Converting second pattern to millisecond in awk - bash

Question

I have file which is having pattern 's' , I need to convert into 'ms' by multiplying by 1000. I am unable to do it. Please help me.
file.txt
First launch 1
App: +1s170ms
First launch 2
App: +186ms
First launch 3
App: +1s171ms
First launch 4
App: +1s484ms
First launch 5
App: +1s227ms
First launch 6
App: +204ms
First launch 7
App: +1s180ms
First launch 8
App: +1s177ms
First launch 9
App: +1s183ms
First launch 10
App: +1s155ms
My code:
awk 'BEGIN { FS="[: ]+"}
/:/ && $2 ~/ms$/{vals[$1]=vals[$1] OFS $2+0;next}
END {
for (key in vals)
print key vals[key]
}' file.txt
Expected output:
App 1170 186 1171 1484 1227 204 1180 1177 1183 1155
Output Coming:
App 1 186 1 1 1 204 1 1 1 1
How to convert in above pattern 's' to 'ms' if second pattern comes .

Answer 1

What I will try to do here is explain it a bit generic and then apply it to your case.
Question: I have a string of the form 123a456b7c8d where the numbers are numeric integral values of any length and the letters are corresponding units. I also have conversion factors to convert from unit a,b,c,d to unit f. How can I convert this to a single quantity of unit f?
Example: from 1s183ms to 1183ms
Strategy:
create per string a set of key-value pairs 'a' => 123,'b' => 456, 'c' => 7 and 'd' => 8
multiply each value with the corect conversion factor
add the numbers together
Assume we use awk and the key-value pairs are stored in array a with the key as an index.
Extract key-value pairs from str:
function extract(str,a, t,k,v) {
delete a; t=str;
while(t!="") {
v=t+0; match(t,/[a-zA-Z]+/); k=substr(t,RSTART,RLENGTH);
t=substr(t,RSTART+RLENGTH);
a[k]=v
}
return
}
convert and sum: here we assume we have an array f which contains the conversion factors:
function convert(a,f, t,k) {
t=0; for(k in a) t+=a[k] * f[k]
return t
}
The full code (for the example of the OP)
# set conversion factors
BEGIN{ f['s']=1000; f['ms'] = 1 }
# print first word
BEGIN{ printf "App:" }
# extract string and print
/^App/ { extract($2,a); printf OFS "%dms", convert(a,f) }
END { printf ORS }
which outputs:
App: 1170ms 186ms 1171ms 1484ms 1227ms 204ms 1180ms 1177ms 1183ms 1155ms

Answer 2

perl -n -e '$s=0; ($s)=/(\d+)s/; ($ms)=/(\d+)ms/;
s/^(\w+):/push #{$vals{$1}}, $ms+$s*1000/e;
eof && print "$_: #{$vals{$_}}\n" for keys %vals;' file`
perl -n doesn't print anything as it loops through the input.
$s and $ms are set to those fields. $s is ensured to reset to zero
s///e is stuffing the %vals hash with a list of numbers in ms for each key, App, in this case.
eof && executes the subsequent code after the end of the file.
print "$_: #{$vals{$_}}\n" for keys %vals is printing the %vals hash as the OP wants.
App: 1170 186 1171 1484 1227 204 1180 1177 1183 1155