I have the following simple script that tries to count the tag encoded with “CB:Z” in SAM/BAM file:
samtools view -h small.bam | grep "CB:Z:" | sed 's/.*CB:Z:([ACGT]*).*/1/' | sort | uniq -c | awk '{print $2 " " $1}'
Typically it needs to process 40 million lines. That codes takes around 1 hour to finish.
This line sed 's/.*CB:Z:([ACGT]*).*/1/'
is very time consuming.
How can I speed it up?
The reason I used the Regex is that the “CB” tag column-wise position is not fixed. Sometimes it’s at column 20 and sometimes column 21.
Example BAM file can be found HERE.
Update
Speed comparison on complete 40 million lines file:
My initial code:
real 21m47.088s user 26m51.148s sys 1m27.912s
James Brown’s with AWK:
real 1m28.898s user 2m41.336s sys 0m6.864s
James Brown’s with MAWK:
real 1m10.642s user 1m41.196s sys 0m6.484s
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Answer
Another awk, pretty much like @tripleee’s, I’d assume:
$ samtools view -h small.bam | awk ' match($0,/CB:Z:[ACGT]*/) { # use match for the regex match a[substr($0,RSTART+5,RLENGTH-5)]++ # len(CB:z:)==5, hence +-5 } END { for(i in a) print i,a[i] # sample output,tweak it to your liking }'
Sample output:
... TCTTAATCGTCC 175 GGGAAGGCCTAA 190 TCGGCCGATCGG 32 GACTTCCAAGCC 76 CCGCGGCATCGG 36 TAGCGATCGTGG 125 ...
Notice: Your sed 's/.*CB:Z:...
matches the last instance where as my awk 'match($0,/CB:Z:[ACGT]*/)...
matches the first.
Notice 2: Quoting @Sundeep in the comments: – – using LC_ALL=C mawk '..'
will give even better speed.