data1 =c(1,2,3,4) data2 =c(2,3,4,5) a =0.05 sigma21 =3 sigma22 =4 n =length(data1) m =length(data2) mean = mean(data1)-mean(data2) mean - qnorm(a/2,lower.tail =FALSE)*sqrt(sigma21/n+sigma22/m)
1
## [1] -3.592789
1
mean + qnorm(a/2,lower.tail =FALSE)*sqrt(sigma21/n+sigma22/m)
1
## [1] 1.592789
方差未知
方差相同
PPT12的40页
1 2 3 4 5 6 7 8 9 10 11
data1 =c(1,2,3,4) data2 =c(2,3,4,5) a =0.05 sigma2 =3 n =length(data1) m =length(data2) mean = mean(data1)-mean(data2) var1 = var(data1) var2 = var(data2) Sw =sqrt(((n-1)*var1+(m-1)*var2)/(n+m-2)) mean - qt(a/2,n+m-2,lower.tail =FALSE)*sqrt(1/n+1/m)*Sw
1
## [1] -3.233715
1
mean + qt(a/2,n+m-2,lower.tail =FALSE)*sqrt(1/n+1/m)*Sw
1
## [1] 1.233715
方差不同
PPT12的41页
1 2 3 4 5 6 7 8 9 10
data1 =c(1,2,3,4) data2 =c(2,3,4,5) a =0.05 n =length(data1) m =length(data2) mean = mean(data1)-mean(data2) var1 = var(data1) var2 = var(data2) v =(var1/n+var2/m)^2/(var1^2/(n^2*(n-1))+var2^2/(m^2*(m-1))) mean - qt(a/2,v,lower.tail =FALSE)*sqrt(var1/n+var2/m)
1
## [1] -3.233715
1
mean + qt(a/2,v,lower.tail =FALSE)*sqrt(var1/n+var2/m)
1
## [1] 1.233715
方差比
PPT12的42页
均值已知
1 2 3 4 5 6 7 8 9 10
data1 =c(1,2,3,4) data2 =c(2,3,4,5) a =0.05 n =length(data1) m =length(data2) miu1 =2 miu2 =3 var1 =sum((data1 - miu1)^2)/n var2 =sum((data2 - miu2)^2)/m var1/(var2*qf(a/2,n,m,lower.tail =FALSE))
1
## [1] 0.1041175
1
var1/(var2*qf(1-a/2,n,m,lower.tail =FALSE))
1
## [1] 9.60453
均值未知
PPT12的43页
1 2 3 4 5 6 7 8
data1 =c(1,2,3,4) data2 =c(2,3,4,5) a =0.05 n =length(data1) m =length(data2) var1 = var(data1) var2 = var(data2) var1/(var2*qf(a/2,n-1,m-1,lower.tail =FALSE))
1
## [1] 0.06477027
1
var1/(var2*qf(1-a/2,n-1,m-1,lower.tail =FALSE))
1
## [1] 15.43918
非正态总体
二项分布
PPT12的44-47页
n>=30
单个总体
1 2 3 4 5
n =60# 尝试次数 m =40# 成功次数 a =0.05 est_p = m/n est_p - qnorm(a/2,lower.tail =FALSE)*sqrt(est_p*(1-est_p)/n)
# prop.test()函数 prop.test(x = m, n = n, conf.level = a, correct =FALSE)# correct = TRUE:进行连续性校正
1 2 3 4 5 6 7 8 9 10 11
## ## 1-sample proportions test without continuity correction ## ## data: m out of n, null probability 0.5 ## X-squared = 6.6667, df = 1, p-value = 0.009823 ## alternative hypothesis: true p is not equal to 0.5 ## 5 percent confidence interval: ## 0.66283960.6704718 ## sample estimates: ## p ## 0.6666667
n =15# 总实验次数 m =6# 成功次数 binom.test(m,n,conf.level =0.95)
1 2 3 4 5 6 7 8 9 10 11
## ## Exact binomial test ## ## data: m and n ## number of successes = 6, number of trials = 15, p-value = 0.6072 ## alternative hypothesis: true probability of success is not equal to 0.5 ## 95 percent confidence interval: ## 0.16336430.6771302 ## sample estimates: ## probability of success ## 0.4
若要求给定的置信区间长度不超过给定的正数𝛿
PPT12的54页
1 2 3 4 5 6
n =60# 尝试次数 m =40# 成功次数 a =0.05 delta =0.1 est_p = m/n qnorm(a/2,lower.tail =F)*est_p*(1-est_p)/(delta^2/4)
1
## [1] 174.219
泊松分布
PPT12的48页
n>=30
1 2 3 4 5 6
n =60 lambda =5 a =0.05 data = rpois(n, lambda) est_lambda =sum(data)/n est_lambda - qnorm(a/2,lower.tail =FALSE)*sqrt(est_lambda/n)
## ## Exact Poisson test ## ## data: m time base: 1 ## number of events = 5, time base = 1, p-value = 0.00366 ## alternative hypothesis: true event rate is not equal to 1 ## 95 percent confidence interval: ## 1.62348611.668332 ## sample estimates: ## event rate ## 5
t.test 默认两组数据间相互独立(默认参数 paired = FALSE),执行独立样本的 t 检验; 默认方差不相等 (默认参数 var.equal = FALSE) ; 默认的备择假设是双侧的(默认参数 alternative = ‘two.sided’),即执行双侧检验
均值
H0:均值等于μ0
1 2 3 4
data =c(1,2,3,4,5) h0 =7 a =0.05 t.test(x = data, y =NULL, alternative ="two.sided", miu = h0, level =1-a)
1 2 3 4 5 6 7 8 9 10 11
## ## One Sample t-test ## ## data: data ## t = 4.2426, df = 4, p-value = 0.01324 ## alternative hypothesis: true mean is not equal to 0 ## 95 percent confidence interval: ## 1.0367574.963243 ## sample estimates: ## mean of x ## 3
+ H0:均值大于μ0
1 2 3 4
data =c(1,2,3,4,5) h0 =7 a =0.05 t.test(x = data, y =NULL, alternative ="greater", miu = h0, level =1-a)
1 2 3 4 5 6 7 8 9 10 11
## ## One Sample t-test ## ## data: data ## t = 4.2426, df = 4, p-value = 0.006618 ## alternative hypothesis: true mean is greater than 0 ## 95 percent confidence interval: ## 1.492557 Inf ## sample estimates: ## mean of x ## 3
+ H0:均值小于μ0
1 2 3 4
data =c(1,2,3,4,5) h0 =7 a =0.05 t.test(x = data, y =NULL, alternative ="less", miu = h0, level =1-a)
1 2 3 4 5 6 7 8 9 10 11
## ## One Sample t-test ## ## data: data ## t = 4.2426, df = 4, p-value = 0.9934 ## alternative hypothesis: true mean is less than 0 ## 95 percent confidence interval: ## -Inf 4.507443 ## sample estimates: ## mean of x ## 3
方差
1 2 3 4 5 6 7 8 9
data =c(1,2,3,4,5) h0 =3 a =0.05 n =length(data) var = var(data) sigma = h0 df = n-1 chisq2 = df*var/sigma q = pchisq(chisq2, df)
泊松分布→正态分布:λ检验
1 2 3 4 5 6 7 8
x_bar =1.5 lambda0 =1 a =0.05 z =(x_bar - lambda0)/sqrt(lambda0 / n) p_value = pnorm(z,0,1,lower.tail =FALSE) up = qnorm(a/2,0,1,lower.tail =FALSE) down = qnorm(a/2,0,1) data.frame(z,p_value,up,down)
1 2
## z p_value up down ## 11.1180340.13177621.959964-1.959964
## ## Shapiro-Wilk normality test ## ## data: data1 ## W = 0.95998, p-value = 0.7857
1
shapiro.test(data2)# 检测是否正态
1 2 3 4 5
## ## Shapiro-Wilk normality test ## ## data: data2 ## W = 0.87357, p-value = 0.11
1
ks.test(data,"pnorm", mean = mean(data), sd = sd(data))# 可以检测多种分布
1 2 3 4 5 6
## ## Exact one-sample Kolmogorov-Smirnov test ## ## data: data ## D = 0.13646, p-value = 0.9998 ## alternative hypothesis: two-sided
方差齐次性检验
1 2 3 4 5
data1 =c(3,5,7,9,10) data2 =c(1,2,4,6,8,4) h0 =1 a =0.05 var.test(x=data1, y=data2, ratio = h0,alternative="two.sided",conf.level=1-a )
1 2 3 4 5 6 7 8 9 10 11
## ## F test to compare two variances ## ## data: data1 and data2 ## F = 1.2487, num df = 4, denom df = 5, p-value = 0.7963 ## alternative hypothesis: true ratio of variances is not equal to 1 ## 95 percent confidence interval: ## 0.169024111.6937047 ## sample estimates: ## ratio of variances ## 1.248731
1
bartlett.test(list(data1,data2))# <0.05则方差是非齐性
1 2 3 4 5
## ## Bartlett test of homogeneity of variances ## ## data: list(data1, data2) ## Bartlett's K-squared = 0.049565, df = 1, p-value = 0.8238
## ## Paired t-test ## ## data: data1 and data2 ## t = -7.926, df = 9, p-value = 2.384e-05 ## alternative hypothesis: true mean difference is not equal to 0 ## 95 percent confidence interval: ## -0.3501458-0.1946542 ## sample estimates: ## mean difference ## -0.2724
不配对均值检验
1 2 3
x <-c(4.05,4.18,5.93,4.3,2.41,5.6,6.61,2.98,5.93,4.18,4.05,3.62) y <-c(4.54,4.63,3.64,5.07,6.44,5.62,6.14,4.81,6.42,5.35) shapiro.test(x)# 先检测是否为正态分布
1 2 3 4 5
## ## Shapiro-Wilk normality test ## ## data: x ## W = 0.93768, p-value = 0.4687
1
shapiro.test(y)
1 2 3 4 5
## ## Shapiro-Wilk normality test ## ## data: y ## W = 0.95199, p-value = 0.6921
1
bartlett.test(list(a=x,b=y))
1 2 3 4 5
## ## Bartlett test of homogeneity of variances ## ## data: list(a = x, b = y) ## Bartlett's K-squared = 1.0346, df = 1, p-value = 0.3091
1
t.test(x,y, alternative="less", var.equal=TRUE)
1 2 3 4 5 6 7 8 9 10 11
## ## Two Sample t-test ## ## data: x and y ## t = -1.6189, df = 20, p-value = 0.06056 ## alternative hypothesis: true difference in means is less than 0 ## 95 percent confidence interval: ## -Inf 0.05093173 ## sample estimates: ## mean of x mean of y ## 4.4866675.266000
## ## Welch Two Sample t-test ## ## data: data1 and data2 ## t = 1.5925, df = 8.199, p-value = 0.149 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -1.1638606.430527 ## sample estimates: ## mean of x mean of y ## 6.8000004.166667
## ## Welch Two Sample t-test ## ## data: data1 and data2 ## t = 4.8148, df = 10.635, p-value = 0.0005944 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## 1.621394.37315 ## sample estimates: ## mean of x mean of y ## 8.3748575.377587
## ## Pearson's Chi-squared test ## ## data: M ## X-squared = 11.951, df = 2, p-value = 0.00254
1 2 3
# 2*2列联表 M <- as.table(rbind(c(81,83),c(19,32)))# 第一行,第二行 chisq.test(M, correct =F)
1 2 3 4 5
## ## Pearson's Chi-squared test ## ## data: M ## X-squared = 2.3028, df = 1, p-value = 0.1291
2*2列联表
n>=40,且每个E>=5:卡方检验
1 2
M <- as.table(rbind(c(12,32),c(22,14)))# 第一行,第二行 chisq.test(M, correct =F)
1 2 3 4 5
## ## Pearson's Chi-squared test ## ## data: M ## X-squared = 9.2774, df = 1, p-value = 0.00232
n>=40, 1<E<5:yates连续性校正
1 2 3 4 5 6
a =30# 1行1 b =45# 1行2 c=3# 2行1 d =50# 2行2 chisq2 =(a+b+c+d)*(abs(a*d-b*c)-0.5*n)^2/((a+b)*(c+d)*(a+c)*(b+d)) chisq2
1
## [1] 19.06814
1 2 3
# 或者用chisq.test函数 M <- as.table(rbind(c(81,83),c(19,32)))# 第一行,第二行 chisq.test(M, correct =T)
1 2 3 4 5
## ## Pearson's Chi-squared test with Yates' continuity correction ## ## data: M ## X-squared = 1.8409, df = 1, p-value = 0.1749
n<=40或E<1:fisher精确检验
1 2
data = matrix(c(1,2,3,4),ncol =2,nrow =2) fisher.test(data)
1 2 3 4 5 6 7 8 9 10 11
## ## Fisher's Exact Test for Count Data ## ## data: data ## p-value = 1 ## alternative hypothesis: true odds ratio is not equal to 1 ## 95 percent confidence interval: ## 0.008512238 20.296715040 ## sample estimates: ## odds ratio ## 0.693793
# H0:A和B一样好,H1:A和B有差异→统计A比B好的数目m n =200 m =72 p =0.3 a =0.05 binom.test(m,n,p,alternative="two.sided", conf.level =1-a)
1 2 3 4 5 6 7 8 9 10 11
## ## Exact binomial test ## ## data: m and n ## number of successes = 72, number of trials = 200, p-value = 0.07558 ## alternative hypothesis: true probability of success is not equal to 0.3 ## 95 percent confidence interval: ## 0.29350470.4307139 ## sample estimates: ## probability of success ## 0.36
1 2 3 4 5 6
# H0:A不优于B,H1:A优于B→统计A比B好的数目m n =200# 总次数 m =72#成功次数 a =0.05 p =0.3 binom.test(m,n, p=0.3,alternative ="greater", conf.level =1-a)# alternative = "greater"→A大于B
1 2 3 4 5 6 7 8 9 10 11
## ## Exact binomial test ## ## data: m and n ## number of successes = 72, number of trials = 200, p-value = 0.03963 ## alternative hypothesis: true probability of success is greater than 0.3 ## 95 percent confidence interval: ## 0.30351471.0000000 ## sample estimates: ## probability of success ## 0.36
配对符号检验
只看A与B的差异的有无,不看差异的具体大小
1 2 3 4 5 6 7 8 9 10 11 12 13
# H0:两个总体分布相同 # 小样本 data =c(1,0,0,0,0,0,0) n_pos =sum(data) n =7# 相等组剔除 p =0.5 p_value =0 i =0 while(i<=n_pos){ p_value = p_value+choose(n,i)*(p)^i*(1-p)^(n-i) i = i+1 } p_value
1
## [1] 0.0625
1 2 3
# 大样本示例 a =0.05 (n_pos-n/2)/sqrt(n/4)#U统计量
1
## [1] -1.889822
1
-qnorm(a/2)
1
## [1] 1.959964
1
qnorm(a/2)
1
## [1] -1.959964
1 2 3 4
# 别听上面的了,不如用wilcox做 before <-c(65,70,68,72,67,69,71,70,66,68) after <-c(68,69,67,75,65,70,72,70,67,68) wilcox.test(before, after, paired =TRUE)# 这里是配对的关键
## ## Wilcoxon signed rank test with continuity correction ## ## data: before and after ## V = 12, p-value = 0.4291 ## alternative hypothesis: true location shift is not equal to 0
data =c(44.21,45.30,46.39,49.47,51.05,53.16,53.26,54.37,57.16,67.37,71.05,87.37) # 目测是否为正态 density = density(data) plot(density)
1
hist(data)
1 2
# 目测不是正态 wilcox.test(data,mu =45.30, alternative ="greater")
1 2
## Warning in wilcox.test.default(data, mu = 45.3, alternative = "greater"): ## 有0时无法计算精确的p值
1 2 3 4 5 6
## ## Wilcoxon signed rank test with continuity correction ## ## data: data ## V = 65, p-value = 0.00255 ## alternative hypothesis: true location is greater than 45.3
配对符号秩检验
求秩
1 2
data =c(3,6,2,4,7,8,1,3,5) rank(data, na.last =TRUE, ties.method ="average")
## Warning in wilcox.test.default(data1, data2, paired = TRUE, alternative = ## "two.sided"): 无法精确计算带连结的p值
1 2 3 4 5 6
## ## Wilcoxon signed rank test with continuity correction ## ## data: data1 and data2 ## V = 4.5, p-value = 0.2463 ## alternative hypothesis: true location shift is not equal to 0
## Call: ## aov(formula = Sepal.Length ~ Species, data = iris) ## ## Terms: ## Species Residuals ## Sum of Squares 63.2121338.95620 ## Deg. of Freedom 2147 ## ## Residual standard error: 0.5147894 ## Estimated effects may be unbalanced
1
summary(aov1)
1 2 3 4 5
## Df Sum Sq Mean Sq F value Pr(>F) ## Species 263.2131.606119.3 <2e-16 *** ## Residuals 14738.960.265 ## --- ## Signif. codes: 0'***'0.001'**'0.01'*'0.05'.'0.1' '1
n =3# 组数 m =5# 每组的样本数 # 有样本具体数值 data =c(method_A,method_B,method_C) mean_list =c(mean(method_A),mean(method_B),mean(method_C)) mean = mean(data) var_list =c(var(method_A),var(method_B),var(method_C))
# 只有样本平均数与方差 mean_list =c(33.62,37.83,35.10,34.37) var_list =c(5.12,4.87,10.98,7.08) mean = mean(mean_list) n =4 m =10
H =12/(N*(N+1))*sum(m*(list-(N+1)/2)^2) p_value = pchisq(H,df = n-1, lower.tail =FALSE) data.frame(H,p_value)
1 2
## H p_value ## 110.3850.005558094
局部分析:两两t检验→矫正p值(克服I类错误膨胀)
1 2 3 4 5 6 7 8 9 10 11
# 创建数据 yield_A <-c(20,23,25,27,30) yield_B <-c(18,20,19,22,23) yield_C <-c(28,30,32,35,38) # 合并数据 yield <-c(yield_A, yield_B, yield_C) # 创建组别向量 group <- factor(rep(c("A","B","C"), each =5))# each = 5表示每个组别有5个数据 # 进行KW检验 result <- kruskal.test(yield ~ group) print(result)
1 2 3 4 5
## ## Kruskal-Wallis rank sum test ## ## data: yield by group ## Kruskal-Wallis chi-squared = 10.441, df = 2, p-value = 0.005405
1
pairwise.t.test(x = yield, g=group, p.adjust.method ="bonferroni", alternative ="two.sided")
1 2 3 4 5 6 7 8 9 10
## ## Pairwise comparisons using t tests with pooled SD ## ## data: yield and group ## ## A B ## B 0.16030 - ## C 0.012250.00031 ## ## P value adjustment method: bonferroni
双因素方差分析
1 2
data("ToothGrowth") str(ToothGrowth)
1 2 3 4
## 'data.frame': 60 obs. of3 variables: ## $ len : num 4.211.57.35.86.41011.211.25.27 ... ## $ supp: Factor w/ 2 levels "OJ","VC": 2222222222 ... ## $ dose: num 0.50.50.50.50.50.50.50.50.50.5 ...
## Errorin rcorr(as.matrix(iris[, 1:4]), type = "pearson"): 没有"rcorr"这个函数
斯皮尔曼spearman秩相关系数
把数据的秩拿出来算一下皮尔逊相关系数 适用范围:有离群点的数据;相比皮尔逊,有一点处理非线性的能力
1 2 3 4
x =c(1,2,3,4) y =c(2,3,4,8) n =length(x) cor(x = x, y = y, use ="everything", method ="spearman")
1
## [1] 1
1
cor.test(x, y, method ="spearman")
1 2 3 4 5 6 7 8 9
## ## Spearman's rank correlation rho ## ## data: x and y ## S = 0, p-value = 0.08333 ## alternative hypothesis: true rho is not equal to 0 ## sample estimates: ## rho ## 1
kendall-τ相关系数
适用范围:非线性数据 处理非线性数据的能力最强,也能处理线性数据,但是效率比较低
1 2 3 4
x =c(1,2,3,4) y =c(2,3,4,8) n =length(x) cor(x = x, y = y, use ="everything", method ="kendall")
1
## [1] 1
1
cor.test(x, y, method ="kendall")
1 2 3 4 5 6 7 8 9
## ## Kendall's rank correlation tau ## ## data: x and y ## T = 6, p-value = 0.08333 ## alternative hypothesis: true tau is not equal to 0 ## sample estimates: ## tau ## 1
线性回归模型
模型的假设LINE:
L线性性:X与Y之间线性关系
I独立性:每个样本(Xi,Yi)之间独立
N正态性:误差服从正态分布,均值为0
E方差齐性:不同X值对应的误差方差相同(ε $\sim$ N(0,σ^2))
最小二乘:点到直线的距离(垂直于x轴)——残差平方和
1 2 3 4 5 6
x =c(0.5,0.5,0.5,1.0,1.0,1.0,1.5,1.5,1.5) y =c(55,61,62,90,95,103,132,129,145) df = data.frame(x,y) model = lm(y~x,data = df)# coefficients部分:intercept是截距,x是系数,t是检验系数是否显著不为0 # residual standard error:sqrt(Q/(n-2))——Q为残差平方和 summary(model)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
## ## Call: ## lm(formula = y ~ x, data = df) ## ## Residuals: ## Min 1Q Median 3Q Max ## -6.889-3.889-1.8893.11110.111 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 20.8895.4053.8650.00617 ** ## x 76.0005.00415.1891.29e-06 *** ## --- ## Signif. codes: 0'***'0.001'**'0.01'*'0.05'.'0.1' '1 ## ## Residual standard error: 6.128 on 7 degrees of freedom ## Multiple R-squared: 0.9706, Adjusted R-squared: 0.9663 ## F-statistic: 230.7 on 1 and 7 DF, p-value: 1.29e-06
1 2
model = lm(x~y,data = df) summary(model)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
## ## Call: ## lm(formula = x ~ y, data = df) ## ## Residuals: ## Min 1Q Median 3Q Max ## -0.11440-0.054460.024120.051620.08993 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) -0.23731050.0856579-2.770.0277 * ## y 0.01277040.000840815.191.29e-06 *** ## --- ## Signif. codes: 0'***'0.001'**'0.01'*'0.05'.'0.1' '1 ## ## Residual standard error: 0.07944 on 7 degrees of freedom ## Multiple R-squared: 0.9706, Adjusted R-squared: 0.9663 ## F-statistic: 230.7 on 1 and 7 DF, p-value: 1.29e-06
1 2 3 4 5 6 7
# 手动计算 lxy =sum((x-mean(x))*(y-mean(y))) lxx =sum((x-mean(x))^2) k = lxy/lxx b = mean(y)-k*mean(x) sigma2 =sum((y-b-k*x)^2)/(n-2)# 牺牲两个自由度,所以是n-2 data.frame(lxy,lxx,k,b,sigma2)
1 2
## lxy lxx k b sigma2 ## 11141.57620.88889131.4444
lstatxlims=c(1,40) lstatgrid=seq(1,40) fit <- loess(medv ~ lstat, data = Boston, span=0.2) plot(Boston$lstat, Boston$medv, xlim = lstatxlims, cex =.5, col ="darkgrey") lines(lstatgrid, predict(fit, data.frame(lstat = lstatgrid)), col ="red", lwd =2) legend("topright", legend =c("Span = 0.2","Span = 0.5"), col =c("red","blue"), lty =1, lwd =2, cex =.8)
# 预测 results<-predict(fit, data.frame(lstat =c(20,30,35)), se =TRUE) points(c(20,30,35),results$fit, col ="green", lwd =3)
逻辑回归
Logistic回归虽然名字里带“回归”,但主要用于两分类问题(即输出只有两种,分别代表两个类别)
1 2 3 4 5
y =c(0.1,0.2,0.3,0.4,0.5,0.6,0.7) x1 =c(14,26,24,12,67,26,47) x2 =c(12,46,23,56,35,12,13) dataset = data.frame(x1,x2,y) model = glm(y ~ x1+x2,family ="binomial", data = dataset)
1
## Warning in eval(family$initialize): non-integer #successes in a binomial glm!
1
summary(model)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
## ## Call: ## glm(formula = y ~ x1 + x2, family = "binomial", data = dataset) ## ## Coefficients: ## Estimate Std. Error z value Pr(>|z|) ## (Intercept) -0.9251832.144832-0.4310.666 ## x1 0.0251040.0437040.5740.566 ## x2 -0.0098130.049882-0.1970.844 ## ## (Dispersion parameter for binomial family taken to be 1) ## ## Null deviance: 1.24939 on 6 degrees of freedom ## Residual deviance: 0.84707 on 4 degrees of freedom ## AIC: 13.973 ## ## Number of Fisher Scoring iterations: 4
1 2
# 能不能用过去五天的交易投资回报率预测今天的涨跌? library(ISLR2)
1
## Error in library(ISLR2): 不存在叫'ISLR2'这个名称的程序包
1
attach(Smarket)
1
## Error: 找不到对象'Smarket'
1 2
glm.fits <- glm( Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Smarket, family ="binomial")
1
## Error in eval(mf, parent.frame()): 找不到对象'Smarket'