确定R中特定级别的变量的平均值

Question

我试图找到特定于我赋予不同变量的级别的变量的平均值（平均值）。

到目前为止，我创建了一个新变量，其中包含与之关联的各种级别：

• 等级1：值<= 0％，
• 第2级：值<1％，和
• 第3级：值> = 1％。

pincome$income_growth <- ifelse(pincome$incomechng <= 0, "level 1",
                                ifelse(pincome$incomechng < 1,"level 2","level 3"))

现在我想确定与上述水平相关的另一个变量的平均值（例如，1级的平均收入（收入增长小于0％）。

我希望这是有道理的，我对R来说非常新手并试图掌握它！

Answer 1

如果你想要基础R，试试by（?by）。如果你开始做更复杂的事情，plyr / dplyr包非常惊人，如果你打算用大量数据集捣乱并且不介意更多的初始学习曲线，data.table包也令人惊叹。

一个reproducible example会很棒。

EG

set.seed(1) # so your random numbers are the same as mine
pincome <- data.frame(incomechng = runif(20, min=-1, max=3))

# what you had was fine too; using ?cut is another way to do it
# have just put it in for demonstration purposes.
# though `cut` uses intervals like (a, b] or [a, b) whereas yours
#  are (-Inf, 0] (0, 1) [1, Inf) which is a little different.    
pincome$income_growth <- cut(pincome$incomechng,
                             breaks=c(-Inf, 0, 1, Inf),
                             labels=paste("level", 1:3))

现在我们可以取每组内的平均值。我已经展示了三种选择;我相信还有更多。

# base R ?by
by(pincome$incomechng, pincome$income_growth, mean)
# pincome$income_growth: level 1
# [1] -0.6848674
# ------------------------------------------
# pincome$income_growth: level 2
# [1] 0.4132334
# ------------------------------------------
# pincome$income_growth: level 3
# [1] 1.772039

# plyr (dplyr has pipe syntax you may prefer but is otherwise the same)
library(plyr)
ddply(pincome, .(income_growth), summarize, avgIncomeGrowth=mean(incomechng))
#   income_growth avgIncomeGrowth
# 1       level 1      -0.6848674
# 2       level 2       0.4132334
# 3       level 3       1.7720395

# data.table
library(data.table)
setDT(pincome)
pincome[, list(avgIncomeGrowth=mean(incomechng)), by=income_growth]
#    income_growth avgIncomeGrowth
# 1:       level 2       0.4132334
# 2:       level 3       1.7720395
# 3:       level 1      -0.6848674

Answer 2

如果你想要一个整齐的解决方案：

library(tidyverse)
pincome %>%
 mutate(income_growth = case_when(incomechng <= 0 ~ "level 1",
                                  incomechng < 1 ~ "level 2",
                                  TRUE ~ "level 3")) %>%
 group_by(income_growth) %>%
 summarize(avgIncomeGrowth = mean(incomechng,na.rm=TRUE))