Question

我的data.frame＆＃34;分析＆＃34;是7个变量的180,010个obs。它的结构的缩写示例如下：

ID <- c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2)
Rating <- c("Poor", "Excellent", "Very Good", "Poor", "Good", "Fair", 
            "Very Good", "Fair", "Poor", "Excellent")
Speed <- c(10, 19, 20, 21, 22, 20, 20, 21, 23, 15)

我希望循环使用＆＃34;分析$ Speed＆＃34;并找到所有等于或高于19且等于或低于25的事件。应该有至少4个或更多符合此标准的连续值 - 如果有3个，则忽略这些值。我希望创建一个新的data.frame＆＃34;输出＆＃34;包含价值及其各自的＆＃34; ID＆＃34;，＆＃34;评级＆＃34;和＆＃34;速度＆＃34;但我不确定该怎么做。

例如，从上面开始：

ID <- c(1, 1, 1, 1, 2, 2, 2, 2)
Rating <- c("Excellent", "Very Good", "Poor", "Good", "Fair", "Very 
            Good", "Fair", "Poor")
Speed <- c(19, 20, 21, 22, 20, 20, 21, 23)

我在编写循环方面的经验非常有限（无），大多数问题都是定量数据或搜索字符串，而我的是混合。

Answer 1

这对我有用：

ID <- c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2)
Rating <- c("Poor", "Excellent", "Very Good", "Poor", "Good", "Fair", 
            "Very Good", "Fair", "Poor", "Excellent")
Speed <- c(10, 19, 20, 21, 22, 20, 20, 21, 23, 15)

ID2 <- c()
Rating2 <- c()
Speed2 <- c()

for (i in 1:length(Speed)) {
  if (Speed[i] >= 25 | Speed[i] <= 19){
    ID2 <- c(ID2,ID[i])
    Rating2 <- c(Rating2,Rating[i])
    Speed2 <- c(Speed2, Speed[i])
  }
}

Output <- data.frame(ID = ID2, Rating = Rating2, Speed = Speed2)

Answer 2

假设'ID'是分组变量而“Speed＆lt; = 19”将改为“Speed＆lt; 19”，我们可以使用ave和rle来获取逻辑索引<连续元素> 3使用条件（“速度＆lt; 19 |速度＆gt; = 25”），并使用该索引对原始数据集进行子集化。

f1 <- function(dat,Var1 , Var2){
   indx <- as.logical(with(dat, ave(Var2, Var1, FUN=function(x) {
        inverse.rle(within.list(rle(x < 19 | x>=25),
              values <- lengths[values] >3 |!values))})))
   dat[indx,]
}

f1(Analysis, ID, Speed)
#   ID    Rating Speed
#2  1 Excellent    19
#3  1 Very Good    20
#4  1      Poor    21
#5  1      Good    22
#6  2      Fair    20
#7  2 Very Good    20
#8  2      Fair    21
#9  2      Poor    23

使用另一个例子（具有超过3个符合条件的连续元素）

f1(AnalysisN, ID, Speed)
#    ID    Rating Speed
#2   1 Excellent    20
#3   1 Excellent    15
#4   1 Excellent    27
#5   1 Excellent    19
#6   2      Poor    22
#7   2      Fair    14
#8   2      Poor    20
#9   2      Fair    22
#12  3 Excellent    11
#13  3 Very Good    18
#14  3      Fair    10
#15  3      Poor    15
#16  4      Fair    19
#17  4 Excellent    23
#18  4      Fair    26
#19  4 Very Good    20
#22  5 Very Good    26
#23  5      Poor    15
#24  5 Excellent    29
#25  5 Excellent    13

数据

Analysis <- structure(list(ID = c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2), 
Rating = c("Poor", 
"Excellent", "Very Good", "Poor", "Good", "Fair", "Very Good", 
"Fair", "Poor", "Excellent"), Speed = c(10, 19, 20, 21, 22, 20, 
20, 21, 23, 15)), .Names = c("ID", "Rating", "Speed"), 
row.names = c(NA, -10L), class = "data.frame")

 set.seed(30)
 AnalysisN <- data.frame(ID= rep(1:5, each=5), 
 Rating= sample(c('Poor', 'Excellent', 'Very Good', 'Fair'), 25, 
 replace=TRUE), Speed =sample(10:30, 25, replace=TRUE), 
    stringsAsFactors=FALSE)

迭代列以满足条件

2 个答案:

数据