[英]Using awk, calculate average of rows based on string in 2nd and 5th column and value in 3rd column, and append result
This is a variant on Using awk, how to convert dates to week and quarter? 这是使用awk的变体,如何将日期转换为星期和季度?
Input data.txt: 输入data.txt:
a;2016-04-25;10;2016-w17;2016-q2
b;2016-04-25;20;2016-w17;2016-q2
c;2016-04-25;30;2016-w17;2016-q2
d;2016-04-26;40;2016-w17;2016-q2
e;2016-07-25;50;2016-w30;2016-q3
f;2016-07-25;60;2016-w30;2016-q3
g;2016-07-25;70;2016-w30;2016-q3
Wanted output.txt: 想要output.txt:
a;2016-04-25;10;2016-w17;2016-q2;50
b;2016-04-25;20;2016-w17;2016-q2;50
c;2016-04-25;30;2016-w17;2016-q2;50
d;2016-04-26;40;2016-w17;2016-q2;50
e;2016-07-25;50;2016-w30;2016-q3;180
f;2016-07-25;60;2016-w30;2016-q3;180
g;2016-07-25;70;2016-w30;2016-q3;180
Hence, calculate the quarterly average of the days which has data and append the result. 因此,请计算具有数据的天数的季度平均值并附加结果。
For 2016-q2 the average is calculated as follows: 对于2016年第二季度,平均值计算如下:
(10+20+30+40)/2 = 50 ("2" is the number_of_unique_dates for that quarter)
For 2016-q3 the average is: 对于2016年第三季度,平均值为:
(50+60+70)/1 = 180
Here is my work in progress which seem quite close to a final solution, but not sure how to get the "number of unique dates" (column 2) and use as divisor? 这是我正在进行的工作,似乎很接近最终解决方案,但是不确定如何获取“唯一日期数”(第2列)并将其用作除数吗?
awk '
BEGIN { FS=OFS=";" }
NR==FNR { s[$5]+=$3; next }
{ print $0,s[$5] / need_num_of_unique_dates_here }
' output.txt output.txt
Any idea how to get the "number of unique dates" per quarter? 知道如何获取每个季度的“唯一日期数”吗?
$ cat tst.awk
BEGIN { FS=OFS=";" }
$5 != p5 { prt(); p5=$5 }
{ lines[++numLines]=$0; dates[$2]; sum+=$3 }
END { prt() }
function prt( lineNr) {
for (lineNr=1; lineNr<=numLines; lineNr++) {
print lines[lineNr], sum/length(dates)
}
delete dates
numLines = sum = 0
}
$ awk -f tst.awk file
a;2016-04-25;10;2016-w17;2016-q2;50
b;2016-04-25;20;2016-w17;2016-q2;50
c;2016-04-25;30;2016-w17;2016-q2;50
d;2016-04-26;40;2016-w17;2016-q2;50
e;2016-07-25;50;2016-w30;2016-q3;125
f;2016-07-25;60;2016-w30;2016-q3;125
g;2016-07-25;70;2016-w30;2016-q3;125
h;2016-04-01;70;2016-w30;2016-q3;125
Another gawk solution: 另一个gawk解决方案:
awk -F';' '{ a[$5][$2]+=$3; r[NR]=$0; q[NR]=$5 }
END {
for (i in a) { s=0; len=length(a[i]);
for (j in a[i]) { s += a[i][j] }
a[i]["avg"] = s/len
}
for (n=1;n<=NR;n++) { print r[n],a[q[n]]["avg"] }
}' OFS=";" file
The output: 输出:
a;2016-04-25;10;2016-w17;2016-q2,50
b;2016-04-25;20;2016-w17;2016-q2,50
c;2016-04-25;30;2016-w17;2016-q2,50
d;2016-04-26;40;2016-w17;2016-q2,50
e;2016-07-25;50;2016-w30;2016-q3,180
f;2016-07-25;60;2016-w30;2016-q3,180
g;2016-07-25;70;2016-w30;2016-q3,180
a[$5][$2]+=$3
- multidimensional array, summing up values for each unique date within a certain quarter a[$5][$2]+=$3
多维数组,将某个季度内每个唯一日期的值相加
len=length(a[i])
- determining the number of unique dates within a certain quarter len=length(a[i])
-确定某个季度内唯一日期的数量
for(j in a[i]){ s+=a[i][j] }
- summing up values for all dates within a quater for(j in a[i]){ s+=a[i][j] }
-对四分之一内所有日期的值求和
a[i]["avg"]=s/len
- calculating average value a[i]["avg"]=s/len
计算平均值
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