从数据帧中按组查找顶部十进制数 [英] Find top deciles from dataframe by group
问题描述
我正在尝试使用函数创建新变量,而不是使用循环在数据中正确使用 lapply
。我曾经使用Stata,并且会使用类似于这里。
由于以编程方式命名变量在R中非常困难或至少令人尴尬(并且您似乎无法使用索引为 assign
),我已经离开了命名过程,直到 lapply
之后。然后我使用来
循环在合并之前进行重命名,并再次合并。有更有效的方法吗?如何更换循环?我应该做某种重组吗?
#可复制数据
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)
#分析每个类别(v1)的客户分布的功能
pf< - function(cat,df){
df < - df [df $ v1 == cat,]
df < - df [order(-df $ v2),]
#将客户转化为最高百分比
nr< - nrow(df)
p10 < - round(nr * .10,0)
cat( 10%中的人数:,p10,\\\
)
p20< - round(nr * .20,0)
p11_20 < - p20-p10
cat(11-20%中的人数,p11_20,\\\
)
#仅顶部组中的客户
df< - df [1:p20,]
#创建一个变量来标识客户在
中的百分比组top_pct< - integer(length = p10 + p11_20)
#识别每个组中的那些
top_pct [1:p10]< - 10
top_pct [(p10 + 1):p20]< - 20
#将这个变量添加到数据框
df $ top_pct< - top_pct
#仅保留custID和新变量
df< - subset(df ,select = c(custID,top_pct))
return(df)
}
##运行客户分发函数
v1Levels< - levels(data $ v1)
res< - lapply(v1Levels,pf,df = data)
#Explore结果
summary res)
#长度类模式
#[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
## ge两个数据帧,但是top_pct作为每个类别的不同变量
#更改新的变量名称
for(i in 1:length(res)){
names (res [[i]])[2]< - paste0(v1Levels [i],_top_pct)
}
#Merge结果
res_m<对于(i in 2:length(res))的
{
res_m< - merge(res_m,res [[i]],by =custID,all = TRUE )
}
打印(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
在R中做这种事情的惯用方式是使用 split
和 lapply
。您的中途是您使用 lapply
;你只需要使用 split
。
lapply(split数据,数据$ v1),函数(df){
截止< - 分位数(df $ v2,c(0.8,0.9))
top_pct< - ifelse(df $ v2& 2],10,ifelse(df $ v2> cutoff [1],20,NA))
na.omit(data.frame(id = df $ custID,top_pct))
})
使用 quantile
查找分位数。 p>
I am attempting to create new variables using a function and lapply
rather than working right in the data with loops. I used to use Stata and would have solved this problem with a method similar to that discussed here.
Since naming variables programmatically is so difficult or at least awkward in R (and it seems you can't use indexing with assign
), I have left the naming process until after the lapply
. I am then using a for
loop to do the renaming prior to merging and again for the merging. Are there more efficient ways of doing this? How would I replace the loops? Should I be doing some sort of reshaping?
#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
The idiomatic way to do this kind of thing in R would be to use a combination of split
and lapply
. You're halfway there with your use of lapply
; you just need to use split
as well.
lapply(split(data, data$v1), function(df) {
cutoff <- quantile(df$v2, c(0.8, 0.9))
top_pct <- ifelse(df$v2 > cutoff[2], 10, ifelse(df$v2 > cutoff[1], 20, NA))
na.omit(data.frame(id=df$custID, top_pct))
})
Finding quantiles is done with quantile
.
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