如何获得R中两个范畴变量的百分比
如何获得R中两个范畴变量的百分比
提问于 2020-03-27 13:20:01
假设样本总数为8，数据帧如下所示。所有健康分数低于3分的人都是健康的，所有健康分数大于3分的人都生病了。地位显示他们的就业状况。
Status<-(Employed,Unemployed,Student,Student,Employed,Unemployed,Unemployed,Housewife)
Health<-(Healthy,Healthy,Healthy,Sick,Sick,Control,Sick,Sick)
df<-(Status,Health)
level(Health)<-("Healthy,"Sick",Control)
level(Status)<-("Employed","Unemployed","Student","Housewife")我希望看到健康，生病或控制的人的百分比属于每个职业类别。我要输出如下所示。(举个例子，P.S值只是假设的)就像在所有就业人员中，有多少百分比是健康的？
Healthy Sick Control
Employed 10% 2% 1%
Unemployed 5% 1% 1%
Student 6% 3% 1%
Housewife 2% 5% 6%我正在使用以下代码。但它只是给了我频率，而不是百分比。我需要百分比。
tab <- with(df, table(df$Health,df$Status))回答 1
Stack Overflow用户
回答已采纳
发布于 2020-03-27 13:28:17
我们可以count每个Status和Health的个体数，group_by状态并计算百分比。为了获得更好的可见性，我们将数据转换为宽格式。
library(dplyr)
df %>%
count(Status, Health) %>%
group_by(Status) %>%
mutate(n = n/sum(n) * 100) %>%
tidyr::pivot_wider(names_from = Health, values_from = n,
values_fill = list(n = 0))
# Status Healthy Sick Control
# <fct> <dbl> <dbl> <dbl>
#1 Employed 50 50 0
#2 Housewife 0 100 0
#3 Student 50 50 0
#4 Unemployed 33.3 33.3 33.3在基数R中，我们可以使用prop.table和table来得到百分比。
prop.table(table(df), 1) * 100数据
df <- structure(list(Status = structure(c(1L, 4L, 3L, 3L, 1L, 4L, 4L,
2L), .Label = c("Employed", "Housewife", "Student", "Unemployed"
), class = "factor"), Health = structure(c(2L, 2L, 2L, 3L, 3L,
1L, 3L, 3L), .Label = c("Control", "Healthy", "Sick"),
class = "factor")), class = "data.frame",row.names = c(NA, -8L))页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/60886862
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