Insert a row and a column in a matrix using awk - linux

Question

I have a gridded dataset with 250 rows x 300 columns in matrix form:
ifile.txt
2 3 4 1 2 3
3 4 5 2 4 6
2 4 0 5 0 7
0 0 5 6 3 8
I would like to insert the latitude values at the first column and longitude values at the top. Which looks like:
ofile.txt
20.00 20.33 20.66 20.99 21.32 21.65
100.00 2 3 4 1 2 3
100.33 3 4 5 2 4 6
100.66 2 4 0 5 0 7
100.99 0 0 5 6 3 8
The increment is 0.33
I can do it for a small size matrix in manually, but I can't able to get any idea how to get my output in my desired format. I was writing a script in the following way, but completely useless.
echo 20 > latitude.txt
for i in `seq 1 250`;do
i1=$(( i + 0.33 )) #bash can't recognize fractions
echo $i1 >> latitude.txt
done
echo 100 > longitude.txt
for j in `seq 1 300`;do
j1=$(( j + 0.33 ))
echo $j1 >> longitude.txt
done
paste longitude.txt ifile.txt > dummy_file.txt
cat latitude.txt dummy_file.txt > ofile.txt

Answer 1

$ cat tst.awk
BEGIN {
lat = 100
lon = 20
latWid = lonWid = 6
latDel = lonDel = 0.33
latFmt = lonFmt = "%*.2f"
}
NR==1 {
printf "%*s", latWid, ""
for (i=1; i<=NF; i++) {
printf lonFmt, lonWid, lon
lon += lonDel
}
print ""
}
{
printf latFmt, latWid, lat
lat += latDel
for (i=1; i<=NF; i++) {
printf "%*s", lonWid, $i
}
print ""
}
$ awk -f tst.awk file
20.00 20.33 20.66 20.99 21.32 21.65
100.00 2 3 4 1 2 3
100.33 3 4 5 2 4 6
100.66 2 4 0 5 0 7
100.99 0 0 5 6 3 8

Answer 2

Following awk may also help you on same.
awk -v col=100 -v row=20 'FNR==1{printf OFS;for(i=1;i<=NF;i++){printf row OFS;row=row+.33;};print ""} {col+=.33;$1=$1;print col OFS $0}' OFS="\t" Input_file
Adding non one liner form of above solution too now:
awk -v col=100 -v row=20 '
FNR==1{
printf OFS;
for(i=1;i<=NF;i++){
printf row OFS;
row=row+.33;
};
print ""
}
{
col+=.33;
$1=$1;
print col OFS $0
}
' OFS="\t" Input_file

Answer 3

Awk solution:
awk 'NR == 1{
long = 20.00; lat = 100.00; printf "%12s%.2f", "", long;
for (i=1; i<NF; i++) { long += 0.33; printf "\t%.2f", long } print "" }
NR > 1{ lat += 0.33 }
{
printf "%.2f%6s", lat, "";
for (i=1; i<=NF; i++) printf "\t%d", $i; print ""
}' file

Answer 4

With perl
$ perl -lane 'print join "\t", "", map {20.00+$_*0.33} 0..$#F if $.==1;
print join "\t", 100+(0.33*$i++), #F' ip.txt
20 20.33 20.66 20.99 21.32 21.65
100 2 3 4 1 2 3
100.33 3 4 5 2 4 6
100.66 2 4 0 5 0 7
100.99 0 0 5 6 3 8
-a to auto-split input on whitespaces, result saved in #F array
See https://perldoc.perl.org/perlrun.html#Command-Switches for details on command line options
if $.==1 for the first line of input
map {20.00+$_*0.33} 0..$#F iterate based on size of #F array, and for each iteration, we get a value based on equation inside {} where $_ will be 0, 1, etc upto last index of #F array
print join "\t", "", map... use tab separator to print empty element and results of map
For all the lines, print contents of #F array pre-fixed with results of 100+(0.33*$i++) where $i will be initially 0 in numeric context. Again, tab is used as separator while joining these values
Use sprintf if needed for formatting, also $, can be initialized instead of using join
perl -lane 'BEGIN{$,="\t"; $st=0.33}
print "", map { sprintf "%.2f", 20+$_*$st} 0..$#F if $.==1;
print sprintf("%.2f", 100+($st*$i++)), #F' ip.txt