[英]how to calculate value differences between two neighboring fields
I am trying to calculate distances between two neighboring fields. 我正在尝试计算两个相邻字段之间的距离。 My input file is like below.
我的输入文件如下所示。
1 11160 11533 11556 11731 11822 11870 12149 12411 12461 12686 12829 13315 13420 ....
1 11160 11533 11556 11731 11822 11870 12149 12411 12461 12686 12829 13315 13420 ....
In the output, I want to keep the first field, and the following field would be the value differences between the current field and the next field, $2=$3-$2
, $3=$4-$3
... 在输出中,我想保留第一个字段,以下字段将是当前字段和下一个字段之间的值差异,
$2=$3-$2
, $3=$4-$3
...
A complete output will be like: 完整的输出将是:
1 373 23 175 91 48 279 262 50 225 143 486 105...
1 373 23 175 91 48 279 262 50 225 143 486 105 ......
How can I do this? 我怎样才能做到这一点?
In my code, each value is printed as a new line also the numbers are reversely printed. 在我的代码中,每个值都打印为一个新行,同时反向打印数字。
BEGIN {FS=" "}
{
out[1]=$1
for (i=2;i<=NF-1;i++)
out[i]=$(i+1)-$i
}
END{
for (i in out)
print out[i]
}
Here is current output 这是当前的输出
373 23 175 91 48 279 262 50 225 143 486 105 1
373 23 175 91 48 279 262 50 225 143 486 105 1
EDIT: Adding code suggested by anubhava sir too in comment section. 编辑:在评论部分添加anubhava先生建议的代码。
awk '{s=$1; for (i=2; i<NF; i++) s = s OFS $(i+1) - $i; print s}' Input_file
Could you please try following. 你可以尝试一下吗?
awk '{printf $1 OFS;for(i=2;i<NF;i++){printf("%d%s",$(i+1)-$i,i==(NF-1)?ORS:OFS)}}' Input_file
Output will be as follows. 输出如下。
1 373 23 175 91 48 279 262 50 225 143 486 105
Explanation: Adding explanation too here. 说明:此处也添加说明。
awk '
{
printf $1 OFS ##Printing first field and OFS(whose value is space by default).
for(i=2;i<NF;i++){ ##Starting for loop from value of 2 to till NF-1 value where NF is number of field in current line.
printf("%d%s",$(i+1)-$i,i==(NF-1)?ORS:OFS) ##Printing diffrence of next field and current field and checking condition for 2nd print if i==NF-1 then new line else print space for that line.
} ##Closing for loop block here.
}
' Input_file ##Mentioning Input_file name here.
awk '{printf("%d",$1); for (i=2;i<NF;i++) printf(" %d",$(i+1)-$i)}' file
Output: 输出:
1 373 23 175 91 48 279 262 50 225 143 486 105
another awk
另一个
awk
$ awk -v RS=' ' 'NR!=2{printf "%s ", $0-p} {p=$0}' file
1 373 23 175 91 48 279 262 50 225 143 486 105
you may want to add a final \\n
. 你可能想要添加一个
\\n
。
Your specification corresponds to displaying all pairwise diffs expect second one that's why there is NR!=2
code. 您的规范对应于显示所有成对差异,期望第二个,这就是为什么有
NR!=2
代码。 First field is compared to 0, so stays the same. 第一个字段与0进行比较,因此保持不变。
You almost have it! 你几乎拥有它! Just need to change the output from
print out[i]
to: 只需要将
print out[i]
更改为:
printf out[i] " "
The printf
will not append a newline after each iteration, and instead you tack on a space. printf
在每次迭代后都不会附加换行符,而是在空格上添加。
The output I recieved from the above adjustment turned into: 我从上述调整中收到的输出变为:
1 373 23 175 91 48 279 262 50 225 143 486 105
As far as the "numbers not making sense", they are as you are expecting, a difference between the consecutive numbers. 至于“没有意义的数字”,它们就像你期望的那样,是连续数字之间的差异。
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