R语言竞赛表中中位年龄的计算方法

68de4m5k 于 2023-01-03 发布在 R语言

关注(0)|答案(3)|浏览(238)

我希望你能帮助我找到一个种族表中的年龄中位数。我的数据示例

Asian Black Hispanic White
  15                   0     0        0     6
  17                   0     0        0     9
  19                   0     0        0     8
  20                   0     0        0    12
  20.8388888888889     2     0        0     0
  20.8583333333333     2     0        0     0
  21                   1     7        1    31
  21.4888888888889     0     0        2     0
  21.5277777777778     0     0        2     0

第一列继续显示年龄为99岁。2我尝试使用下面的lapply函数来这样做，但是我只得到了亚洲人种的中位数。

RaceAge= tapply(age,RACE, median)
RaceAge
   Asian    Black Hispanic    White 
      43       NA       NA       NA

非常感谢。

来源：https://stackoverflow.com/questions/74936119/a-method-to-calculate-the-median-age-in-a-race-table-in-r

3条答案

按热度按时间

f0brbegy1#

我们可以使用

library(dplyr)
library(spatstat)
as.data.frame.matrix(tbl1) %>% 
  rownames_to_column('age') %>%
  summarise(across(-age, ~ weighted.median(age, w = .x, type = 1)))

输出

Asian Black Hispanic White
1    26    17       19  21.8

或者另一个选择是

df2 <- transform(as.data.frame(tbl1),
    AgeNiki = as.numeric(as.character(AgeNiki)))
by(df2, df2$Var2, FUN = function(x) weighted.median(x$AgeNiki,
     w = x$Freq, type = 1))
 df2$Var2: Asian
[1] 26
--------------------------------------------------------------------------------------------------------------------- 
df2$Var2: Black
[1] 17
--------------------------------------------------------------------------------------------------------------------- 
df2$Var2: Hispanic
[1] 19
--------------------------------------------------------------------------------------------------------------------- 
df2$Var2: White
[1] 21.8

数据

tbl1 <- structure(c(0L, 0L, 0L, 0L, 2L, 2L, 1L, 0L, 0L, 0L, 0L, 5L, 0L, 
0L, 0L, 5L, 7L, 13L, 11L, 4L, 0L, 0L, 0L, 0L, 0L, 0L, 7L, 0L, 
0L, 0L, 0L, 3L, 3L, 0L, 0L, 0L, 8L, 6L, 31L, 8L, 0L, 0L, 0L, 
0L, 0L, 0L, 1L, 2L, 2L, 2L, 2L, 0L, 0L, 0L, 0L, 1L, 4L, 5L, 26L, 
28L, 6L, 9L, 8L, 12L, 0L, 0L, 31L, 0L, 0L, 0L, 0L, 36L, 30L, 
1L, 1L, 29L, 28L, 35L, 89L, 66L), .Dim = c(20L, 4L), .Dimnames = list(
    AgeNiki = c("15", "17", "19", "20", "20.8388888888889", "20.8583333333333", 
    "21", "21.4888888888889", "21.5277777777778", "21.8944444444444", 
    "21.9138888888889", "22", "23", "23.1", "23.2166666666667", 
    "24", "25", "26", "27", "28"), c("Asian", "Black", "Hispanic", 
    "White")), class = "table")

赞(0）回复(0）举报 2023-01-03

esbemjvw2#

data <- data.frame(age = c(15, 17, 19, 20, 20.8388888888889, 20.8583333333333, 
21, 21.4888888888889, 21.5277777777778), Asian = c(0L, 0L, 0L, 
0L, 2L, 2L, 1L, 0L, 0L), Black = c(0L, 0L, 0L, 0L, 0L, 0L, 7L, 
0L, 0L), Hispanic = c(0L, 0L, 0L, 0L, 0L, 0L, 1L, 2L, 2L), White = c(6L, 
9L, 8L, 12L, 0L, 0L, 31L, 0L, 0L))

apply(data[-1], 2, \(x) median(rep(data$age,x)))
#>    Asian    Black Hispanic    White 
#> 20.85833 21.00000 21.48889 20.00000

赞(0）回复(0）举报 2023-01-03

wgmfuz8q3#

您可以将DescTools::Untable()与tapply()一起使用：

with(DescTools::Untable(tbl1, colnames = c("age", "race")),
     tapply(as.numeric(levels(age))[age], race, median))

#   Asian    Black Hispanic    White
#    26.0     17.0     19.0     21.8

或与aggregate()：

aggregate(age ~ race, DescTools::Untable(tbl1, colnames = c("age", "race")),
          \(x) median(as.numeric(levels(x))[x]))

#       race  age
# 1    Asian 26.0
# 2    Black 17.0
# 3 Hispanic 19.0
# 4    White 21.8

数据tbl1是@akrun的答案。

赞(0）回复(0）举报 2023-01-03

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R语言竞赛表中中位年龄的计算方法

3条答案

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