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Count of Multiple Fields(After Validation) in UNIX

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# 1  
Old 01-01-2013
Count of Multiple Fields(After Validation) in UNIX
Hi ,
i have scenario where i have to get count of each field ,if the field have value equal to "A"
For Example :

Code:
Field_01     Field_02 
   A               A 
   A               B 
   B               A 
   A               A

OUTPUT :

Code:
Field_01    Field_count    Filed_02     Filed_count 
A                  3                    A                3 
A                  3                    B                3 
B                  3                    A                3 
A                  3                    A                3

Thanks
Anudeep
Last edited by Scrutinizer; 01-01-2013 at 11:27 AM.. Reason: code tags
# 2  
Old 01-01-2013
try:
Code:
awk -v s="A" '
{for (i=1; i<=NF; i++) {a[NR,i]=$i ; if ($i==s) c[i]++}}
END {
  for (i=1; i<=NR; i++) {
    for (j=1; j<=NF; j++) {
       printf a[i,j] " " (c[j] ? c[j]:0) " ";
    }
    print "";
  }
}' input

# 3  
Old 01-01-2013
Pls use code tags as advised.
I'm not sure I interpret your requirement correctly, as your desired output does NOT relate to your req.- phrasing (B showing a count of three even though there are only 2 Bs in file, and, moreover, B should not have been included at all!), and you don't specify what to with the header.
However, taking some assumptions, this might do what you want:
Code:
awk -v s="A" '
     NR==1 {for (i=1; i<=NF; i++)  {printf "%s\t%s_count\t", $i, $i} printf "\n"; next}
     {for (i=1; i<=NF; i++) {a[NR,i]=$i ; c[$i,i]+=$i==s}}
     END {for (i=2; i<=NR; i++)
        {for (j=1; j<=NF; j++)
           {printf "%s\t%s\t", a[i,j], c[a[i,j],j];
           }
           print "";
        }
         }
    ' file
Field_01    Field_01_count    Field_02    Field_02_count    
A    3    A    3    
A    3    B    0    
B    0    A    3    
A    3    A    3

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