Wsrtjtyk

I have a file that looks like this :

header,d0,d1,d2,d3, ...

s1,0,5,2,8, ...

s2,0,8,2,4, ...

s3,0,7,3,4, ...

s4,0,3,2,1, ...

...

I want to remove any column with all zeros like d0

I can manually inspect for columns with all zeros and find d0 and execute

cut -d "," -f 1,3- file> file_revised  

The desired output is

header,d1,d2,d3, ...

s1,5,2,8, ...

s2,8,2,4, ...

s3,7,3,4, ...

s4,3,2,1, ...

...

But since I have so many columns, it is hard to inspect manually.

How can I automatically remove columns with all zeros?

Thank you.

$ cat file

header,d0,d1,d2,d3

s1,0,5,2,8

s2,0,8,2,4

s3,0,7,3,4

s4,0,3,2,1

$

$ cat tst.awk

NR==1 {

    for (i=1; i<=NF; ++i)

        a[i]

    next

}

NR==FNR {

    for (i in a)

        if ($i != "0")

            delete a[i]

    next

}

{

    sep = ""

    out = ""

    for (i=1; i<=NF; ++i) {

        if (i in a)

            continue

        out = out sep $i

        sep = FS

    }

    print out

}

$

$ awk -F, -f tst.awk file file

header,d1,d2,d3

s1,5,2,8

s2,8,2,4

s3,7,3,4

s4,3,2,1

Here is one that gathers the fields to print to a variable (p="$1,$3" ... etc.) and uses system to call awk to print p:

$ awk '

BEGIN { FS=OFS="," }

NR==1 {

    for(i=1;i<=NF;i++)               # gather all field numbers to c

        c[i]

    next }

{

    for(i in c)                      # test all fields that still are all zeros

        if($i!=0)

            delete c[i] }

END {                                # after testing all the records

    for(i=1;i<=NF;i++)

        if(!(i in c))

            p=p (p==""?"":OFS) "$" i # make list of list of fields to print

    p="print " p                     # p="print $1,$3,$4,$5,$6"

    system("awk 47BEGIN{FS=OFS=","}{" cmd "}47 " FILENAME)

}' file

Output:

header,d1,d2,d3, ...

s1,5,2,8, ...

s2,8,2,4, ...

s3,7,3,4, ...

s4,3,2,1, ...

If all fields are all zeros, p="print" and the whole file gets printed.

Provided that the first column does not contain all zeros, this awk script should do the job

awk -F',' '(NR==FNR && NR >1){for(i = 1; i <= NF; i++)

                              {a[i] = a[i]+$i}}

           (FNR!=NR){out=$1

                     for(i = 2; i<= NF; i++){

                      if(a[i]!=0){out=out","$i}

                     }

                    print out

                    }' file_name file_name

Note that the sript takes the name of the input file file_name twice!

For example, for the input:

header,d0,d

s1,0,5,2,8,

s2,0,8,2,4,

s3,0,7,3,4,

s4,0,3,2,1,

the script yields as output

header,d

s1,5,2,8

s2,8,2,4

s3,7,3,4

s4,3,2,1

maybe you can use sed command like below:

$ sed 's/b0,b//g' test.txt

header,d0,d1,d2,d3

s1,5,2,8

s2,8,2,4

s3,7,3,4

s4,3,2,1

Using Perl

> cat sumin.txt

header,d0,d1,d2,d3

s1,0,5,2,8

s2,0,8,2,4

s3,0,7,3,4

s4,0,3,2,1

> cat rem_zero.sh

perl -F, -lane '

@FH=@F if $.==1;



if($.>1)

{

$F[$_] and $nz[$_]||=1 for 0..$#F;

push(@L,[@F]);

}



END {

@cols = grep $nz[$_], 0..$#nz;

print join(",",@FH[@cols]);

for my $line (@L) { print "@{$line}[@cols]" }

}



' $1

> rem_zero.sh sumin.txt

header,d1,d2,d3

s1 5 2 8

s2 8 2 4

s3 7 3 4

s4 3 2 1

>