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awk to grab the most found lines

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# 1  
Old 02-24-2013
awk to grab the most found lines
Code:
awk 'NR>=5034&&/Max/{++c}c==3{o=$0 RS $0 RS $0; print o; c=0}' logfile

i would like to modify the above awk command to instead show lines that are the most frequent:

for instance, given a log file containing:

Code:
Warning MaxClient is having issues - please
MaxClients java error...yams potato turkey
MaxClients error Fatal exception known known
MaxClients error Fatal exception known known
Could not complete. Error found. MaxClient initiated
MaxClients error Fatal exception known known
Could not complete. Error found. MaxClient initiated

I want the awk statement to show only the lines that occur the most. The content of the log should not matter. if I egrep for everything in the log, how do i modify the awk statement to show the lines that are the most frequent?

Code:
awk 'NR>=5034&&/./{++c}c==3{o=$0 RS $0 RS $0; print o; c=0}' logfile

# 2  
Old 02-24-2013
Try:
Code:
sort logfile | uniq -c | sort -nrk1 | head

This User Gave Thanks to bartus11 For This Post:
SkySmart
# 3  
Old 02-24-2013
Code:
awk 'NR>=5034{a[$0]++}END{for(i in a) { if(a[i]>max) { max=a[i]; v=i } } print v }' logfile

This User Gave Thanks to Yoda For This Post:
SkySmart
# 4  
Old 02-24-2013
Quote:
Originally Posted by bipinajith
Code:
awk 'NR>=5034{a[$0]++}END{for(i in a) { if(a[i]>max) { max=a[i]; v=i } } print v }' logfile

can you please explain this statement?
# 5  
Old 02-24-2013
Code:
awk ' NR >= 5034 {
        a[$0]++                         # Store current record in array and count occurrances
}
END {                                   # END block
        for (i in a) {
                if (a[i] > max) {       # If a[i] ( number of occurrances ) > max
                        max = a[i]      # Set max = a[i] ( number of occurances )
                        v = i           # Set v = i ( array element )
                }
        }
        print v                         # Print v
}' file

This User Gave Thanks to Yoda For This Post:
SkySmart
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