Grep a word that contains minimum 5 or 6 same characters

Shell Programming and Scripting
1

Hi,

I am looking for a solution to grep for minimum 5 or 6 characters in a file, otherwise ignore.

Example

1121221222
2212121211
1221122122
2121222222
2222112222
1211221121

So it greps 5 X 1 or 6 X 1

2212121211
1211221121

Thanks for you help

2

try awk:

awk 'gsub(c,c)>=5' c="1" infile
3

Thanks for the reply.

I think I should have asked different.
I meant minimum 5 or 6 but not more than 6.

awk solution is great

Thanks

---------- Post updated at 12:23 PM ---------- Previous update was at 12:21 PM ----------

Thank,

Figured it out myself

awk 'gsub(c,c)==5' c="1"

awk 'gsub(c,c)==6' c="1"

Thanks this forum is awesome :b::b::b::b::b::b::b:

4

in one pass:

awk 'gsub(c,c)==5 || gsub(c,c)==6' c="1" infile
5

And, slightly faster:

awk '(n=gsub(c,c))==5 || n==6' c="1" infile

And, if someone wants to try any of these awk suggestions on a Solaris/SunOS system, change awk to /usr/xpg4/bin/awk or nawk .

6

How about

awk -F1 'NF==6;NF==7' file
2212121211
1211221121
1 Like
7

Not necessarily clearer but this should work too:

awk 'gsub(c,c)~/^[56]$/' c=1 file

or

awk -F1 'NF~/^[67]$/' file