频数表和列联表，以及进一步处理分析 -- R
目的
数据框包含了一些分类变量，问？
1.如何统计分类变量的分布次数 -- 频数表
2.如何统计多分类变量的分布次数 -- 频联表
3.单个分类变量的分类结果是否满足理论分类结果 -- 拟合优度问题
4.多个分类变量的分类结果是否相关干扰 -- 分类变量（多因素）独立性检验
数据
library(vcd)
data(Arthritis)
head(Arthritis)
# ID Treatment Sex Age Improved
#1 57 Treated Male 27 Some
#2 46 Treated Male 29 None
#3 77 Treated Male 30 None
#4 17 Treated Male 32 Marked
#5 36 Treated Male 46 Marked
#6 23 Treated Male 58 Marked
# 频数表
table(Arthritis$Treatment)
# Placebo Treated
43 41
# 频联表
table(Arthritis$Treatment,Arthritis$Improved)
# None Some Marked
# Placebo 29 7 7
# Treated 13 7 21
代码
- 操作频数表
# 把频数表变成百分比占比
prop.table(table(Arthritis$Treatment))
# Placebo Treated
# 0.5119 0.4881
prop.table(table(Arthritis$Treatment)) *100
# Placebo Treated
# 51.19 48.81
# 把列联表变成百分比形式
prop.table(table(Arthritis$Treatment,Arthritis$Improved))
# None Some Marked
# Placebo 0.345238 0.083333 0.08333
# Treated 0.154762 0.083333 0.250000
# 按行求百分比
prop.table(table(Arthritis$Treatment,Arthritis$Improved),1)
# None Some Marked
# Placebo 0.67442 0.16279 0.16279
# Treated 0.31707 0.17073 0.51220
# 按列求百分比
prop.table(table(Arthritis$Treatment,Arthritis$Improved),2)
# None Some Marked
# Placebo 0.69048 0.50000 0.25000
# Treated 0.30952 0.50000 0.75000
# 给列联表添加行列计数
addmargins(table(Arthritis$Treatment,Arthritis$Improved))
# None Some Marked Sum
# Placebo 29 7 7 43
# Treated 13 7 21 41
# Sum 42 14 28 84
addmargins(table(Arthritis$Treatment,Arthritis$Improved),1)
# None Some Marked
# Placebo 29 7 7
# Treated 13 7 21
# Sum 42 14 28
addmargins(table(Arthritis$Treatment,Arthritis$Improved),2)
# None Some Marked Sum
# Placebo 29 7 7 43
# Treated 13 7 21 41
prop.table(addmargins(table(Arthritis$Treatment,Arthritis$Improved)))
# None Some Marked Sum
# Placebo 0.086310 0.020833 0.020833 0.127976
# Treated 0.038690 0.020833 0.062500 0.122024
# Sum 0.125000 0.041667 0.083333 0.250000
# 单个分类变量的分类结果是否满足理论分类结果
table(Arthritis$Improved)
# None Some Marked
# 42 14 28
x <- Arthritis$Improved
# 模拟一个理论分布
y <- c(rep("None",30),rep("Some",30),rep("Marked",24))
# 卡方检验
chisq.test(x,y)
# Pearson's Chi-squared test
# data: x and y
# X-squared = 6.78, df = 4, p-value = 0.15
# 原假设 H0: 实际频次分布和理论频次分布不相似
# 因为 p >= 0.05，接受原假设，也就是数据中的频次分布和理论上的分布（y）不相似
#模拟一个 和实际分布很相似的数据
y <- c(rep("None",39),rep("Some",16),rep("Marked",29))
chisq.test(x,y)
# Pearson's Chi-squared test
# data: x and y
# X-squared = 16.8, df = 4, p-value = 0.0021
# 此时p <= 0.05，需要拒绝原假设接受备择假设，也就是实际频次分布和理论频次分布相似
# 多个分类变量的分类结果是否相关干扰 -- 分类变量（多因素）独立性检验
table(Arthritis$Treatment,Arthritis$Improved)
# None Some Marked
# Placebo 29 7 7
# Treated 13 7 21
#问 treatment 方式对improved的频次分布有影响嘛？或者说两个分类变量独立吗？
# 卡方检验，直接把频联表丢给chisq.test( )函数即可 <==== 参数检验
mytable <- table(Arthritis$Treatment,Arthritis$Improved)
chisq.test(mytable)
# Pearson's Chi-squared test
# data: mytable
# X-squared = 13.1, df = 2, p-value = 0.0015
# p <= 0.05 拒绝原假设：相互独立，也就是treatment 方式对improved的频次分布有影响
# Fisher's精确检验（Fisher's exact test） <=========非参数检验
fisher.test(mytable)
# Fisher's Exact Test for Count Data
# data: mytable
# p-value = 0.0014
# alternative hypothesis: two.sided
# p <= 0.05 拒绝原假设：相互独立，也就是treatment 方式对improved的频次分布有影响
# 上面看到两个分类变量是相互影响的
# 问：如何度量它们之间的相关性强度呢？
library(vcd)
mytable <- table(Arthritis$Treatment,Arthritis$Improved)
assocstats(mytable)
# X^2 df P(> X^2)
# Likelihood Ratio 13.530 2 0.0011536
# Pearson 13.055 2 0.0014626
# Phi-Coefficient : NA
# Contingency Coeff.: 0.367
# Cramer's V : 0.394
vcdӉ中的assocstats()函数可以计算二维列联表的phi系数，列联系数，Cramer‘s V系数
总体来说，较大的数值意味着较强的相关性