Question

让我举个例子：

person | salary
----------------
1      | 30'000
2      | 10'000
3      | 15'000
4      | 25'000
5      | 80'000
6      | 56'000
...    | ...

获得此结果的步骤是订购工资，然后创建一个新表，从开始到相应的行分配行/人的份额，以及从开始到工资的总和的份额各行（总工资）。

然后我们必须为人们选择最接近20％的行，我们知道他们赚了多少钱。

这是一个非常标准的问题 - 但由于我不知道如何口头提及它，我不能谷歌。

所以我很感激，如果有人能告诉我该怎么称呼它，以及如何在R中计算和绘制这个最简单的东西 - 所以没有循环和东西。我的直觉告诉我至少有5个包和10个函数来解决这个问题。也许类似于带有固定分位数的summary（）。

因此，我们假设上表可用作数据框：

salaries <- data.frame(person = c(1,2,3,...), salary = c(30000,...))

Answer 1

使用SLID - 包中的car收入数据集：

library(car)

dat <- SLID[!is.na(SLID$wage),]       # Remove missing values
dat$income <- dat$wage*40*50          # "translate" the wages to their full time annual earnings equivalent.
dat$id <- seq(1,nrow(dat))       

# Create a data.frame with a person ID and their annual income:
keep <- data.frame(id = seq(1, nrow(dat)), 
                   income = dat$income)
keep <- keep[order(keep$income, decreasing = TRUE),]  # Descending ordering according to income
keep$accum <- cumsum(keep$income)                     # Cumulative sum of the descending incomes
keep$pct <- keep$accum/sum(keep$income)*100           # % of the total income

keep$check <- keep$pct<80                      # Check where the % is smaller than 80%
threshold <- min(which(keep$check == FALSE))   # First line where % is larger than 80%
border <- threshold/nrow(keep)*100             # Check which percentile that was
border <- round(border, digits = 2)
paste0(border, "% of the people earn 80% of the income")

#[1] "62.41% of the people earn 80% of the income"

正如我们所期望的那样，经典的80-20规则将显示＆＃34; 20％的人获得80％的收入＆＃34;。此规则不适用于此处，您可以看到..

颠倒的论点：

# The 20% of the people earn X % of total income:

linenr <- ceiling(1/5*nrow(keep))
outcome2 <- round(keep$pct[linenr], digits = 2)
paste0(outcome2, "% of total income is earned by the top 20% of the people")

# [1] "36.07% of total income is earned by the top 20% of the people"

请注意，此处显示的数字不代表现实世界:)

此外，Wikipedia还有关于帕累托原则的更多信息，也称为80-20规则。似乎这个规则出现在多种设置中，例如商业，经济和数学。

“前20％的人赚80％的钱”......这种结果使用R

1 个答案: