Question

我试图了解如何汇总我的输出。我创建了一些虚拟数据，它们近似于我的实际数据，即：数百个group1,3个level2组，以及几十个验证逻辑。抱歉，如果这看起来很简单，我已经狩猎和啄了很多，并且不得不说作为R的新手，各种各样的工具（应用系列，ddply，聚合，表，重塑等）在那里既美妙又有点可怕:)）

 #create data
group1 <- paste("Group", rep(LETTERS[1:7], sep=''))
group2 <- c("UNC", "UNC", "SS", "LS", "LS", "SS", "UNC")
valid1 <- c("Y", "N", NA, "N", "Y", "Y", "N")
valid2 <- c("N", "N", "Y", "N", "N", "Y", "N")
valid3 <- c(1.4, 1.2, NA, 0.7, 0.3, NA, 1.7)
valid4 <- c(0.4, 0.3, 0.53, 0.66, 0.3, 0.3, 0.71)
valid5 <- c(8.5, 11.2,NA, NA, 8.3, NA, 11.7)

testdata <- data.frame(cbind(group, group2, valid1, valid2, valid3, valid4, valid5))

valid <- function(testdata){
  for(i in group)
    val1 <- ifelse(valid1=="Y", 1,0)
     val2 <- ifelse(valid2=="Y", 1,0)
      val3 <- ifelse(valid3>=1.0, 1,0)
      val4 <- ifelse(valid4<=0.5, 1,0)
       val5 <- ifelse(valid5>=10.0, 1,0)

  test.out <- data.frame(cbind(group1,group2, val1, val2, val3, val4, val5))

}
validtry <- valid(testdata)'

然后，我需要将这些逻辑转换为数字，以便将它们相加：

#make validations numeric
# why doesn't this work:
# validtry[,3:7] <- as.numeric(validtry[,3:7])
#but these do
validtry[,3] <- as.numeric(validtry[,3])
validtry[,4] <- as.numeric(validtry[,4])
validtry[,5] <- as.numeric(validtry[,5])
validtry[,6] <- as.numeric(validtry[,6])
validtry[,7] <- as.numeric(validtry[,7])
######

#summarize validtry
#sum on both groups
aggregate(validtry[,3:7], by=list(validtry$group1, validtry$group2), sum, na.rm=T)

#sum on one group
aggregate(validtry[,3:7], by=list(validtry$group2), sum, na.rm=T)

所以，最后两个让我接近，但我想我需要一些不同的东西？我试图对两组的行和列进行求和。我熟悉tapply，但似乎并没有得到它。

提前感谢!!

Answer 1

目前尚不清楚预期产量。我的猜测是：

 testdata <- data.frame(group1, group2, valid1, valid2, valid3, valid4, valid5)
 str1 <- c("valid1=='Y'", "valid2=='Y'", "valid3>=1.0", "valid4 <=0.5", "valid5>=10.0")
 validtry <- testdata

 #Though I used eval(parse(...)), it is not that recommended 
 validtry[,-(1:2)] <- lapply(str1, function(x) 1*with(testdata, eval(parse(text=x))))

 library(reshape2) 
 lst <-  lapply(validtry[3:7], function(x)
       dcast(data.frame(validtry[1:2], x), group1~group2, value.var="x", sum, na.rm=TRUE))

 lst[[1]]
 #   group1 LS SS UNC
 #1 Group A  0  0   1
 #2 Group B  0  0   0
 #3 Group C  0  0   0
 #4 Group D  0  0   0
 #5 Group E  1  0   0
 #6 Group F  0  1   0
 #7 Group G  0  0   0

聚合超过2组

1 个答案: