bash - How can I get common rows that exist in at lest two files or more? -


i have 7 test files. following

file 1

chr     start   end     strand chr1    10525   10525   + chr1    10542   10542   + chr1    10571   10571   + chr1    10577   10577   + chr2    10589   10589   + chr2    565262  565262  + chr2    565397  565397  + chr3    567239  567239  + chr3    567312  567312  + chr4    567348  567348  +

how can rows common in @ least 2 files in following format

chr     start   end     strand  file1   file2   file3   file4   file5   file6   file7 chr1    10525   10525   +   0   1   0   0   0   1   1 chr1    10542   10542   +   1   1   1   1   1   0   0 chr1    10571   10571   +   0   1   0   1   1   0   0 chr3    10577   10577   +   1   1   0   0   0   1   0 chr3    10589   10589   +   0   0   1   0   1   0   1 chr4    565262  565262  +   1   0   0   1   1   1   1

"1" row exist in given file , "0" rows on exist in given file. not want show rows not common in files.

using awk:

awk '     fnr==1{ #header line:         fn[++i]=filename; # record filenames          fn[0]=$0; # & file header     }      (fnr>1){ # lines other header lines         list[$0]++; # record line         file_list[$0 filename]++; # record file has line     }      end{         for(t=0;t<=i;t++) printf "%s\t", fn[t]; # print header & file names         print ""; # quick hack printing newline.         for(t in list){ # every line occurred in of files             if (list[t]>=2){ # if count >= 2                 printf "%s\t", t; # print line                 for(j=1;j<=i;j++) {                     printf "%d\t", file_list[t fn[j]]; # print per file occurrence count.                 }                 print "" # print newline.             }         }     }' file{1..7}

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