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qq56521 发表于 2015-2-11 10:48 
可以讲一下你是怎么做到的吗? - pddata <- read.table("D:\\pdd.csv", header=TRUE, sep=",")
- p1 <- data.frame(pddataRO,pddataRO, pddataTOOL, pddataCT,pddataC_T, pddataI_T,
- pddataFTIME,pddataF_TIME, pddataFP_Num)
- p2 <- na.omit(p1)
- p3 <- p2[which(p2pddata.RO == "Reg" & p2pddata.RO == "Reg" & p2pddata.FP_Num >= 12.8),]
- p3 <- transform(p3, fdate = substr(p3$pddata.F_TIME, 1, 10))
- vep <- levels(p3$F_DATE)
- ve <- sort(vep, decreasing = TRUE)
- agf <- as.list(rep(1:3))
- aggnf <- as.list(rep(1:3))
- aggnaf <- as.list(rep(1:3))
- ao <- as.list(rep(1:3))
- aon <- as.list(rep(1:3))
- aod <- as.list(rep(1:3))
- tools <- c("atool","btool","ctool")
- for (i in 1:3)
- {
- agf[[i]] <- aggregate(p3[6], by=list(p3[,7],p3[,i+1]),mean)
- aggnf[[i]] <- aggregate(p3[6], by=list(p3[,7],p3[,i+1]),length)
- aggnaf[[i]] <- cbind(agf[[i]],aggnf[[i]][3])
- names(aggnaf[[i]])[1:4] <- c("ftime","atool")
- names(aggnaf[[i]])[2] <- tools[i]
- ao[[i]] <- aggnaf[[i]][order(aggnaf[[i]][1],decreasing = TRUE),]
- ao[[i]][[2]] <- factor(ao[[i]][[2]],levels=c("01-atool-001","01-atool-002","01-atool-003","01-atool-004","01-atool-005","01-atool-006","01-atool-007","01-btool-001","01-btool-002","01-btool-003","01-btool-004","01-btool-005","01-btool-006"))
- aon[[i]] <- na.omit(ao[[i]])
- aod[[i]] <- aon[[i]][which(aon[[i]][1] == ve[1] | aon[[i]][1] == ve[2] | aon[[i]][1] == ve[3] | aon[[i]][1] == ve[4]),]
- print(aod[[i]])
- }
复制代码你参考一下吧,这个代码比较早了,还是可以优化的
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