摘要: 仅用于记录R语言学习过程：
内容提要：
描述性统计；t检验；数据转换；方差分析；卡方检验；回归分析与模型诊断；生存分析；COX回归
写在正文前的话，关于基础知识，此篇为终结篇，笔记来自医学方的课程，仅用于学习R的过程。
正文：
描述性统计
n 如何去生成table1 用table()函数，快速汇总频数
u 生成四格表：table(行名，列名)
> table(tips$sex,tips$smoker)
No Yes
Female 54 33
Male 97 60
u addmargins()函数：对生成的table表格进行计算
> table(esoph$agegp,esoph$ncases)
0 1 2 3 4 5 6 8 9 17
25-34 14 1 0 0 0 0 0 0 0 0
35-44 10 2 2 1 0 0 0 0 0 0
45-54 3 2 2 2 3 2 2 0 0 0
55-64 0 0 2 4 3 2 2 1 2 0
65-74 1 4 2 2 2 2 1 0 0 1
75+ 1 7 3 0 0 0 0 0 0 0
> tt <- table(esoph$agegp,esoph$ncases)
> addmargins(tt,margin = c(1,2)) # margin 1表示行，2表示列
0 1 2 3 4 5 6 8 9 17 Sum
25-34 14 1 0 0 0 0 0 0 0 0 15
35-44 10 2 2 1 0 0 0 0 0 0 15
45-54 3 2 2 2 3 2 2 0 0 0 16
55-64 0 0 2 4 3 2 2 1 2 0 16
65-74 1 4 2 2 2 2 1 0 0 1 15
75+ 1 7 3 0 0 0 0 0 0 0 11
Sum 29 16 11 9 8 6 5 1 2 1 88
n xtabs()函数：在数据集中取子集，生成表格
> hightip <- tips[,'tip'] > mean(tips[,'tip'] )
> as.data.frame(xtabs(~tips$sex + hightip,subset = tips$smoker=='No'))
#as.data.frame转换成数据框；~后面的公式类似table括号中的内容，为分类变量；~左边需添加的是连续型变量；有一个子集subset可进行提取
tips.sex hightip Freq
1 Female FALSE 31
2 Male FALSE 49
3 Female TRUE 23
4 Male TRUE 48
> as.data.frame(xtabs(~tips$sex + hightip,subset = tips$day %in% c('Sun','Sat')))
tips.sex hightip Freq
1 Female FALSE 21
2 Male FALSE 53
3 Female TRUE 25
4 Male TRUE 64
示例2：
> xtabs(ncontrols~agegp + alcgp,data = esoph)
alcgp
agegp 0-39g/day 40-79 80-119 120+
25-34 61 45 5 5
35-44 89 80 20 10
45-54 78 81 39 15
55-64 89 84 43 26
65-74 71 53 29 8
75+ 27 12 2 3
> addmargins(xtabs(ncontrols~agegp + alcgp,data = esoph),margin = c(1,2))
alcgp
agegp 0-39g/day 40-79 80-119 120+ Sum
25-34 61 45 5 5 116
35-44 89 80 20 10 199
45-54 78 81 39 15 213
55-64 89 84 43 26 242
65-74 71 53 29 8 161
75+ 27 12 2 3 44
Sum 415 355 138 67 975
#cbind(数值型，数值型)
> xtabs(cbind(ncases,ncontrols)~agegp + alcgp,data = esoph)
, , = ncases
alcgp
agegp 0-39g/day 40-79 80-119 120+
25-34 0 0 0 1
35-44 1 4 0 4
45-54 1 20 12 13
55-64 12 22 24 18
65-74 11 25 13 6
75+ 4 4 2 3
, , = ncontrols
alcgp
agegp 0-39g/day 40-79 80-119 120+
25-34 61 45 5 5
35-44 89 80 20 10
45-54 78 81 39 15
55-64 89 84 43 26
65-74 71 53 29 8
75+ 27 12 2 3
n ftable()函数：扁平表格，接受xtabs对象，进行调整
> ftable(xtabs(cbind(ncases,ncontrols)~agegp +alcgp,data = esoph))
ncases ncontrols
agegp alcgp
25-34 0-39g/day 0 61
40-79 0 45
80-119 0 5
120+ 1 5
35-44 0-39g/day 1 89
40-79 4 80
80-119 0 20
120+ 4 10
45-54 0-39g/day 1 78
40-79 20 81
80-119 12 39
120+ 13 15
55-64 0-39g/day 12 89
40-79 22 84
80-119 24 43
120+ 18 26
65-74 0-39g/day 11 71
40-79 25 53
80-119 13 29
120+ 6 8
75+ 0-39g/day 4 27
40-79 4 12
80-119 2 2
120+ 3 3
n 数据汇总可结合前面学习的函数，如：
u summary(数据集)函数
u describe（数据集）函数（psych包里的）
u describe.data.frame(数据集)函数 （Hmisc包里的）
u apply()家族等
t检验
n 假设检验:服从正态分布，方差齐性（如果不齐，可以用t’检验），
n 示例：
u 生成随机数据，并检验是否符合正态分布：（shapiro.test()函数）
> data1 <- sample(1:100,50)
>
> #检验正态性 shapiro.test()函数
> shapiro.test(data1)
Shapiro-Wilk normality test
data: data1
W = 0.96424, p-value = 0.1338 #不能拒绝原假设，服从正态分布
除此之外，还有以下5个函数可用于检验是否符合正态分布：
library(nortest)
lillie.test(data1)
ad.test(data1)
cvm.test(data1)
pearson.test(data1)
sf.test(data1)
如：
> lillie.test(data1)
Lilliefors (Kolmogorov-Smirnov) normality test
data: data1
D = 0.064492, p-value = 0.8693
> ad.test(data1)
Anderson-Darling normality test
data: data1
A = 0.40918, p-value = 0.333
> cvm.test(data1)
Cramer-von Mises normality test
data: data1
W = 0.054212, p-value = 0.4437
> pearson.test(data1)
Pearson chi-square normality test
data: data1
P = 4, p-value = 0.7798
> sf.test(data1)
Shapiro-Francia normality test
data: data1
W = 0.97496, p-value = 0.3075
u 生成服从正态分布的数据：rnorm()函数
> #生成服从正态分布的数据
> data3 <- rnorm(100,3,5)
> data4 <- rnorm(200,3.4,8)
>
> shapiro.test(data3)
Shapiro-Wilk normality test
data: data3
W = 0.99369, p-value = 0.9258
>
> shapiro.test(data4)
Shapiro-Wilk normality test
data: data4
W = 0.98946, p-value = 0.149
u 检验方差齐性：var.test()函数：仅适用于两组
> var.test(data3,data4)
F test to compare two variances
data: data3 and data4
F = 0.40348, num df = 99, denom df = 199, p-value = 1.088e-06
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.2893355 0.5739700
sample estimates:
ratio of variances
0.4034794 #p值小于0.05，说明方差不齐性
u 进行t检验函数：t.test()函数
u 两组均数t检验
> t.test(data3,data4,var.equal = F) #方差不齐则var.equal设置为F，默认也为FALSE，若其设置为TRUE，如为FALSE,则会使用t’检验
Welch Two Sample t-test
data: data3 and data4
t = -1.3681, df = 281.41, p-value = 0.1724 #说明两组均数无显著统计学差异
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-2.587004 0.465494
sample estimates:
mean of x mean of y
2.468549 3.529304
u 样本均数与总体均数t检验
> #样本均数与总体均数的比较
> t.test(data3,mu =3.2)
One Sample t-test
data: data3
t = -1.4117, df = 99, p-value = 0.1612 #样本与总体无统计学差异
alternative hypothesis: true mean is not equal to 3.2
95 percent confidence interval:
1.440424 3.496675
sample estimates:
mean of x
2.468549
u 配对数据的t检验
data3 <- rnorm(200,3,5)
> data4 <- rnorm(200,3.4,5)
>
> #配对数据
> #进行配对t检验
> t.test(data3,data4,paired = TRUE)
Paired t-test
data: data3 and data4
t = 0.59079, df = 199, p-value = 0.5553 #p大于0.05，说明无显著差异
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.6946177 1.2888581
sample estimates:
mean of the differences
0.2971202
数据变换
n 数据不满足正态分布时该如何处理：有时可采用取log()、sqrt()、1/x等等方式，
n runif()平均分布函数
> mydata <- runif(100,min = 1,max = 2)
> shapiro.test(mydata)
Shapiro-Wilk normality test
data: mydata
W = 0.93149, p-value = 6.05e-05
> shapiro.test(log(mydata))
Shapiro-Wilk normality test
data: log(mydata)
W = 0.93082, p-value = 5.54e-05
> shapiro.test(sqrt(mydata))
Shapiro-Wilk normality test
data: sqrt(mydata)
W = 0.93236, p-value = 6.794e-05
> shapiro.test(1/mydata)
Shapiro-Wilk normality test
data: 1/mydata
W = 0.9195, p-value = 1.325e-05
#几种方式均不可行。
n boxcox转换---对公式的，加载MASS包，运用里面的boxcox()函数，#boxcox()转换，核心为找到lambda的值进行相应的转换
bc <- boxcox(Volume ~ log(Height) + log(Girth),data = trees,
lambda = seq(-0.25,0.25,length =10))
#Volume为拟操作的变量，Height和Girth为用此两个数据进行估计，但要取log，trees为数据集，lambda后面的公式为取最佳值
u 用公式：(x^lambda-1)/lambda 进行数据转换（lambda 不等于1），新的Volume_bc就服从正态分布了。
> Volume_bc <- (trees$Volume^lambda-1)/lambda
> shapiro.test(Volume_bc)
Shapiro-Wilk normality test
data: Volume_bc
W = 0.96431, p-value = 0.3776
u 可用qqnorm(Volume_bc)和qqline(Volume_bc)c查看图，是否符合正态分布
n boxcox转换---对数值的，加载forecast包，利用包中的BoxCox.lambda()函数
> BoxCox.lambda(trees$Volume)
[1] -0.4954451 #lambda的值 ，采用1/sqrt(x)
分以下几种情况：
ü lambda接近-1时 1/x
ü -0.5 1/sqrt(x)
ü 0 ln(x) 或log(x)
ü 0.5 sqrt()
ü 1 不用变换了
完整示例：
> BoxCox.lambda(trees$Volume)
[1] -0.4954451
> new_volum <- 1/sqrt(trees$Volume)
> shapiro.test(new_volum)
Shapiro-Wilk normality test
data: new_volum
W = 0.94523, p-value = 0.1152
n 利用car包中powerTransform得到lambda值，再进行正态分布分析
> library(car)
> powerTransform(trees$Volume)
Estimated transformation parameter
trees$Volume
-0.07476608
> new_volum <- trees$Volume^-0.07476608
> shapiro.test(new_volum)
Shapiro-Wilk normality test
data: new_volum
W = 0.96428, p-value = 0.3768
方差分析
n 用于多组均值的比较。两组均值的t检验与方差分析等价，但是对于>=3组的均数比较，t检验不适用，需用方差分析。
n 假设检验，需满足的条件：
u 正态性
u 方差齐性
u 独立性
n 方差齐性的检验
u 安装multcomp包，加载cholesterol数据集(里面包含response组和trt组)
u 方差齐性的检验：
l R语言的内置包：bartlett.test()
> bartlett.test(response~trt,data = cholesterol)
Bartlett test of homogeneity of variances
data: response by trt
Bartlett's K-squared = 0.57975, df = 4, p-value =
0.9653
> #p = 0.9653 方差是齐性的
>
> #正态性检验
> shapiro.test(cholesterol$response)
Shapiro-Wilk normality test
data: cholesterol$response
W = 0.97722, p-value = 0.4417
l R语言的内置包：fligner.test() 与bartlett.test()检验原理不同，但公式一样
> #方差齐性检验
> fligner.test(response~trt,data = cholesterol)
Fligner-Killeen test of homogeneity of variances
data: response by trt
Fligner-Killeen:med chi-squared = 0.74277, df = 4,
p-value = 0.946
l car包中的ncvTest()函数：
> ncvTest(lm(response~trt,data = cholesterol))
Non-constant Variance Score Test
Variance formula: ~ fitted.values
Chisquare = 0.1338923 Df = 1 p = 0.71443
l car包中的leveneTest()函数：
> leveneTest(response~trt,data = cholesterol)
Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 4 0.0755 0.9893
45
n 方差分析
u 单因素方差分析：aov()函数
> aov(response~trt,data = cholesterol)
Call:
aov(formula = response ~ trt, data = cholesterol)
Terms:
trt Residuals
Sum of Squares 1351.3690 468.7504
Deg. of Freedom 4 45
Residual standard error: 3.227488
Estimated effects may be unbalanced
> summary(fit)
Df Sum Sq Mean Sq F value Pr(>F)
trt 4 1351.4 337.8 32.43 9.82e-13 ***
Residuals 45 468.8 10.4
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#用gplots包中的plotmeans可视化
plotmeans(response~trt,data = cholesterol)
u 单因素方差分析：oneway.test()函数
> oneway.test(response~trt,data = cholesterol)
One-way analysis of means (not assuming equal
variances)
data: response and trt
F = 30.709, num df = 4.00, denom df = 22.46, p-value =
8.227e-09
> oneway.test(response~trt,data = cholesterol,var.equal = T)
One-way analysis of means
data: response and trt
F = 32.433, num df = 4, denom df = 45, p-value =
9.819e-13
u 两两比较：TukeyHSD()函数
> TukeyHSD(fit)
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = response ~ trt, data = cholesterol)
$trt
diff lwr upr p adj
2times-1time 3.44300 -0.6582817 7.544282 0.1380949
4times-1time 6.59281 2.4915283 10.694092 0.0003542
drugD-1time 9.57920 5.4779183 13.680482 0.0000003
drugE-1time 15.16555 11.0642683 19.266832 0.0000000
4times-2times 3.14981 -0.9514717 7.251092 0.2050382
drugD-2times 6.13620 2.0349183 10.237482 0.0009611
drugE-2times 11.72255 7.6212683 15.823832 0.0000000
drugD-4times 2.98639 -1.1148917 7.087672 0.2512446
drugE-4times 8.57274 4.4714583 12.674022 0.0000037
drugE-drugD 5.58635 1.4850683 9.687632 0.0030633
#可视化
plot(TukeyHSD(fit))
u 多因素方差分析：aov()函数
> data('ToothGrowth')
> head('ToothGrowth')
len supp dose
1 4.2 VC 0.5
2 11.5 VC 0.5
3 7.3 VC 0.5
4 5.8 VC 0.5
5 6.4 VC 0.5
6 10.0 VC 0.5
> levels(ToothGrowth$supp)
[1] "OJ" "VC"
> levels(ToothGrowth$dose)
NULL
> levels(as.factor(ToothGrowth$dose))
[1] "0.5" "1" "2"
> table(ToothGrowth$supp,ToothGrowth$dose)
0.5 1 2
OJ 10 10 10
VC 10 10 10
#公式的写法 关注交互作用
> fangcha <- aov (len~supp+dose,data = ToothGrowth)
> summary(fangcha)
Df Sum Sq Mean Sq F value Pr(>F)
supp 1 205.4 205.4 11.45 0.0013 **
dose 1 2224.3 2224.3 123.99 6.31e-16 ***
Residuals 57 1022.6 17.9
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#交互作用
> fangcha <- aov (len~supp*dose,data = ToothGrowth)
> summary(fangcha)
Df Sum Sq Mean Sq F value Pr(>F)
supp 1 205.4 205.4 12.317 0.000894 ***
dose 1 2224.3 2224.3 133.415 < 2e-16 ***
supp:dose 1 88.9 88.9 5.333 0.024631 * #交互作用不能忽视
Residuals 56 933.6 16.7
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#可视化 交互作用图
>interaction.plot(ToothGrowth$dose,ToothGrowth$supp,ToothGrowth$len,type = 'b',
+ pch = c(1,10),col = c('red','green'))
u 单因素协方差分析：aov()函数
> data('litter')
> head(litter)
dose weight gesttime number
1 0 28.05 22.5 15
2 0 33.33 22.5 14
3 0 36.37 22.0 14
4 0 35.52 22.0 13
5 0 36.77 21.5 15
6 0 29.60 23.0 5
> facn <- aov(weight~gesttime + dose + gesttime:dose,data = litter)
> summary(facn)
Df Sum Sq Mean Sq F value Pr(>F)
gesttime 1 134.3 134.30 8.289 0.00537 ** #协变量确实有影响，该如何去除？
dose 3 137.1 45.71 2.821 0.04556 *
gesttime:dose 3 81.9 27.29 1.684 0.17889
Residuals 66 1069.4 16.20
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
u 去除协变量的影响：加载effects包，用其中的effect()函数
> effect('dose',facn)
NOTE: dose is not a high-order term in the model
dose effect
dose
0 5 50 500
32.30722 28.50625 30.61078 29.29005
#常见研究设计的表达式
卡方检验
n 对分类变量的检验方法
n 分类：
u 拟合优度检验：用chisq.test()函数
（针对样本数据的分布与已知总体分布是否一致）
> men <- c(11,120,60,45)
> women <- c(20,102,39,30)
> df <- as.data.frame(rbind(men,women))
> colnames(df) <- c('AB','O','A','B')
> chisq.test(men)
Chi-squared test for given probabilities
data: men
X-squared = 105.46, df = 3, p-value < 2.2e-16
> chisq.test(men,p = c(0.1,0.5,0.2,0.2)) #p 中为总体人群中各血型的比例
Chi-squared test for given probabilities
data: men
X-squared = 10.335, df = 3, p-value = 0.01592
u 卡方齐性检验：用于比较不同分类水平下，各个类型的比例是否一致。
> chisq.test(df) #行变量与列变量的相关性检验
Pearson's Chi-squared test
data: df
X-squared = 6.8607, df = 3, p-value = 0.07647 #男女不同血型的分布一致
u 卡方独立性检验：
> chisq.test(df)
Pearson's Chi-squared test
data: df
X-squared = 6.8607, df = 3, p-value = 0.07647 # 行变量和列变量无关联，血型分布和性别无关
u CMH检验：分层检验 三维，行变量为无序，列变量为有序数据；判断是否有混杂因素
#采用mantelhaen.test()函数
> Rabbits <-
+ array(c(0,0,6,5,
+ 3,0,3,6,
+ 6,2,0,4,
+ 5,6,1,0,
+ 2,5,0,0),
+ dim = c(2,2,5),
+ dimnames = list(
+ Delay = c('None','1.5h'),
+ Response = c('Cured','Died'),
+ Penicillin.Level = c('1/8','1/4','1/2','1','4')))
> Rabbits
, , Penicillin.Level = 1/8
Response
Delay Cured Died
None 0 6
1.5h 0 5
, , Penicillin.Level = 1/4
Response
Delay Cured Died
None 3 3
1.5h 0 6
, , Penicillin.Level = 1/2
Response
Delay Cured Died
None 6 0
1.5h 2 4
, , Penicillin.Level = 1
Response
Delay Cured Died
None 5 1
1.5h 6 0
, , Penicillin.Level = 4
Response
Delay Cured Died
None 2 0
1.5h 5 0
> mantelhaen.test(Rabbits)
Mantel-Haenszel chi-squared test with continuity
correction
data: Rabbits
Mantel-Haenszel X-squared = 3.9286, df = 1, p-value =
0.04747
alternative hypothesis: true common odds ratio is not equal to 1
95 percent confidence interval:
1.026713 47.725133
sample estimates:
common odds ratio
7 #可以判定盘尼西林是否为混杂因素
u CMH检验：有序分类的卡方检验：连续型变量
#采用mantelhaen.test()函数
> Satisfaction <-
+ as.table(array(c(1,2,0,0,3,3,1,2,
+ 11,17,8,4,2,3,5,2,
+ 1,0,0,0,1,3,0,1,
+ 2,5,7,9,1,1,3,6),
+ dim = c(4,4,2),
+ dimnames =
+ list(Income =
+ c('<5000','5000-15000',
+ '15000-25000','>25000'),
+ 'Job Satisfaction' =
+ c('V_D','L_S','M_S','V_S'),
+ Gender = c('Female','Male'))))
> Satisfaction
, , Gender = Female
Job Satisfaction
Income V_D L_S M_S V_S
<5000 1 3 11 2
5000-15000 2 3 17 3
15000-25000 0 1 8 5
>25000 0 2 4 2
, , Gender = Male
Job Satisfaction
Income V_D L_S M_S V_S
<5000 1 1 2 1
5000-15000 0 3 5 1
15000-25000 0 0 7 3
>25000 0 1 9 6
> mantelhaen.test(Satisfaction)
Cochran-Mantel-Haenszel test
data: Satisfaction
Cochran-Mantel-Haenszel M^2 = 10.2, df = 9, p-value =
0.3345 #工资水平与满意度无关
u 配对四格表的卡方检验：采用mcnemar.test()函数
> paired <- as.table(matrix(c(157,24,69,18),nrow = 2,dimnames = list(case =c('A','B'),control = c('A','B'))))
> paired
control
case A B
A 157 69
B 24 18
> mcnemar.test(paired)
McNemar's Chi-squared test with continuity correction
data: paired
McNemar's chi-squared = 20.817, df = 1, p-value =
5.053e-06
> chisq.test(paired)
Pearson's Chi-squared test with Yates' continuity
correction
data: paired
X-squared = 1.9244, df = 1, p-value = 0.1654
u 的
回归分析与模型诊断
前提：是否适合做线性回归，是否满足正态分布
只要因变量是连续型变量即可做线性回归，因变量是分类变量需要做Logistic回归；自变量是连续型或者分类变量无影响
n 一元回归分析：lm(因变量~自变量)
u x为连续型变量
> x<- seq(1,5,len =100)
> noise <- rnorm(n=100,mean = 0,sd = 1)
> beta0 = 1
> beta1 <-2
> y <- beta0 + beta1*x + noise
> plot(y~x) #看一下是否适合做线性回归
> model <- lm(y~x)
> summary(model)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-2.49732 -0.64794 0.00215 0.75355 3.06579
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.77938 0.28733 2.712 0.00789 **
x 2.05561 0.08927 23.027 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.041 on 98 degrees of freedom
Multiple R-squared: 0.844, Adjusted R-squared: 0.8424 #模型可解释度
F-statistic: 530.3 on 1 and 98 DF, p-value: < 2.2e-16 #模型可靠性
u x为分类变量
> x <- factor(rep(c(0,1,2),each = 20))
> y <- c(rnorm(20,0,1),rnorm(20,1,1),rnorm(20,2,1))
> model <- lm(y~x)
> summary(model)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-2.50566 -0.82826 0.01137 0.87966 2.41338
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.05338 0.24331 0.219 0.827
x1 0.81708 0.34409 2.375 0.021 *
x2 1.99903 0.34409 5.810 2.94e-07 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.088 on 57 degrees of freedom
Multiple R-squared: 0.3745, Adjusted R-squared: 0.3525
F-statistic: 17.06 on 2 and 57 DF, p-value: 1.558e-06
> exp( 1.99903) #转化成OR值
[1] 7.381892
n 模型诊断
u 非正态性：shapiro.test()判断
> #检验残差是否服从正态分布
> data(LMdata,package = 'rinds')
> model <- lm(y~x,data = LMdata$NonL)
> #找残差
>
> res1 <- residuals(model)
> shapiro.test(res1) #残差不符合正态分布，需要重新做
Shapiro-Wilk normality test
data: res1
W = 0.93524, p-value = 1e-04 #残差不符合正态分布，需要重新做
u 非线性
> #2.非线性
> model2 <- lm(y~x,data = LMdata$NonL)
> plot(model2)
#y 与x2的关系是否成线性
model2 <- lm(y~x+ I(x^2),data = LMdata$NonL)
summary(model2)
#踢掉x
> model3 <- update(model2,y~.-x)
> summary(model3)
Call:
lm(formula = y ~ 1, data = LMdata$NonL)
Residuals:
Min 1Q Median 3Q Max
-19.456 -12.907 -2.464 10.725 28.697
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 21.829 1.429 15.28 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 14.29 on 99 degrees of freedom
> model2 <- lm(y~x+ I(x^2),data = LMdata$NonL)
> summary(model2)
Call:
lm(formula = y ~ x + I(x^2), data = LMdata$NonL)
Residuals:
Min 1Q Median 3Q Max
-2.32348 -0.60702 0.00701 0.56018 2.27346
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.98702 0.62216 1.586 0.116
x 0.11085 0.45405 0.244 0.808
I(x^2) 1.97966 0.07456 26.552 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.907 on 97 degrees of freedom
Multiple R-squared: 0.9961, Adjusted R-squared: 0.996
F-statistic: 1.224e+04 on 2 and 97 DF, p-value: < 2.2e-16
> model3 <- update(model2,y~.-x)
> summary(model3)
Call:
lm(formula = y ~ I(x^2), data = LMdata$NonL)
Residuals:
Min 1Q Median 3Q Max
-2.34289 -0.60692 0.01722 0.58186 2.29601
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.13378 0.15963 7.103 1.97e-10 ***
I(x^2) 1.99760 0.01271 157.195 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.9026 on 98 degrees of freedom
Multiple R-squared: 0.996, Adjusted R-squared: 0.996
F-statistic: 2.471e+04 on 1 and 98 DF, p-value: < 2.2e-16
> AIC(model,model2,model3)
df AIC
model 3 478.4558
model2 4 269.2121
model3 3 267.2736 #拟合效果越来越好
plot(model3$residuals~LMdata$NonL) #在0左右分布
u 异方差：考虑的是噪声对模型的影响
model4 <- lm(y~x,data = LMdata$Hetero)
plot(model4$residuals~LMdata$Hetero$x)
#处理方法：使用加权最小二乘法：对于不同的样本点给予不同的权重
> summary(model5)
Call:
lm(formula = y ~ x, data = LMdata$Hetero, weights = 1/x^2)
Weighted Residuals:
Min 1Q Median 3Q Max
-2.25132 -0.68409 -0.03924 0.63997 2.61098
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.2611 0.5139 2.454 0.0159 *
x 1.8973 0.2317 8.190 9.98e-13 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.025 on 98 degrees of freedom
Multiple R-squared: 0.4063, Adjusted R-squared: 0.4003
F-statistic: 67.07 on 1 and 98 DF, p-value: 9.977e-13
> summary(model4)
Call:
lm(formula = y ~ x, data = LMdata$Hetero)
Residuals:
Min 1Q Median 3Q Max
-8.3371 -1.6383 -0.1533 1.4075 12.3423
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.6175 1.0019 1.615 0.11
x 1.7671 0.3113 5.677 1.4e-07 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 3.63 on 98 degrees of freedom
Multiple R-squared: 0.2475, Adjusted R-squared: 0.2398
F-statistic: 32.23 on 1 and 98 DF, p-value: 1.396e-07
#处理方法：使用迭代重复加权最小二乘法 采用nlme包中的gls()函数，找到最合适的weights公式，得到最小AIC
> glsmodel <- gls(y~x,weights = varFixed(~x),data = LMdata$Hetero)
> summary(glsmodel)
Generalized least squares fit by REML
Model: y ~ x
Data: LMdata$Hetero
AIC BIC logLik
517.5286 525.2835 -255.7643
Variance function:
Structure: fixed weights
Formula: ~x
Coefficients:
Value Std.Error t-value p-value
(Intercept) 1.421324 0.7091780 2.004184 0.0478
x 1.832543 0.2603653 7.038351 0.0000
Correlation:
(Intr)
x -0.908
Standardized residuals:
Min Q1 Med Q3
-2.17451000 -0.55322766 -0.03454023 0.51060224
Max
2.93969900
Residual standard error: 1.890127
Degrees of freedom: 100 total; 98 residual
u 自相关：利用lmtest包中的DW检验函数：dwtest()函数
> model6 <- lm(y~x,data = LMdata$AC)
> dwtest(model6)
Durbin-Watson test
data: model6
DW = 0.65556, p-value = 2.683e-12 #存在自相关，不能使用最小二乘法，可使用gls()函数，实现广义最小二乘法，注意gls中的correlation 参数，设置为corAR1()
alternative hypothesis: true autocorrelation is greater than 0
> glsmodel2 <- gls(y~x,correlation = corAR1(),data = LMdata$AC)
> summary(glsmodel2)
Generalized least squares fit by REML
Model: y ~ x
Data: LMdata$AC
AIC BIC logLik
268.7617 279.1016 -130.3809
Correlation Structure: AR(1)
Formula: ~1
Parameter estimate(s):
Phi
0.7041665
Coefficients:
Value Std.Error t-value p-value
(Intercept) 1.335059 0.7792019 1.713367 0.0898
x 2.072936 0.2405513 8.617438 0.0000
Correlation:
(Intr)
x -0.926
Standardized residuals:
Min Q1 Med Q3
-1.667086282 -0.571601900 0.001724114 0.568360354
Max
2.306177988
Residual standard error: 1.25329
Degrees of freedom: 100 total; 98 residual
> summary(model6)
Call:
lm(formula = y ~ x, data = LMdata$AC)
Residuals:
Min 1Q Median 3Q Max
-2.10131 -0.72894 -0.03329 0.66138 2.77461
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.2626 0.3303 3.822 0.000232 ***
x 2.1118 0.1026 20.578 < 2e-16 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.197 on 98 degrees of freedom
Multiple R-squared: 0.8121, Adjusted R-squared: 0.8101
F-statistic: 423.4 on 1 and 98 DF, p-value: < 2.2e-16
u 异常值：离群点，杠杆点、高影响点
model7 <- lm(y~x,data = LMdata$Outlier)
plot(y~x,data = LMdata$Outlier)
abline(model7) #画出回归直线
#利用car包中的infuencePlot()函数进行判断，圆圈越大，对模型影响越大，做线性回归模型时需踢掉该点，用update函数
model8 <- update(model7,y~x,subset = -32,data = LMdata$Outlier)
plot(y~x,data = LMdata$Outlier) #比较去除前后的差异
abline(model7,col = 'red') #比较去除前后的差异
abline(model8,col = 'green') #比较去除前后的差异
u 多重共线性的判断：自变量之间存在线性关系
#x1,x2,x3与y之间p值无显著差异，但是总体上的p值和R值非常显著，说明x1,x2,x3之间可能是存在相关性的，但与y没有相关性，不能进行回归分析，需要用函数进行检验确认：vif()函数计算方差膨胀因子
> model9 <- lm(y~x1+x2+x3,data = LMdata$Mult)
> summary(model9)
Call:
lm(formula = y ~ x1 + x2 + x3, data = LMdata$Mult)
Residuals:
Min 1Q Median 3Q Max
-1.29100 -0.26369 0.00141 0.32529 0.91182
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.3848 0.5813 0.662 0.5095
x1 7.2022 4.8552 1.483 0.1412
x2 -14.0916 12.1385 -1.161 0.2486
x3 8.2312 4.8559 1.695 0.0933 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.499 on 96 degrees of freedom
Multiple R-squared: 0.9984, Adjusted R-squared: 0.9984
F-statistic: 2.057e+04 on 3 and 96 DF, p-value: < 2.2e-16
#vif计算方差膨胀因子，一般而言，大于10则认为存在共线性
> vif(model9)
x1 x2 x3
7560.819 214990.752 222630.742
#运用自动回归判断是x中的哪些变量存在共线性：step()函数
> model10 <- step(model9)
Start: AIC=-135.11
y ~ x1 + x2 + x3
Df Sum of Sq RSS AIC
- x2 1 0.33560 24.241 -135.71
<none> 23.905 -135.11
- x1 1 0.54795 24.453 -134.84
- x3 1 0.71550 24.621 -134.16
Step: AIC=-135.71
y ~ x1 + x3
Df Sum of Sq RSS AIC
<none> 24.2 -135.71
- x1 1 189.2 213.4 79.81
- x3 1 15276.4 15300.7 507.05
> summary(model10)
Call:
lm(formula = y ~ x1 + x3, data = LMdata$Mult)
Residuals:
Min 1Q Median 3Q Max
-1.24046 -0.27519 -0.02751 0.32824 0.91344
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.38158 0.58230 0.655 0.514
x1 1.56614 0.05692 27.514 <2e-16 ***
x3 2.59393 0.01049 247.241 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.4999 on 97 degrees of freedom
Multiple R-squared: 0.9984, Adjusted R-squared: 0.9984
F-statistic: 3.074e+04 on 2 and 97 DF, p-value: < 2.2e-16
n Logistic回归
与线性回归的区别：y为分类变量。
u 使用glm()函数：下载HSAUR2包的数据集plasma
> library(HSAUR2)
> data('plasma')
> head(plasma)
fibrinogen globulin ESR
1 2.52 38 ESR < 20
2 2.56 31 ESR < 20
3 2.19 33 ESR < 20
4 2.18 31 ESR < 20
5 3.41 37 ESR < 20
6 2.46 36 ESR < 20
> fit1 <- glm(ESR~fibrinogen+globulin,data = plasma,
+ family = binomial()) #family选择二分类
> summary(fit1)
Call:
glm(formula = ESR ~ fibrinogen + globulin, family = binomial(),
data = plasma)
Deviance Residuals:
Min 1Q Median 3Q Max
-0.9683 -0.6122 -0.3458 -0.2116 2.2636
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -12.7921 5.7963 -2.207 0.0273 *
fibrinogen 1.9104 0.9710 1.967 0.0491 *
globulin 0.1558 0.1195 1.303 0.1925
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 30.885 on 31 degrees of freedom
Residual deviance: 22.971 on 29 degrees of freedom
AIC: 28.971
Number of Fisher Scoring iterations: 5
> exp(coef(fit1)['fibrinogen'])
fibrinogen
6.755579
> exp(1.9104)
[1] 6.755791 #ORR值
> exp(confint(fit1,parm = 'fibrinogen'))
Waiting for profiling to be done...
2.5 % 97.5 % #95%的可信区间
1.404131 73.000836
n 生存分析与COX回归
生存分析
u 用coin包中的glima数据集演示，用survival包进行生存分析，利用其中的survfit()函数做出生存曲线图
> g3 <- subset(glioma,histology =='Grade3')
> fit <- survfit(Surv(time,event)~group,data = g3) #注意公式的写法
> plot(fit,lty = c(2,3),col = c(2,1))
> legend('bottomright',legend = c('Control','Treatment'),lty = c(2,1),col = c(2,1))
u 用survdiff()函数计算两组间生存差异
> survdiff(Surv(time,event)~group,data = g3) #接受传入的形式
Call:
survdiff(formula = Surv(time, event) ~ group, data = g3)
N Observed Expected (O-E)^2/E (O-E)^2/V
group=Control 6 4 1.49 4.23 6.06
group=RIT 11 2 4.51 1.40 6.06
Chisq= 6.1 on 1 degrees of freedom, p= 0.0138
u 用survdiff()函数计算两组间生存差异
> logrank_test(Surv(time,event)~group,data = g3,distribution='exact')
# 注意公式的写法，选择精确分布
Exact Two-Sample Logrank Test
data: Surv(time, event) by group (Control, RIT)
Z = -2.1711, p-value = 0.02877
alternative hypothesis: true theta is not equal to 1
#讨论不同histology的对照组与治疗组是否有差异
logrank_test(Surv(time,event)~group|histology,data = glioma,distribution = approximate(B =1000))
Approximative Two-Sample Logrank Test
data: Surv(time, event) by
group (Control, RIT)
stratified by histology
Z = -3.6704, p-value < 2.2e-16
alternative hypothesis: true theta is not equal to 1
COX回归
u 用GBSG2数据集
> install.packages('TH.data')
Error in install.packages : Updating loaded packages
> data('GBSG2',package = 'TH.data')
> head(GBSG2)
horTh age menostat tsize tgrade pnodes progrec estrec time cens
1 no 70 Post 21 II 3 48 66 1814 1
2 yes 56 Post 12 II 7 61 77 2018 1
3 yes 58 Post 35 II 9 52 271 712 1
4 yes 59 Post 17 II 4 60 29 1807 1
5 no 73 Post 35 II 1 26 65 772 1
6 no 32 Pre 57 III 24 0 13 448 1
> plot(survfit(Surv(time,cens)~horTh,data = GBSG2),lty = c(2,1),col = c(2,1), mark.Time = T) #如果需要标记生存时间，在后面加上mark.Time = T
> legend('bottomright',legend = c('yes','no'),lty = c(2,1),col = c(2,1))
#画出了生存曲线图
u 生存分析：使用survival包中的coxph()函数
> coxreg <- coxph(Surv(time,cens)~.,data = GBSG2)
> summary(coxreg)
Call:
coxph(formula = Surv(time, cens) ~ ., data = GBSG2)
n= 686, number of events= 299
coef exp(coef) se(coef) z Pr(>|z|)
horThyes -0.3462784 0.7073155 0.1290747 -2.683 0.007301 **
age -0.0094592 0.9905854 0.0093006 -1.017 0.309126
menostatPost 0.2584448 1.2949147 0.1834765 1.409 0.158954
tsize 0.0077961 1.0078266 0.0039390 1.979 0.047794 *
tgrade.L 0.5512988 1.7355056 0.1898441 2.904 0.003685 **
tgrade.Q -0.2010905 0.8178384 0.1219654 -1.649 0.099199 .
pnodes 0.0487886 1.0499984 0.0074471 6.551 5.7e-11 ***
progrec -0.0022172 0.9977852 0.0005735 -3.866 0.000111 ***
estrec 0.0001973 1.0001973 0.0004504 0.438 0.661307
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
exp(coef) exp(-coef) lower .95 upper .95
horThyes 0.7073 1.4138 0.5492 0.9109
age 0.9906 1.0095 0.9727 1.0088
menostatPost 1.2949 0.7723 0.9038 1.8553
tsize 1.0078 0.9922 1.0001 1.0156
tgrade.L 1.7355 0.5762 1.1963 2.5178
tgrade.Q 0.8178 1.2227 0.6439 1.0387
pnodes 1.0500 0.9524 1.0348 1.0654
progrec 0.9978 1.0022 0.9967 0.9989
estrec 1.0002 0.9998 0.9993 1.0011
Concordance= 0.692 (se = 0.018 )
Rsquare= 0.142 (max possible= 0.995 )
Likelihood ratio test= 104.8 on 9 df, p=0
Wald test = 114.8 on 9 df, p=0
Score (logrank) test = 120.7 on 9 df, p=0
#画树
install.packages('party')
tree <- ctree(Surv(time,cens)~.,data = GBSG2)
plot(tree)