在shell脚本中以一个间隔查找持续时间及其在数据集之间的平均值 [英] Find the durations and their average between the dataset in an interval in shell script
问题描述
这与我的老问题有关我的数据集为:
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
我想找出6个值间隔中的0个值之间的不同持续时间及其平均值.
I would like to find out different duration and their average between the 0 values in 6 values interval.
我的愿望输出是:
ofile.txt
6 5.33
1 2
1 2
1 2
5 4.2
1 3
3 4
哪里
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
根据上一个问题的答案，我正在尝试以下方法:
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'
推荐答案
再试一次，因为第二组6行包含2 0 2 0 0 2，因此其输出应如果是这种情况，则为1 2, 1 2, 0 0,1 2(我认为应该如此)，然后尝试执行以下操作.
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'
输出如下:
6 5.33
1 2.00
1 2.00
1 2.00
5 4.20
1 3.00
3 4.00
下面的编辑可能会帮助解决与实际问题稍有不同的类似问题，因此请将其保留在后期.
EDITs below may help in similar kind of problems which are bit different from this actual problem, so keeping them here in post.
:如果不想在Input_file中出现零时进行RESET计数，请尝试执行以下操作.这样只会连续查找6行，并且不会重置其计数.
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
:能否请您仅使用提供的示例尝试进行以下测试，测试和编写.
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
代码处理的是什么:
What does code take care of:
- 要注意逻辑，如果连续的两行都具有
0值，它将打印该行的先前计数和平均值. -
这还将处理诸如以下的边缘情况:
- 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-如果某行不是以0结尾，它将通过我创建的found标志检查是否有一些值要打印.
a- In case a line is either NOT ending with a 0 it will check if some values are there to print by found flag I created.
b-如果任何Input_file的最后一行未除以6，则这种情况也将被END块通过found标志对其进行检查的逻辑所覆盖.
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 found flag.
说明: :为上述代码添加了详细说明.
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.
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