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Grep a word that contains minimum 5 or 6 same characters

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# 1  
Old 03-09-2016
Grep a word that contains minimum 5 or 6 same characters
Hi,

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

Example

Code:
1121221222
2212121211
1221122122
2121222222
2222112222
1211221121

So it greps 5 X 1 or 6 X 1

Code:
2212121211
1211221121

Thanks for you help

Moderator's Comments:
Mod Comment Please use code tags for code and data
Last edited by Scrutinizer; 03-10-2016 at 03:42 AM.. Reason: typo; mod: code tags
# 2  
Old 03-09-2016
try awk:
Code:
awk 'gsub(c,c)>=5' c="1" infile

# 3  
Old 03-09-2016
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

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

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

Thanks this forum is awesome SmilieSmilieSmilieSmilieSmilieSmilieSmilie
# 4  
Old 03-09-2016
in one pass:
Code:
awk 'gsub(c,c)==5 || gsub(c,c)==6' c="1" infile

# 5  
Old 03-10-2016
And, slightly faster:
Code:
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  
Old 03-10-2016
How about
Code:
awk -F1 'NF==6;NF==7' file
2212121211
1211221121

These 2 Users Gave Thanks to RudiC For This Post:
Don Cragun
# 7  
Old 03-10-2016
Not necessarily clearer but this should work too:
Code:
awk 'gsub(c,c)~/^[56]$/' c=1 file

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

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