Question

我试图在R中编写一个函数来计算给定收入和人口份额的基尼评分（收入不平等系数）。这就是我想要做的事情：

incomes <- c(1175,1520,1865,2210,2555) # incomes
population <- rep(1/5,5)*100           # population shares (5 times 1/5)

income <- incomes*population/sum(incomes*population) # income * frequency / total income
data <- as.data.frame(cbind(incomes,income,population/100))
names(data) <- c("incomes","income","population")

data <- data[order(as.numeric(data$incomes)),] # sort by percentage of income

for (i in 1:length(income)){
    data$richer[i] <- 1-sum(data$population[1:i])
}
data$score <- data$income * (data$population + 2 * data$richer)
gini <- round(1-sum(data$score),4) # gini

这一切都运作良好。但现在我想绘制收入分配图，为此我创建了一个新的数据集：

data$population2 <- data$richer + data$population # cumulative
x <- as.data.frame(matrix(data=NA,ncol=1,nrow=20))
names(x) <- c("population2")
x$population2 <- rev(seq(0.05,1,0.05))

data.graph <- join(x, data, by = "population2")

所以＆＃39;数据$ population2＆＃39;变量的值为1,0.8,0.6,0.4,0.2，x $ population2的值为1,0.95,0.9,0.85,0.8等，直到0.05。但是，join函数只加入值为1,0.8,0.2，而不是0.6和0.4的值！任何人都可以帮助我吗？

Answer 1

欢迎来到first circle of R hell。：）

乍一看，data$population2中的所有值看起来都应该与x$population2匹配：

> x$population2
 [1] 1.00 0.95 0.90 0.85 0.80 0.75 0.70 0.65 0.60 0.55 0.50 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05
> data$population2
[1] 1.0 0.8 0.6 0.4 0.2

但事实并非如此：

> x$population2[9]
[1] 0.6
> data$population2[3]
[1] 0.6

> data$population2[3] == x$population2[9]
[1] FALSE
> all.equal(data$population2[3], x$population2[9]) 
[1] TRUE
# all.equal tolerates numerical differences smaller than 1.5e-8 by default

> print(x$population2[9], digits = 20)
[1] 0.60000000000000009
> print(data$population2[3], digits = 20)
[1] 0.59999999999999987

以下内容适用于示例案例，但我要注意不要在每个场景中应用它，而不考虑四舍五入的小数位数是否合适。通常，使用字符键执行连接更安全：

library(plyr); library(dplyr)

join(x %>% mutate(population2 = round(population2, 3)), 
     data%>% mutate(population2 = round(population2, 3)), 
     by = "population2")

   population2 incomes    income population richer      score
1         1.00    1175 0.1260054        0.2    0.8 0.22680965
2         0.95      NA        NA         NA     NA         NA
3         0.90      NA        NA         NA     NA         NA
4         0.85      NA        NA         NA     NA         NA
5         0.80    1520 0.1630027        0.2    0.6 0.22820375
6         0.75      NA        NA         NA     NA         NA
7         0.70      NA        NA         NA     NA         NA
8         0.65      NA        NA         NA     NA         NA
9         0.60    1865 0.2000000        0.2    0.4 0.20000000
10        0.55      NA        NA         NA     NA         NA
11        0.50      NA        NA         NA     NA         NA
12        0.45      NA        NA         NA     NA         NA
13        0.40    2210 0.2369973        0.2    0.2 0.14219839
14        0.35      NA        NA         NA     NA         NA
15        0.30      NA        NA         NA     NA         NA
16        0.25      NA        NA         NA     NA         NA
17        0.20    2555 0.2739946        0.2    0.0 0.05479893
18        0.15      NA        NA         NA     NA         NA
19        0.10      NA        NA         NA     NA         NA
20        0.05      NA        NA         NA     NA         NA

作为旁注，在前面的步骤中使用for循环，您可以执行以下操作：

library(dplyr)

# use this
data <- data %>% mutate(richer = 1-cumsum(population))

# instead of this
for (i in 1:length(income)){
  data$richer[i] <- 1-sum(data$population[1:i])
}

对于循环操作在R中相对较慢（在较大的数据集中很明显）。 R针对矢量化操作进行了优化。

加入（plyr库），奇怪的错误

1 个答案: