AWK - check column data

Linux and Unix-Like Emergency UNIX and Linux Support
1

Hi,

I have a data in a file .

Infile:

1 e 1.2  1.6 
5 f  2.3  3.6
3 g 1.2  2.6
6 i  2.3  3.6
8 o 1.2  3.6

output:

1 e 1.2  1.6 
5 f  2.3  3.6
3 g 1.1  2.6
6 i  2.2  3.5
8 o 1.0  3.4

In column #3 the first occurence of value is unchanged. the next same value occurence will be reduced by 0.1, and the next occurence will be reduced by 0.2,....it should continue according to the repeats.

Anybody has answer,

Thanks in advance
Vasanth

2 
awk '{ 
  for (i=2; ++i<=NF;)
    c[i,$i] = f[i,$i]++ ? c[i,$i] += 0.1 : 0 
  }
{ 
  for (i=2; ++i<=NF;) 
    $i = sprintf("%.1f",$i -= c[i,$i])
  }
 9' infile
3

Hello ,
first of all Happy New year :slight_smile:
try using hashes (err maybe in perl :wink: )
would come back to you .
:slight_smile:
Regards.

4

Dear radoulov,

I am getting error as ' unexpected.. please can u explain ur program...

Dear gaurav1086,

Thanks. same wishes to all...
Perl i can,t use. awk ans???

Thanks in advance,
Vasanth

5

What error do you get?
If you're on Solaris, you should use gawk (if available), nawk or /usr/xpg4/bin/awk.

6

i am using cygwin.

Please, can u explain the above progam.

i will try my best to solve from my side..

Thanks
vasanth

7

It should work on Cygwin.
Please post the error.

8

Hello
Here you go

gaurav@localhost:~$ perl -n -e 'split(/\s+/,$_);foreach $i (@_){if($i!~/[a-z]/){print $i-0.1*$n{$i}," ";$n{$i}++}else{print $i," ";}}print "\n"' infile
1 e 1.2 1.6 
5 f 2.3 3.6 
3 g 1.1 2.6 
6 i 2.2 3.5 
8 o 1 3.4 
gaurav@localhost:~$

cheers ,
Happy New year.
:slight_smile:

9

It's not clear to me what should be the output for an input like this one:

1 e 1.2 1.6
5 f 1.6 1.2

This:

% perl -n -e 'split(/\s+/,$_);foreach $i (@_){if($i!~/[a-z]/){print $i-0.1*$n{$i}," ";$n{$i}++}else{print $i," ";}}print "\n"' infile
1 e 1.2 1.6 
5 f 1.5 1.1

Or this (unchanged);

% awk '{
  for (i=2; ++i<=NF;)
    c[i,$i] = f[i,$i]++ ? c[i,$i] += 0.1 : 0
  }
{
  for (i=2; ++i<=NF;)
    $i = sprintf("%.1f",$i -= c[i,$i])
  }
 9' infile
1 e 1.2 1.6
5 f 1.6 1.2
10

I think he wants to reduce the value the no. of times it occurs * 0.1 .
Incidently the same value doesnt occur in the different columns in the dataset provided by him.

11

Dear Radoulov,

Sorry,It's working fine.... Thanks a lot...

Please can u give a comment. how this program works..

Thanks
vasanth

---------- Post updated at 12:12 PM ---------- Previous update was at 12:04 PM ----------

Dear Radoulov,

If the input file contains more than 4 column, then the output, after fifth column prints as -0.0,-0.1,etc...

The other should be printed as it is..

Thanks in advance..
VAsanth

12

This should fix it.

awk '{ 
  for (i=2; ++i<=4;)
    c[i,$i] = f[i,$i]++ ? c[i,$i] += 0.1 : 0 
  }
{ 
  for (i=2; ++i<=4;) 
    $i = sprintf("%.1f",$i -= c[i,$i])
  }
 9' infile
13

Dear radoulov,

Yaa... Its worked fine...

Still if i need to check pairs of third column($3) and fourth column($4), with the next line upto end... wht will be the program..

Eg:

Input file :

w e 4.2 5
t j 5 6
y u 4.2 5
y i 5 5

output:

w e 4.2 5
t j 5 6
y u 4.1 5
y i 5 5

Thanks in advance

14 
awk '$3 -= 0.1 * c[$3,$4]++' infile

Thanks to gaurav1086 for the algorithm, I'm not good at math :slight_smile:

---------- Post updated at 10:22 PM ---------- Previous update was at 10:07 PM ----------

It should be:

awk '($3 -= 0.1 * c[$3,$4]++)||1' infile

Otherwise you'll loose the records where $3 = 0 :slight_smile:

15 
#!/usr/bin/awk

BEGIN {
   col3[""]=0;
   col4[""]=0;
}

{
   if(col3[$3] <= 0){
       col3[$3] = 1;
       data3 = $3;
   }
   else {
      col3[$3]++;
      data3 = ($3-0.1*(col3[$3]-1)); 
   }

   if(col4[$4] <= 0) {
      col4[$4] = 1;
      data4 = $4;
   }
   else {
      col4[$4]++;
      data4 = ($4-0.1*(col4[$4]-1));
   }

   print $1 $2 data3 data4;

}

This is not tested..

16

another approach:-

nawk 'a[$3$4]++ { $3 -= 0.1}1' test

but kindly radoulov in your solution

awk '($3 -= 0.1 * c[$3,$4]++)||1' infile

why you used multiplication instead of

awk '($3 -= 0.1 && c[$3,$4]++)||1' infile
17

aaiaz:-

using multiplication and not && is for the subtraction value (amount) that we have to subtract after pattern repetition so you have to use
"*" not "&&"

by the way ...you code will not work proparly because each time you subtract
"0.1" although in the second pattern you have to subtract "0.2".

BR

:D:D:D

18

Thanks man for the details ...I had understand now.

:b::b::b: