我有一个数据集与许多缺less值为-999。 部分数据是
input.txt 30 -999 10 40 23 44 -999 -999 31 -999 54 -999 -999 -999 -999 -999 -999 10 23 2 5 3 8 8 7 9 6 10 and so on 我想计算每个5,6,6行间隔的平均值，而不考虑缺失值。
欲望输出是
ofile.txt 25.75 (ie consider first 5 rows and take average without considering missing values, so (30+10+40+23)/4) 43 (ie consider next 6 rows and take average without considering missing values, so (44+31+54)/3) -999 (ie consider next 6 and take average without considering missing values. Since all are missing, so write as a missing value -999) 8.6 (ie consider next 5 rows and take average (10+23+2+5+3)/5) 8 (ie consider next 6 rows and take average)
如果它是有规律的间隔（让我们说5），我可以做到这一点
awk '!/\-999/{sum += $1; count++} NR%5==0{print count ? (sum/count) :-999;sum=count=0}' input.txt
我在这里定期询问类似的问题计算平均值，而不考虑shell脚本中的缺失值？ 但在这里，我要求解决不规则的时间间隔。
用AWK
awk -vf="5" 'f&&f--&&$0!=-999{c++;v+=$0} NR%17==0{f=5;r++} !f&&NR%17!=0{f=6;r++} r&&!c{print -999;r=0} r&&c{print v/c;r=v=c=0} END{if(c!=0)print v/c}' input.txt
产量
25.75 43 -999 8.6 8
分解
f&&f--&&$0!=-999{c++;v+=$0} #add valid values and increment count NR%17==0{f=5;r++} #reset to 5,6,6 pattern !f&&NR%17!=0{f=6;r++} #set 6 if pattern doesnt match r&&!c{print -999;r=0} #print -999 if no valid values r&&c{print v/c;r=v=c=0} #print avg END{ if(c!=0) #print remaining values avg print v/c }
$ cat tst.awk function nextInterval( intervals) { numIntervals = split("5 6 6",intervals) intervalsIdx = (intervalsIdx % numIntervals) + 1 return intervals[intervalsIdx] } BEGIN { interval = nextInterval() noVal = -999 } $0 != noVal { sum += $0 cnt++ } ++numRows == interval { print (cnt ? sum / cnt : noVal) interval = nextInterval() numRows = sum = cnt = 0 } $ awk -f tst.awk file 25.75 43 -999 8.6 8