grep pattern on multiple columns but one

I have a tab delimiter file with ~10,000 columns.
I want to filter rows where some columns contain pattern. But I don't want to have the grep checking first column.

$ cut -f2,1691-1725 myfile.txt
1402138 2331    2331
1402422 181     630
1402795 1401            1401
1405425 2331
1405771 1727    1727
1406169 2331    2331
1406475 2252    2252
1408259 1744    1744

I tried this :

cut -f2,1691-1725 myfile.txt | grep -E '^140|^141|^143|^144|^145|^146|^148|^149' > myoutput.txt

but it keeps all because grep also applies on first column

Desired output :

1402795 1401            1401

I search a way with awk but can't find easy one without listing each columns

Thank you

From what I understand from your question, and from what I can see from your provided input and output, you have the following requirement:

Question: Provide a command that, given a set of column-numbers, returns all rows where each column is empty and/or matches a given regular expression

When talking about columns and regular expressions, the best answer is clearly awk:

awk 'BEGIN{a[2]; for(i=1691;i<=1725;++i) a[i]}
     BEGIN{ere="^14[01345689]"}
     BEGIN{FS=OFS="\t"}
     {f=1;for(i=1;i<=NF;++i) if(i in a) if(!($i ~ ere)) f=($i=="")}
     f' file

How does it work:

1. BEGIN{a[2]; for(i=1691;i<=1725;++i) a[i]}: Define an associative array a which is indexed by the columns of interest. In this case, those are 2, 1691-1725
2. BEGIN{ere="^14[01345689]"}: Define the regular expression that each column should match
3. BEGIN{FS=OFS="\t"}: Define the input and output field separator to be the <tab>-character
4. f=1;for(i=1;i<=NF;++i) if(i in a) if(!($i ~ ere)) f=($i==""):
   Each time a line is read, perform the following actions:
   1. f=1: Set the variable f equal to 1. This will be used as a flag to determine whether a line should be printed
   2. for(i=1;i<=NF;++i) ...: loop over all fields
   3. ... if(i in a)...: and check if the field is a field of interest
   4. ... if(!($i ~ ere)) f=($i==""): if the field is a field of interest and if the field does not match the regular expression, then set the flag f to ZERO or not ZERO depending if the field is empty or not.
5. f: If the flag f is not zero, print the current line. (This is a shorthand for (f!=0){print $0}

If your data always starts at the exact same character column, you could use ^.{n} to skip n characters, then start matching.

$ cat myfile.txt
1402138 2331    2331
1402422 181     630
1402795 1401            1401
1405425 2331
1405771 1727    1727
1406169 2331    2331
1406475 2252    2252
1408259 1744    1744

$ cat myfile.txt | grep -E '^.{24}(140|141|143|144|145|146|148|149)'
1402795 1401            1401

The pattern above first matches any 24 characters (except newline) since the start of the line, then tries to match (140|141|143|144|145|146|148|149) starting from character position 25.

In this scenario (140|141|143|144|145|146|148|149) could be simplified to 14[01345689] as well.

For filtering on the values of a certain column, you might better use awk (grep is better suited for filtering the entire line), something like this:

prompt> cat file.txt
Name Value1 Value2
a 1 2
b 2 5
c 5 1

prompt> awk -F " " '{if ($2==1) print $0}' file.txt
a 1 2