parsing rows

Shell Programming and Scripting
1

Hi,

I have a file that looks like this (tab seperated):

Barry  -3  -4  -5  -10  -4  6  -8  20  -6  NaN  NaN  NaN
Brend -2   4   -3  -7   -3  8   -9  -10  -6 NaN  NaN  NaN  NaN  NaN
Harry  -10 -9 -40  6  -7  3  -7  -2  -5  NaN NaN NaN NaN NaN NaN NaN

I want to print the first column (name) then print the column where theres value equal to or greatr than 0 (along with its position where the counts start from right to left AND also could how many NaN's there are per row.

so the output would look like this

name position|value #of NaN's

Barry 4|6  2|20  3
Brend 7|4  4|8  5
Harry 5|6  7|3  7

hope i was clear

Gisele

2

second and third line's positions in your sample output are wrong. Here is the code:

$ awk '{printf $1}
{for (i=2;i<=10;i++) if ($i>=0) {printf " "11-i"|"$i} }
{for (j=11;j<=NF;j++) tol++} {print " " tol}
{tol=0} ' urfile

Barry 4|6 2|20 3
Brend 8|4 4|8 5
Harry 6|6 4|3 7
3 
while(<DATA>){
	my @tmp=split;
	my @t1 = grep { $_ eq "NaN" } @tmp;
	print @tmp[0]," ";
	@tmp=@tmp[1..$#tmp-$#t1-1];
	for (my $i=0;$i<=$#tmp;$i++){
		print $#tmp+1-$i,"|",$tmp[$i]," " if $tmp[$i]>0;
	}
	print " ",$#t1+1,"\n";
}
__DATA__
Barry  -3  -4  -5  -10  -4  6  -8  20  -6  NaN  NaN  NaN
Brend -2   4   -3  -7   -3  8   -9  -10  -6 NaN  NaN  NaN  NaN  NaN
Harry  -10 -9 -40  6  -7  3  -7  -2  -5  NaN NaN NaN NaN NaN NaN NaN
4

hey

$ awk '{printf $1}
{for (i=2;i<=10;i++) if ($i>=0) {printf " "11-i"|"$i} }
{for (j=11;j<=NF;j++) tol++} {print " " tol}
{tol=0} ' urfile

works but I have decimal places for some of the #'s so the output comes out a bit different

so the order is all wrong

ex. of input

joe	-2	-4	-6	-3	-1	9	3	NaN	NaN	NaN

output is

joe 4|9 3|3 2|NaN 1|NaN 1
5 
awk '{s=0;p=2;z=0
      for (i=2;i<=NF;i++) {
        if ($i !~ /[-.0-9]/) {s++; if (z==0) z=i}
        else if ($i>0) {$(p++)=i;$(p++)=$i}
      }
      printf "%s %s|%s  %s|%s  %s\n",$1,z-$2,$3,z-$4,$5,s}' infile

Barry 4|6  2|20  3
Brend 8|4  4|8  5
Harry 6|6  4|3  7
joe 2|9  1|3  3