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awk or other way to find out number of occurrence of 7th character

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# 8  
Old 02-03-2017
Hi All, Thanks for the responses, Don , Rudi C, all thanks,

Rudi C, yes that is the only Snippet .., and I went ahead and got this.. ,

{if(substr($1,7,1)=="p"){p++}}END{print "p_hosts = "p}
{if(substr($1,7,1)=="d"){d++}}END{print "d_hosts = "d}
{if(substr($1,7,1)=="o"){o++}}END{print "o_hosts = "o}
{if(substr($1,7,1)=="m"){m++}}END{print "m_hosts = "m}
{if((substr($1,7,1)!~"p") && (substr($1,7,1)!~"d") &&  (substr($1,7,1)!~"o") &&  (substr($1,7,1)!~"m")  ){n++}}END{print "Non_p_d_o_m_hosts = "n+0}

$ awk -f count.awk datafile 
p_hosts = 17
d_hosts = 1
o_hosts = 1
m_hosts = 8
Non_p_d_o_m_hosts = 0

generated the counts now of the pattern..., Thanks all,
# 9  
Old 02-03-2017
Well, after rveri posted his attempt, all hidden / unapproved posts were unhidden / approved.
# 10  
Old 02-03-2017
awk '
{$0 ~ /^......[dpom]/ ? a[substr($0, 7, 1)]++ : x++}
END {for (i in a) print i "_hosts = " a[i]; print "Non_p_d_o_m_hosts = " x++} ' infile

# 11  
Old 02-04-2017
Quicky on the command line:
cut -c7 file | sort | uniq -c

   1 d
   8 m
   1 o
  17 p

This User Gave Thanks to Scrutinizer For This Post:

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