Shell Programming and Scripting

try this

Code:

awk ' 
$1 ~ "^[0-9][0-9]*$" && $NF == 0   { grp1=grp1+$(NF-1); tot1=tot1+1 }
$1 ~ "^[0-9][0-9]*c$" && $NF == 0 { grp2=grp2+$(NF-1); tot2=tot2+1 }
END {
     printf("Group1\nAverage = %f \n",grp1/tot1)
     printf("Group2c\nAverage = %f\n",grp2/tot2)
}' file

Code:

average_c = [] #to catch "c" groups
average  = []  #to catch non c groups
data = open("test.txt").readlines()
for i in range(len(data)):
        if "x= 1" in data[i]:
		nextline_splitted = data[i+1].split()
		if nextline_splitted[0].endswith("c"):
			 average_c.append( int( nextline_splitted[-2] ) )
		else:
			average.append( nextline_splitted[-2] ) 

print "Group 1: " , ''.join(average)
print "Group 2: " , sum(average_c)/2

Output:

Code:

D:\yhlee\test>python test.py
Group 1:  2826
Group 2:  3804

I'll try to explain all awk and sed code in this thread: if($0~"x= 1") $0 represents whole input line feeded to awk, here ~ is match operator, so it becomes, if x= 1 found anywhere in the input line then, change the value of flag variable to 1 and print the whole line, else{ if(flag==1) else if awk is unable to find x= 1 in the input line then check for the value of flag variable, if its equal to 1 that means previous line contained x= 1 somewhere in it, since we have to print 2 consecutive lines of input in case of a match in the first line, so print 2nd consecutive line also and a "\n" and change flag's value to 0 to repeat the above process. $0 ~ /^.*x= 1.*/ Match the input line for pattern x= 1, if match found, { print; getline; print $0"\n" } print command without any arguments prints the whole input line, so it'll print the line where pattern x= 1 is found, then getline command takes very next line into buffer and print $0"\n" will print that line and a "\n" charachter. sed -n '/^.*x= 1.*/-n flag will supress the sed's default printing, if regex /^.*x= 1.*/ found then N command will append the next line to the pattern space, G command will append a "\n" to the pattern space and p will print all the lines in pattern space and empty it. $1 ~ "^[0-9][0-9]*$" Check if first field of input line contains numerics[0-9] only, NF system variable tells us total no. of fields available in the input line, so $NF means value of last field, so whole expression($1 ~ "^[0-9][0-9]*$" && $NF == 0) states that check if first field contains numerics only and last field is equal to 0 then { grp1=grp1+$(NF-1); tot1=tot1+1 } $(NF-1) means value of 2nd last field, so it'll add value of 2nd last field to variable grp1, and group1's count (tot1) will be incremented, $1 ~ "^[0-9][0-9]*c$" it tells awk to match first field for the pattern containing numerics and ending in a c like 0001c in your case, rest all is same, and End rule is printing the average of both groups by dividing total of a group with its count.

Regards,
Tayyab

Hi Tayyab,

You are such a professional at this!! Thanks alot for your very comprehensive explanation! Just another simple question for you. I notice tht sometimes you have this "^" character for eg in ($1 ~ "^[0-9][0-9]*$" && $NF == 0). What does this^trying to mean in the statement?

^ means start of the string if it is specified at the start of the regular expression

$ means end of the string if it is specified at the end of the regular expression

Hi, I have some problems executing this code.

Error msg is
$ testing1
Unmatched '

My code is below exactly the same as above and is working in a solaris environment. PLs help.

#!/bin/csh

nawk '
$1 ~ "^[0-9][0-9]*$" && $NF == 0 { grp1=grp1+$(NF-1); tot1=tot1+1 }
$1 ~ "^[0-9][0-9]*c$" && $NF == 0 { grp2=grp2+$(NF-1); tot2=tot2+1 }
END {
printf("Group1\nAverage = %f \n",grp1/tot1)
printf("Group2c\nAverage = %f\n",grp2/tot2)
}' xxx.txt

awk '
$1 ~ "^[0-9][0-9]*$" && $NF == 0 { grp1=grp1+$(NF-1); tot1=tot1+1 }
$1 ~ "^[0-9][0-9]*c$" && $NF == 0 { grp2=grp2+$(NF-1); tot2=tot2+1 }
END {
printf("Group1\nAverage = %f \n",grp1/tot1)
printf("Group2c\nAverage = %f\n",grp2/tot2)
}' file

I think you might have missed above '.