从数据帧中按组查找最高十分位数

Question

我试图使用函数创建新变量,lapply而不是使用循环在数据中正常工作.我曾经使用Stata,并且会用类似于此处讨论的方法解决这个问题.

由于在R中以编程方式命名变量是如此困难或至少是尴尬(并且似乎你不能使用索引assign),我已经将命名过程留到了之后lapply.然后我使用for循环在合并之前进行重命名,然后再用于合并.有更有效的方法吗？我该如何更换循环？我应该做某种重塑吗？

#Reproducible data
data <- data.frame("custID" = c(1:10, 1:20),
    "v1" = rep(c("A", "B"), c(10,20)), 
    "v2" = c(30:21, 20:19, 1:3, 20:6), stringsAsFactors = TRUE)

#Function to analyze customer distribution for each category (v1)
pf <- function(cat, df) {

        df <- df[df$v1 == cat,]
        df <- df[order(-df$v2),]

    #Divide the customers into top percents
    nr <- nrow(df)
    p10 <- round(nr * .10, 0)
    cat("Number of people in the Top 10% :", p10, "\n")
    p20 <- round(nr * .20, 0)
    p11_20 <- p20-p10
    cat("Number of people in the 11-20% :", p11_20, "\n")

    #Keep only those customers in the top groups
    df <- df[1:p20,]

    #Create a variable to identify the percent group the customer is in
    top_pct <- integer(length = p10 + p11_20)

    #Identify those in each group
    top_pct[1:p10] <- 10
    top_pct[(p10+1):p20] <- 20

    #Add this variable to the data frame
    df$top_pct <- top_pct

    #Keep only custID and the new variable
    df <- subset(df, select = c(custID, top_pct))

    return(df)

}


##Run the customer distribution function
v1Levels <- levels(data$v1)
res <- lapply(v1Levels, pf, df = data)

#Explore the results
summary(res)

    #      Length Class      Mode
    # [1,] 2      data.frame list
    # [2,] 2      data.frame list

print(res)

    # [[1]]
    #   custID top_pct
    # 1      1      10
    # 2      2      20
    # 
    # [[2]]
    #    custID top_pct
    # 11      1      10
    # 16      6      10
    # 12      2      20
    # 17      7      20



##Merge the two data frames but with top_pct as a different variable for each category

#Change the new variable name
for(i in 1:length(res)) {
    names(res[[i]])[2] <- paste0(v1Levels[i], "_top_pct")
}

#Merge the results
res_m <- res[[1]]
for(i in 2:length(res)) {
    res_m <- merge(res_m, res[[i]], by = "custID", all = TRUE)
}

print(res_m)

    #   custID A_top_pct B_top_pct
    # 1      1        10        10
    # 2      2        20        20
    # 3      6        NA        10
    # 4      7        NA        20

Answer 1

坚持你的Stata本能并使用单一数据集:

require(data.table)
DT <- data.table(data)

DT[,r:=rank(v2)/.N,by=v1]

您可以通过键入来查看结果DT.

从这里,你可以组种内v1排名,r如果你想.遵循Stata成语......

DT[,g:={
  x = rep(0,.N)
  x[r>.8] = 20
  x[r>.9] = 10
  x
}]

这就像gen两个replace ... if陈述一样.再次,您可以看到结果DT.

最后,您可以使用

DT[g>0]

这使

   custID v1 v2     r  g
1:      1  A 30 1.000 10
2:      2  A 29 0.900 20
3:      1  B 20 0.975 10
4:      2  B 19 0.875 20
5:      6  B 20 0.975 10
6:      7  B 19 0.875 20

这些步骤也可以链接在一起:

DT[,r:=rank(v2)/.N,by=v1][,g:={x = rep(0,.N);x[r>.8] = 20;x[r>.9] = 10;x}][g>0]

(感谢@ExperimenteR :)

要重新排列OP中所需的输出,并v1使用列的值,请使用dcast:

dcast(
  DT[,r:=rank(v2)/.N,by=v1][,g:={x = rep(0,.N);x[r>.8] = 20;x[r>.9] = 10;x}][g>0], 
  custID~v1)

目前,dcast需要data.tableGithub提供的最新版本(我认为).

Answer 2

您不需要该功能pf即可实现所需的功能。尝试dplyr/tidyr组合

library(dplyr)
library(tidyr)
data %>% 
    group_by(v1) %>% 
    arrange(desc(v2))%>%
    mutate(n=n()) %>% 
    filter(row_number() <= round(n * .2)) %>% 
    mutate(top_pct= ifelse(row_number()<=round(n* .1), 10, 20)) %>%
    select(custID, top_pct) %>% 
    spread(v1,  top_pct)
#  custID  A  B
#1      1 10 10
#2      2 20 20
#3      6 NA 10
#4      7 NA 20