate solution that does not use a pattern match. Not that it will matter much.
use strict;
use warnings;
my $gene;
while ( my $line = <> ) {
if ( $line =~ /^>(.+)/ ) {
$gene = $1;
next;
}
chomp $line;
printf "%s: %s\n",
$gene,
( grep { $_ eq 'GTG' } split /(...)/, $line ) / length $line;
}
Output:
Gene1: 0.00483091787439614
Gene2: 0
Gene3: 0.00302114803625378
It is essentially similar to Sobrique's answer, but assumes that the gene lines contain the right characters. It splits up the gene string into a list of three-character pieces and takes the ones that are literally GTG.
The splitting works by abusing the fact that split uses a pattern as the delimiter, and that it will also capture the delimiter if a capture group is used. Here's an example.
my #foo = split /(...)/, '1234567890';
p #foo; # from Data::Printer
__END__
[
[0] "",
[1] 123,
[2] "",
[3] 456,
[4] "",
[5] 789,
[6] 0
]
The empty elements get filter out by grep. It might not be the most efficient way, but it gets the job done.
You can run it by calling perl foo.pl horribly-large-gene-sequence.file.

Answer 4

Well, you have an R solution. I've hacked something together in perl because you tagged it:
#!/usr/bin/env perl
use strict;
use warnings;
my $target = 'GTG';
local $/ = "\n>";
while ( <> ) {
my ($gene) = m/(Gene\d+)/;
my #hits = grep { /^$target$/ } m/ ( [GTCA]{3} ) /xg;
print "$gene: ".( scalar #hits), "\n";
}
This doesn't give the same results as your input though:
Gene1: 1
Gene2: 0
Gene3: 1
I'm decomposing your string into 3 element lists, and looking for ones that specifically match. (And I haven't divided by length, as I'm not entirely clear if that's the actual string length in letters, or some other metric).
Including length matching - we need to capture both name and string:
#!/usr/bin/env perl
use strict;
use warnings;
local $/ = "\n>";
while (<>) {
my ($gene, $gene_str) = m/(Gene\d+)\n([GTCA]+)/m;
my #hits = grep { /^GTG$/ } $gene_str =~ m/ ( [GTCA]{3} ) /xg;
print "$gene: " . #hits . "/". length ( $gene_str ), " = ", #hits / length($gene_str), "\n";
}
We use <> which is the 'magic' filehandle, and tells perl to read from either STDIN or a file specified on command line. Much like sed or grep does.
With your input:
Gene1: 1/207 = 0.00483091787439614
Gene2: 0/74 = 0
Gene3: 1/331 = 0.00302114803625378

Answer 5

Here is a function I created based on your requirement. I am pretty sure there are alternate ways better than this but this solves the problem.
require(stringi)
input_gene_list<- list(gene1= "GTGGGGGTTTGTGGGGGTG", gene2= "GTGGGGGTTTGTGGGGGTG", gene3= "GTGGGGGTTTGTGGGGGTG")
gene_counter<- function(gene){
x<- gene
y<- gsub(pattern = "GTG",replacement = "GTG ", x = x, perl=TRUE)
if(str_count(y,pattern = "GTG")) {
gene_count<- unlist(gregexpr(pattern = " ", y))
counter<- 0
for(i in 1:length(gene_count)){
if((gene_count[i] %% 3) == 1) counter=counter+1
}
return(counter/nchar(x))
}
}
output_list<- lapply(input_gene_list, function(x) gene_counter(x))
result<- t(as.data.frame(output_list))
result
[,1]
gene1 0.1052632
gene2 0.1052632
gene3 0.1052632
Also share your thoughts on it! Thanks!