使用R中的逻辑函数将循环替换为应用族函数(或dplyr) [英] replace loops with apply family functions (or dplyr), using logical functions in R

问题描述

我创建了这个代表性的数据框，该框使用for循环分配条件类别.

I have created this representative data frame that assigns condition categories using a for loop.

df <- data.frame(Date=c("08/29/2011", "08/29/2011", "08/30/2011", "08/30/2011", "08/30/2011", "08/29/2012", "08/29/2012", "01/15/2012", "08/29/2012"),
             Time=c("09:45", "10:00", "13:00", "13:30", "10:14", "9:09", "11:23", "17:06", "12:20"),
             Diff = c(0.2,4.3,6.5,15.0, 16.5, 31, 30.2, 21.9, 1.9))

df1<- df %>%
  mutate(Accuracy=ifelse(Diff<=3, "Excellent", "TBD"))

for(i in 1:nrow(df1)){
  if(df1$Diff[i]>3&&df1$Diff[i]<=10){
    df1$Accuracy[i]<-"Good"} 
  if(df1$Diff[i]>10&&df1$Diff[i]<=15){
    df1$Accuracy[i]<-"Fair"} 
  if(df1$Diff[i]>15&&df1$Diff[i]<=30){
    df1$Accuracy[i]<-"Poor"}
  if(df1$Diff[i]>30){
    df1$Accuracy[i]<-"Unacceptable"}
}

我的实际数据集非常大，并且读取表明for循环通常不是用R编写代码的最有效方法.我相信我可以通过为每个条件创建一个逻辑向量来做同样的事情，并且在每个向量内TRUE是满足每个条件.然后，我可以通过子集df1 $ Accuracy [Good]<-"Good"来分配值.但是，我无法弄清楚如何使用Apply系列函数或dplyr函数创建逻辑向量. (但是，也欢迎使用任何避免for循环的解决方案.)如果for循环是更好的方法，那么这也将有所帮助.

My actual dataset is very large and reading indicates for loops are usually not the most efficient way to code in R. I believe I can do the same thing by creating a logical vector for each condition, and within each vector TRUE is when each condition is met. Then, I can assign the values by subsetting, df1$Accuracy[Good]<-"Good" for example. However, I can not figure out how to create the logical vector using the apply family functions or dplyr functions. (But, any solution that avoids for loops is also welcome.) If for loops are the better way to go, that would also be helpful to know.

这是我失败的尝试.这些返回不正确的NA或不正确的逻辑向量.我不了解的许多事情之一是lapply如何知道要遍历列或行.

Here are my failed attempts. These return incorrect NA's or incorrect logical vectors. One of the many things I do not understand is how lapply knows to go over columns or rows.

Good<-apply(df1, 1, function(x) ifelse(df1$Diff[x]>3&& df1$Diff[x]<=10, TRUE, FALSE)) #logical, TRUE where condition is true 
Good<-unlist(lapply(df1$Diff,  function(x) {(ifelse(df1$Diff[x]>3&& df1$Diff[x]<=10, TRUE, FALSE))}))

更新:嵌套的ifelse语句可以使用，但是仍然欢迎有关使用apply的任何建议.

Update: Nested ifelse statements will work, but any suggestions on how to use apply are still welcome.

mutate(Accuracy=ifelse(pDiff<=3, "Excellent", 
                         ifelse(pDiff>3&pDiff<=10, "Good",
                                ifelse(pDiff>10&pDiff<=15, "Fair",
                                       ifelse(pDiff>15&pDiff<30, "Poor",
                                              ifelse(Diff>30, "Unpublishable", "TBD"))))))  

推荐答案

您可以使用dplyr中的case_when:

df1<- df %>%
mutate(Accuracy= case_when(
  .$Diff <=  3 ~ "Excellent",
  .$Diff <=  10  ~ "Good",
  .$Diff <=  15  ~ "Fair",
  .$Diff <=  30  ~ "Poor",
  .$Diff >   30  ~ "Unpublishable",
  TRUE  ~"TBD")
)

 df1
        Date  Time Diff      Accuracy
1 08/29/2011 09:45  0.2     Excellent
2 08/29/2011 10:00  4.3          Good
3 08/30/2011 13:00  6.5          Good
4 08/30/2011 13:30 15.0          Fair
5 08/30/2011 10:14 16.5          Poor
6 08/29/2012  9:09 31.0 Unpublishable
7 08/29/2012 11:23 30.2 Unpublishable
8 01/15/2012 17:06 21.9          Poor
9 08/29/2012 12:20  1.9     Excellent

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