R 从逻辑回归列表中提取系数的所有标准误差
我想从逻辑回归模型列表中提取标准误差 这是逻辑回归函数,其设计方式使我可以一次运行多个分析: 但如何提取系数的标准误差?我似乎在str(glmkort)中找不到它? 以下是年龄的str(glmkort),我在这里查找标准错误:R 从逻辑回归列表中提取系数的所有标准误差,r,logistic-regression,coefficients,standard-error,R,Logistic Regression,Coefficients,Standard Error,我想从逻辑回归模型列表中提取标准误差 这是逻辑回归函数,其设计方式使我可以一次运行多个分析: 但如何提取系数的标准误差?我似乎在str(glmkort)中找不到它? 以下是年龄的str(glmkort),我在这里查找标准错误: str(glmkort) List of 6 $ AGE :List of 30 ..$ coefficients : Named num [1:2] -1.17201 -0.00199 .. ..- attr(*, "names")= chr [
str(glmkort)
List of 6
$ AGE :List of 30
..$ coefficients : Named num [1:2] -1.17201 -0.00199
.. ..- attr(*, "names")= chr [1:2] "(Intercept)" "x"
..$ residuals : Named num [1:40] -1.29 -1.29 -1.29 -1.29 4.39 ...
.. ..- attr(*, "names")= chr [1:40] "1" "2" "3" "4" ...
..$ fitted.values : Named num [1:40] 0.223 0.225 0.225 0.225 0.228 ...
.. ..- attr(*, "names")= chr [1:40] "1" "2" "3" "4" ...
..$ effects : Named num [1:40] 3.2662 -0.0282 -0.4595 -0.4464 2.042 ...
.. ..- attr(*, "names")= chr [1:40] "(Intercept)" "x" "" "" ...
..$ R : num [1:2, 1:2] -2.64 0 -86.01 14.18
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : chr [1:2] "(Intercept)" "x"
.. .. ..$ : chr [1:2] "(Intercept)" "x"
..$ rank : int 2
..$ qr :List of 5
.. ..$ qr : num [1:40, 1:2] -2.641 0.158 0.158 0.158 0.159 ...
.. .. ..- attr(*, "dimnames")=List of 2
.. .. .. ..$ : chr [1:40] "1" "2" "3" "4" ...
.. .. .. ..$ : chr [1:2] "(Intercept)" "x"
.. ..$ rank : int 2
.. ..$ qraux: num [1:2] 1.16 1.01
.. ..$ pivot: int [1:2] 1 2
.. ..$ tol : num 1e-11
.. ..- attr(*, "class")= chr "qr"
..$ family :List of 12
.. ..$ family : chr "binomial"
.. ..$ link : chr "logit"
.. ..$ linkfun :function (mu)
.. ..$ linkinv :function (eta)
.. ..$ variance :function (mu)
.. ..$ dev.resids:function (y, mu, wt)
.. ..$ aic :function (y, n, mu, wt, dev)
.. ..$ mu.eta :function (eta)
.. ..$ initialize: expression({ if (NCOL(y) == 1) { if (is.factor(y)) y <- y != levels(y)[1L] n <- rep.int(1, nobs) y[weights == 0] <- 0 if (any(y < 0 | y > 1)) stop("y values must be 0 <= y <= 1") mustart <- (weights * y + 0.5)/(weights + 1) m <- weights * y if (any(abs(m - round(m)) > 0.001)) warning("non-integer #successes in a binomial glm!") } else if (NCOL(y) == 2) { if (any(abs(y - round(y)) > 0.001)) warning("non-integer counts in a binomial glm!") n <- y[, 1] + y[, 2] y <- ifelse(n == 0, 0, y[, 1]/n) weights <- weights * n mustart <- (n * y + 0.5)/(n + 1) } else stop("for the binomial family, y must be a vector of 0 and 1's\n", "or a 2 column matrix where col 1 is no. successes and col 2 is no. failures") })
.. ..$ validmu :function (mu)
.. ..$ valideta :function (eta)
.. ..$ simulate :function (object, nsim)
.. ..- attr(*, "class")= chr "family"
..$ linear.predictors: Named num [1:40] -1.25 -1.24 -1.24 -1.24 -1.22 ...
.. ..- attr(*, "names")= chr [1:40] "1" "2" "3" "4" ...
..$ deviance : num 42.7
..$ aic : num 46.7
..$ null.deviance : num 42.7
..$ iter : int 4
..$ weights : Named num [1:40] 0.173 0.174 0.174 0.174 0.176 ...
.. ..- attr(*, "names")= chr [1:40] "1" "2" "3" "4" ...
..$ prior.weights : Named num [1:40] 1 1 1 1 1 1 1 1 1 1 ...
.. ..- attr(*, "names")= chr [1:40] "1" "2" "3" "4" ...
..$ df.residual : int 38
..$ df.null : int 39
..$ y : Named num [1:40] 0 0 0 0 1 0 1 0 0 0 ...
.. ..- attr(*, "names")= chr [1:40] "1" "2" "3" "4" ...
..$ converged : logi TRUE
..$ boundary : logi FALSE
..$ model :'data.frame': 40 obs. of 2 variables:
.. ..$ ldata$DFREE: int [1:40] 0 0 0 0 1 0 1 0 0 0 ...
.. ..$ x : int [1:40] 39 33 33 32 24 30 39 27 40 36 ...
.. ..- attr(*, "terms")=Classes 'terms', 'formula' length 3 ldata$DFREE ~ x
.. .. .. ..- attr(*, "variables")= language list(ldata$DFREE, x)
.. .. .. ..- attr(*, "factors")= int [1:2, 1] 0 1
.. .. .. .. ..- attr(*, "dimnames")=List of 2
.. .. .. .. .. ..$ : chr [1:2] "ldata$DFREE" "x"
.. .. .. .. .. ..$ : chr "x"
.. .. .. ..- attr(*, "term.labels")= chr "x"
.. .. .. ..- attr(*, "order")= int 1
.. .. .. ..- attr(*, "intercept")= int 1
.. .. .. ..- attr(*, "response")= int 1
.. .. .. ..- attr(*, ".Environment")=<environment: 0x017a5674>
.. .. .. ..- attr(*, "predvars")= language list(ldata$DFREE, x)
.. .. .. ..- attr(*, "dataClasses")= Named chr [1:2] "numeric" "numeric"
.. .. .. .. ..- attr(*, "names")= chr [1:2] "ldata$DFREE" "x"
..$ call : language glm(formula = ldata$DFREE ~ x, family = binomial)
..$ formula :Class 'formula' length 3 ldata$DFREE ~ x
.. .. ..- attr(*, ".Environment")=<environment: 0x017a5674>
..$ terms :Classes 'terms', 'formula' length 3 ldata$DFREE ~ x
.. .. ..- attr(*, "variables")= language list(ldata$DFREE, x)
.. .. ..- attr(*, "factors")= int [1:2, 1] 0 1
.. .. .. ..- attr(*, "dimnames")=List of 2
.. .. .. .. ..$ : chr [1:2] "ldata$DFREE" "x"
.. .. .. .. ..$ : chr "x"
.. .. ..- attr(*, "term.labels")= chr "x"
.. .. ..- attr(*, "order")= int 1
.. .. ..- attr(*, "intercept")= int 1
.. .. ..- attr(*, "response")= int 1
.. .. ..- attr(*, ".Environment")=<environment: 0x017a5674>
.. .. ..- attr(*, "predvars")= language list(ldata$DFREE, x)
.. .. ..- attr(*, "dataClasses")= Named chr [1:2] "numeric" "numeric"
.. .. .. ..- attr(*, "names")= chr [1:2] "ldata$DFREE" "x"
..$ data :<environment: 0x017a5674>
..$ offset : NULL
..$ control :List of 3
.. ..$ epsilon: num 1e-08
.. ..$ maxit : num 25
.. ..$ trace : logi FALSE
..$ method : chr "glm.fit"
..$ contrasts : NULL
..$ xlevels : Named list()
..- attr(*, "class")= chr [1:2] "glm" "lm"
$ BECK :List of 30
使用
?glm
## Dobson (1990) Page 93: Randomized Controlled Trial :
counts <- c(18,17,15,20,10,20,25,13,12)
outcome <- gl(3,1,9)
treatment <- gl(3,3)
glm.D93 <- glm(counts ~ outcome + treatment, family=poisson())
## copy twice to a list to illustrate
lmod <- list(mod1 = glm.D93, mod2 = glm.D93)
从调用summary()
或者我们可以让summary()
计算标准错误(以及更多),然后使用lappy()
或sapply()
应用一个匿名函数,该函数提取coef(summary(x))
并获取第二列(其中存储标准错误)
给
> lapply(lmod, function(x) coef(summary(x))[,2])
$mod1
(Intercept) outcome2 outcome3 treatment2 treatment3
0.1708987 0.2021708 0.1927423 0.2000000 0.2000000
$mod2
(Intercept) outcome2 outcome3 treatment2 treatment3
0.1708987 0.2021708 0.1927423 0.2000000 0.2000000
而sapply()
将给出:
> sapply(lmod, function(x) coef(summary(x))[,2])
mod1 mod2
(Intercept) 0.1708987 0.1708987
outcome2 0.2021708 0.2021708
outcome3 0.1927423 0.1927423
treatment2 0.2000000 0.2000000
treatment3 0.2000000 0.2000000
根据您想要执行的操作,您可以通过一次调用提取系数和标准错误:
> lapply(lmod, function(x) coef(summary(x))[,1:2])
$mod1
Estimate Std. Error
(Intercept) 3.044522e+00 0.1708987
outcome2 -4.542553e-01 0.2021708
outcome3 -2.929871e-01 0.1927423
treatment2 1.337909e-15 0.2000000
treatment3 1.421085e-15 0.2000000
$mod2
Estimate Std. Error
(Intercept) 3.044522e+00 0.1708987
outcome2 -4.542553e-01 0.2021708
outcome3 -2.929871e-01 0.1927423
treatment2 1.337909e-15 0.2000000
treatment3 1.421085e-15 0.2000000
但是您可能更喜欢单独使用它们?您应该能够从
summary()
中提取它。因此,您需要编写一个lappy
,在列表的每个项目上调用summary()
。实际上coef(summary(…)
应该这样做。请参见?summary.glmYes,coef(summary(…)
谢谢!正是我需要的!现在,我将坚持你的第二个也是最简单的答案,但我肯定会回到第一部分,你会解释如何“手动”进行更多的计算:)谢谢
## Dobson (1990) Page 93: Randomized Controlled Trial :
counts <- c(18,17,15,20,10,20,25,13,12)
outcome <- gl(3,1,9)
treatment <- gl(3,3)
glm.D93 <- glm(counts ~ outcome + treatment, family=poisson())
## copy twice to a list to illustrate
lmod <- list(mod1 = glm.D93, mod2 = glm.D93)
> lapply(lmod, function(x) sqrt(diag(vcov(x))))
$mod1
(Intercept) outcome2 outcome3 treatment2 treatment3
0.1708987 0.2021708 0.1927423 0.2000000 0.2000000
$mod2
(Intercept) outcome2 outcome3 treatment2 treatment3
0.1708987 0.2021708 0.1927423 0.2000000 0.2000000
lapply(lmod, function(x) coef(summary(x))[,2])
> lapply(lmod, function(x) coef(summary(x))[,2])
$mod1
(Intercept) outcome2 outcome3 treatment2 treatment3
0.1708987 0.2021708 0.1927423 0.2000000 0.2000000
$mod2
(Intercept) outcome2 outcome3 treatment2 treatment3
0.1708987 0.2021708 0.1927423 0.2000000 0.2000000
> sapply(lmod, function(x) coef(summary(x))[,2])
mod1 mod2
(Intercept) 0.1708987 0.1708987
outcome2 0.2021708 0.2021708
outcome3 0.1927423 0.1927423
treatment2 0.2000000 0.2000000
treatment3 0.2000000 0.2000000
> lapply(lmod, function(x) coef(summary(x))[,1:2])
$mod1
Estimate Std. Error
(Intercept) 3.044522e+00 0.1708987
outcome2 -4.542553e-01 0.2021708
outcome3 -2.929871e-01 0.1927423
treatment2 1.337909e-15 0.2000000
treatment3 1.421085e-15 0.2000000
$mod2
Estimate Std. Error
(Intercept) 3.044522e+00 0.1708987
outcome2 -4.542553e-01 0.2021708
outcome3 -2.929871e-01 0.1927423
treatment2 1.337909e-15 0.2000000
treatment3 1.421085e-15 0.2000000