Count columns that are non-empty per line

Applications Programming
1

The file is similar to the attached.
Thousands of columns, delimiter is tab, with many columns containing free text and space separated.
I want to get the count of columns with non-empty entries.

eg.Col1=10, Col6=5, Col8=1
awk preferred

2

Any attempts / ideas / thoughts from your side?

3 
awk -F "\t" '{ for(N=1; N<=NF; N++) if($N==""); {print NF}}'  ex.txt

This gives me NF=14, but I want the each col. in the file as header and another line with the number of non-empty lines for each.
So, not sure I am approaching this right.

4

How about

awk -F "\t" '
       {for (N=1; N<=NF; N++) if ($N!="") CNT[N]++}
END    {for (N=1; N<=NF; N++) print "col" N ":", CNT[N]
       }
' /tmp/ex.txt 
col1: 10
col2: 10
col3: 10
col4: 2
col5: 2
col6: 6
col7: 5
col8: 2
col9: 4
col10: 1
col11: 3
col12: 3
col13: 1
col14: 10
1 Like
5

Yes, that worked perfectly. Thank you!
So, let me understand your approach.
First, you are looking for Non-empty cells, and then making a count of such cells?

($N!="") CNT[N]++

Then you are printing

 print "col" N ":", CNT[N]

the col number and the count?

A request: Assuming the file has a header, how to modify such that instead of Col number, the ColHeader is printed?
many Thanks

6

It always helps to post samples, so people don't have to dream up some but can start working on a solution.

Would that be a one line header? No empty fields in it?

And yes, I count up a counter (array) per field, and, in the END section, print those.

7

The sample you posted earlier seems to have a header line. And DOS line terminators (<CR> = \015 = 0x0D = \r = ^M ). Try

awk -F "\t" '
        {sub ("\r","")
        }
NR == 1 {N = split ($0, HD); next
        }
        {for (N=1; N<=NF; N++) if ($N!="") CNT[N]++}
END     {for (N=1; N<=NF; N++) print HD[N] ":", CNT[N]
        }
' /tmp/ex.txt 
ID1: 9
ID2: 9
ID3: 9
Status1: 1
Status2: 1
Status3: 5
Status4: 4
Status5: 1
Status6: 3
..: 
..: 2
..: 2
..: 
StatusX: 2
1 Like
8

Thank you!