Grep for a line containing only 5 or 6 numbers

Question

How would you grep for a line containing only 5 or 6 numbers? Something like this.

case 1 (has leading space)

           10      2       12      1       13

case 2 (no leading space)

   1       2       3       4       5        6

I thought something like this would work.

grep -E '[0-9]{5}'

Answer 1

grep -E '[0-9]{5}' is looking for numbers with at least 5 digits. What you need is 5 numbers with at least one digit:

grep -E '[0-9]+([^0-9]+[0-9]+){4}'

• [0-9]+ - a number of at least one digit
• [^0-9]+[0-9]+ - a number with at least one digit, preceded by at least one non-digit character. We then repeat this 4 times to get 5 numbers separated by non-digits.
• If the requirement is exactly 5, you might want to surround this regex with [^0-9] so that the entire line is matched (with the anchors, of course).
• Depending on what you want here (does 1,2,3,4,6 qualify?), you might look at other separators. For example, a proper scientific notation real number would look like: [+-]?(([0-9]+(\.[0-9]+)?)|([0-9]?\.[0-9]+))([eE][+-][0-9]+)? So separators may not include ., e, etc. They may only be whitespace, as mikeserv notes. Or they maybe commas, if it's a CSV record. Or depending on the locale, a comma would be the decimal separator. Vary [^0-9] as per your need.

Answer 2

I would go with something a little more powerful than grep. This can do it in perl:

perl -ne 'print if s/\d+/$&/g == 5' your_file

The regex substitution replaces all groups of one or more digits (\d+) with themselves ($&): it does nothing. It is used merely for side effect since the s/// operator returns the number of times it managed to substitute for its regex. Thus, the line is printed only if s/// found 5 groups of digits.

Answer 3

Another perl:

$ perl -MList::Util=first -Tnle '
  s/^\s+|\s+$//g;
  @e = split /\s+/;
  print if @e == 5 || @e == 6 and !first {/\D/} @e;
' file
10      2       12      1       13

Explanation

• s/^\s+|\s+$//g trim the line.

• @e = split /\s+/ split the line into array @e.

• We will print the line if:

  • array @e contains 5 or 6 elements.
  • And None of its elements contain non-digit characters (\D match non-digit characters).

Answer 4

grep -E '^(\s*[0-9]+\s+){4,5}[0-9]+\s*$'

Answer 5

awk '{l=$0; n = gsub(/[0-9]+/, "", l)}; n == 5 || n == 6' 

(same principle as in Joseph's answer)

Answer 6

A way with awk that is customisable for different numbers of fields.
Also whitespace does not matter.

awk 'NF~/^(5|6)$/{x=0;for(i=1;i<=NF;i++)x+=($i~/^[0-9]+$/)}x==NF' file

This checks the number of fields is 5 or 6 although more numbers of fields could be added if your requirements ever change.

Then it sets a counter to 0

Then loops checking each field is a number and if it is adds 1 to the counter

If the counter equals the number of fields it prints the line.

example

input

  1       2       3       4       5        6
        2       3       4       5        6
3       4       5        6
blah  1       2       3       4       5
      4       3324       4       53        6

output

  1       2       3       4       5        6
        2       3       4       5        6
      4       3324       4       53        6

Answer 7

I know I have not solved using sed or awk or any of shell commands, However tcl did work out well.

I have kept the script simple and user friendly. elements like 4a abc etc will be taken care

command

script.tcl file

Script

#!/usr/bin/tclsh

if {$argv == ""} {
        puts "please enter the arguement"
        exit ;
}

set tar_fl [lindex $argv 0]
if {![file exists $tar_fl]} {
        puts "$tar_fl doesnot exist"
        exit ;
}

set flptr [open $tar_fl r]

while {[gets $flptr line] >=0 } {
        if {[llength $line] !=5} {continue ;}
        if {[llength $line ] == 5} {
                if {[lsearch -regexp  $line {[^0-9]}]> -1} {continue;}
                puts $line
        }

}

close $flptr

Output

           10      2       12      1       13
1       2       3       4       5