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cat > temp000180255798957892187719
awk ‘{x[NR]=$0; s+=$0; n++} END{a=s/n; for (i in x){ss += (x[i]-a)^2} sd = sqrt(ss/n); print "SD = "sd}‘ temp0001SD = 30.3857
Shell awk 求标准差
原文地址：http://www.cnblogs.com/emanlee/p/4556324.html
