问题
This is related to my older question Find the durations and their maximum between the dataset in an interval in shell script
I have a dataset as:
ifile.txt
2
3
2
3
2
20
2
0
2
0
0
2
1
2
5
6
7
0
3
0
3
4
5
I would like to find out different duration and their average between the 0 values in 6 values interval.
My desire output is:
ofile.txt
6 5.33
1 2
1 2
1 2
5 4.2
1 3
3 4
Where
6 is the number of counts until next 0 within 6 values (i.e. 2,3,2,3,2,20) and 5.33 is the average value among them;
1 is the number of counts until next 0 within next 6 values (i.e. 2,0,2,0,0,2) and 2 is the average;
Next 1 and 2 are within same 6 values;
5 is the number of counts until next 0 within next 6 values (i.e. 1,2,5,6,7,0) and 4.2 is the average among them;
And so on
As per the answer to my previous question, I was trying with this:
awk '
$0!=0{
count++
sum=sum+$0
found=""
}
$0==0{
print count,max
count=max=0
next
}
FNR%6==0{
print count,max
count=max=0
found=1
}
END{
if(!found){
print count,max
}
}
' Input_file | awk '!/^ /' | awk '$1 != 0'
回答1:
EDIT3: One more try since 2nd set of 6 lines have 2 0 2 0 0 2 so its output should be 1 2, 1 2, 0 0,1 2 if this is the case(which I believe ideally should be) then try following.
awk '
{
occur++
}
{
count=$0!=0?++count:count
sum+=$0
}
$0==0 || occur==6{
printf("%d %0.2f\n",count,count?sum/count:prev)
prev=count?sum/count:0
prev_count=count
count=sum=prev=prev_count=""
if(occur==6){
occur=""
}
}
END{
if(occur){
printf("%d %0.2f\n",count?count:prev_count,count?sum/count:prev)
}
}
' Input_file | awk '$1 != 0'
Output will be as follows:
6 5.33
1 2.00
1 2.00
1 2.00
5 4.20
1 3.00
3 4.00
EDITs below may help in similar kind of problems which are bit different from this actual problem, so keeping them here in post.
EDIT2: In case you don't want to RESET count whenever a zero occurs in Input_file then try following. This will continuously look for only 6 lines and will NOT RESET its count.
awk '
{
occur++
}
$0!=0{
count++
sum+=$0
found=prev_count=prev=""
}
$0==0 && occur!=6{
printf("%d,%0.2f\n",count?count:prev_count,count?sum/count:prev)
prev=count?sum/count:0
prev_count=count
count=sum=""
found=1
next
}
occur==6{
printf("%d,%0.2f\n",count,count?sum/count:prev)
prev=count?sum/count:0
prev_count=count
count=sum=occur=""
found=1
}
END{
if(!found){
printf("%d,%0.2f\n",count?count:prev_count,count?sum/count:prev)
}
}
' Input_file
EDIT1: Could you please try following, tested and written with provided samples only.
awk '
{
occur++
}
$0!=0{
count++
sum+=$0
found=prev_count=prev=""
}
$0==0{
printf("%d,%0.2f\n",count?count:prev_count,count?sum/count:prev)
prev=count?sum/count:0
prev_count=count
count=sum=occur=""
found=1
next
}
occur==6{
printf("%d,%0.2f\n",count,count?sum/count:prev)
prev=count?sum/count:0
prev_count=count
count=sum=occur=""
found=1
}
END{
if(!found){
printf("%d,%0.2f\n",count?count:prev_count,count?sum/count:prev)
}
}
' Input_file
What does code take care of:
- It takes care of logic where if any continuous 2 lines are having
0value then it will print previous count and average values for that line. This will also take care of edge cases like:
a- In case a line is either NOT ending with a
0it will check if some values are there to print byfoundflag I created.b- In case of any Input_file's last line is NOT divided by 6 then also this case will be covered by END block's logic of checking it by
foundflag.
Explanation: Adding a detailed explanation for above code.
awk ' ##Starting awk program from here.
{
occur++
}
$0!=0{ ##Checking condition if a line is NOT having zero value then do following.
count++ ##Increment variable count with 1 each time it comes here.
sum+=$0 ##Creating variable sum and keep adding current line value in it.
found=prev_count=prev="" ##Nullifying variables found, prev_count, prev here.
} ##Closing BLOCK for condition $0!=0 here.
$0==0{ ##Checking condition if a line is having value zero then do following.
printf("%d,%0.2f\n",count?count:prev_count,count?sum/count:prev) ##Printing count and count/sum here, making sure later is NOT getting divided by 0 too.
prev=count?sum/count:0 ##Creating variable prev which will be sum/count or zero in case count variable is NULL.
prev_count=count ##Creating variable prev_count whose value is count.
count=sum=occur="" ##Nullify variables count and sum here.
found=1 ##Setting value 1 to variable found here.
next ##next will skip all further statements from here.
} ##Closing BLOCK for condition $0==0 here.
occur==6{ ##Checking if current line is fully divided with 6 then do following.
printf("%d,%0.2f\n",count,count?sum/count:prev) ##Printing count and count/sum here, making sure later is NOT getting divided by 0 too.
prev=count?sum/count:0 ##Creating variable prev which will be sum/count or zero in case count variable is NULL.
prev_count=count ##Creating variable prev_count whose value is count.
count=sum=occur="" ##Nullifying variables count and sum here.
found=1 ##Setting value 1 to variable found here.
} ##Closing BLOCK for condition FNR%6==0 here.
END{ ##Starting END block for this awk program here.
if(!found){ ##Checking condition if variable found is NULL then do following.
printf("%d,%0.2f\n",count?count:prev_count,count?sum/count:prev) ##Printing count and count/sum here, making sure later is NOT getting divided by 0 too.
}
}
' Input_file ##Mentioning Input_file name here.
回答2:
Another which didn't turn out quite the one-liner I was going for:
$ tail -n +2 file | # strip header with tail to not disturb NR
awk '
{
s+=$0 # sum them up
c++ # keep count
}
$0==0 || NR%6==0 { # act if zero or every 6th record
if($0==0) # remove zero effect on c
c--
if(s>0) # avoid division by zero
print c,s/c # output
s=c=0 # rinse and repeat
}
END { # the end-game if NR%6!=0
if($0==0)
c--
if(s>0)
print c,s/c
}'
Output:
6 5.33333
1 2
1 2
1 2
5 4.2
1 3
3 4
The tail removes the header in the file before feeding the data to awk, the idea is that the header won't show in the NR. If you don't have a header, just awk ... file.
来源：https://stackoverflow.com/questions/59621874/find-the-durations-and-their-average-between-the-dataset-in-an-interval-in-shell