R从本地文件计算平均值和中位数

Ben 发表于 Dev

Ben

我是R的新手，所以这很明显。

我到目前为止的代码：

rm(list=ls())

kdata = read.table("data_fra_klassen_v20.txt",header = TRUE,)

library(openxlsx)

kdata = read.xlsx("data_fra_klassen_v20.xlsx") 


head(kdata)

这是数据集：

  gender    shoe    height  colour
  Man        43      176    Green
  Woman      36      166    Brown
  Man        43      182    Other
  Man        36      151    Brown
  Woman      43      183    Blue
  Man        44      184    Blue
  Woman      38      164    Brown
  Woman      37      160    Brown
  Man        41      175    Brown

我正在尝试寻找均值和每种性别的中位数。

我在想也许是这样的：

heightmen = kdata$height[kdata$gender=="Man"]
mean(heightmen)

但是，似乎找不到任何值。

sm925

您可以使用dplyrR中的包来执行此操作：

使用mutate：

library(dplyr)
df %>% 
  group_by(gender)%>% 
  mutate(mean_height = mean(height))%>% 
  mutate(median_height = median(height)) %>% 
  select(gender, mean_height, median_height) %>% 
  unique()

或使用summarise：

df %>%
  group_by(gender) %>%
  summarise(mean_height = mean(height), median_height = median(height))


# A tibble: 2 x 3
# Groups:   gender [2]
#   gender mean_height median_height
#<fct>        <dbl>         <dbl>
#1 Man           174.           176
#2 Woman         168.           165

数据

df <- structure(list(gender = structure(c(1L, 2L, 1L, 1L, 2L, 1L, 2L, 2L, 1L), .Label = c("Man", "Woman"), class = "factor"), shoe = c(43L,36L, 43L, 36L, 43L, 44L, 38L, 37L, 41L), height = c(176L, 166L,182L, 151L, 183L, 184L, 164L, 160L, 175L), colour = structure(c(3L,2L, 4L, 2L, 1L, 1L, 2L, 2L, 2L), .Label = c("Blue", "Brown", "Green", "Other"), class = "factor")), class = "data.frame", row.names = c(NA,-9L))

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