我正在使用插入符号包进行逻辑回归分析。
数据输入为18x6矩阵
除predict()函数外，一切都很好。
R告诉我type参数应该是raw或prob，但raw只是吐出最后一列的精确副本（二项式变量的值）。 prob给了我以下错误：
“dimnames中的错误（out）[[2]]＆lt; - modelFit $ obsLevels： 'dimnames'[2]的长度不等于数组范围 另外：警告信息： 'newdata'有7行，但找到的变量有18行“
install.packages("pbkrtest")
install.packages("caret")
install.packages('e1071', dependencies=TRUE)
#install.packages('caret', dependencies = TRUE)
require(caret)
library(caret)
A=matrix(
c(
64830,18213,4677,24761,9845,17504,22137,12531,5842,28827,51840,4079,1000,2069,969,9173,11646,946,66161,18852,5581,27219,10159,17527,23402,11409,8115,31425,55993,0,0,1890,1430,7873,12779,627,68426,18274,5513,25687,10971,14104,19604,13438,6011,30055,57242,0,0,2190,1509,8434,10492,755,69716,18366,5735,26556,11733,16605,20644,15516,5750,31116,64330,0,0,1850,1679,9233,12000,500,73128,18906,5759,28555,11951,19810,22086,17425,6152,28469,72020,0,0,1400,1750,8599,12000,500,1,1,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1
),
nrow = 18,
ncol = 6,
byrow = FALSE) #"bycol" does NOT exist
################### data set as vectors
a<-c(64830,18213,4677,24761,9845,17504,22137,12531,5842,28827,51840,4079,1000,2069,969,9173,11646,946)
b<-c(66161,18852,5581,27219,10159,17527,23402,11409,8115,31425,55993,0,0,1890,1430,7873,12779,627)
c<-c(68426,18274,5513,25687,10971,14104,19604,13438,6011,30055,57242,0,0,2190,1509,8434,10492,755)
d<-c(69716,18366,5735,26556,11733,16605,20644,15516,5750,31116,64330,0,0,1850,1679,9233,12000,500)
e<-c(73128,18906,5759,28555,11951,19810,22086,17425,6152,28469,72020,0,0,1400,1750,8599,12000,500)
f<-c(1,1,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1)
######################
n<-nrow(A);
K<-ncol(A)-1;
Train <- createDataPartition(f, p=0.6, list=FALSE) #60% of data set is used as training.
training <- A[ Train, ]
testing <- A[ -Train, ]
nrow(training)
#this is the logistic formula:
#estimates from logistic regression characterize the relationship between the predictor and response variable on a log-odds scale
mod_fit <- train(f ~ a + b + c + d +e, data=training, method="glm", family="binomial")
mod_fit
#this isthe exponential function to calculate the odds ratios for each preditor:
exp(coef(mod_fit$finalModel))
predict(mod_fit, newdata=training)
predict(mod_fit, newdata=testing, type="prob")
答案 0 :(得分：0)
我不太清楚，但是A是（a，b，c，d，e，f）的矩阵。所以你不需要创建两个对象。
install.packages("pbkrtest")
install.packages("caret")
install.packages('e1071', dependencies=TRUE)
#install.packages('caret', dependencies = TRUE)
require(caret)
library(caret)
A=matrix(
c(
64830,18213,4677,24761,9845,17504,22137,12531,5842,28827,51840,4079,1000,2069,969,9173,11646,946,66161,18852,5581,27219,10159,17527,23402,11409,8115,31425,55993,0,0,1890,1430,7873,12779,627,68426,18274,5513,25687,10971,14104,19604,13438,6011,30055,57242,0,0,2190,1509,8434,10492,755,69716,18366,5735,26556,11733,16605,20644,15516,5750,31116,64330,0,0,1850,1679,9233,12000,500,73128,18906,5759,28555,11951,19810,22086,17425,6152,28469,72020,0,0,1400,1750,8599,12000,500,1,1,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1
),
nrow = 18,
ncol = 6,
byrow = FALSE) #"bycol" does NOT exist
A <- data.frame(A)
colnames(A) <- c('a','b','c','d','e','f')
A$f <- as.factor(A$f)
Train <- createDataPartition(A$f, p=0.6, list=FALSE) #60% of data set is used as training.
training <- A[ Train, ]
testing <- A[ -Train, ]
nrow(training)
要预测变量，您必须输入解释变量，而不是预测变量
mod_fit <- train(f ~ a + b + c + d +e, data=training, method="glm", family="binomial")
mod_fit
#this isthe exponential function to calculate the odds ratios for each preditor:
exp(coef(mod_fit$finalModel))
predict(mod_fit, newdata=training[,-which(colnames(training)=="f")])
predict(mod_fit, newdata=testing[,-which(colnames(testing)=="f")])
答案 1 :(得分：0)
简短回答，您不应在f等式中包含已解释的变量predict。所以你应该这样做：
predict(mod_fit, newdata=training[, -ncol(training])
predict(mod_fit, newdata=testing[, -ncol(testing])
警告消息'newdata' had 11 rows but variables found have 18 rows的问题是因为您使用整个数据集（18个观察值）运行回归，但仅使用其中的一部分（11或7）进行预测。
编辑：为了简化数据创建和我们可以执行的glm流程：
library(caret)
A <- data.frame(a = c(64830,18213,4677,24761,9845,17504,22137,12531,5842,28827,51840,4079,1000,2069,969,9173,11646,946),
b = c(66161,18852,5581,27219,10159,17527,23402,11409,8115,31425,55993,0,0,1890,1430,7873,12779,627),
c = c(68426,18274,5513,25687,10971,14104,19604,13438,6011,30055,57242,0,0,2190,1509,8434,10492,755),
d = c(69716,18366,5735,26556,11733,16605,20644,15516,5750,31116,64330,0,0,1850,1679,9233,12000,500),
e = c(73128,18906,5759,28555,11951,19810,22086,17425,6152,28469,72020,0,0,1400,1750,8599,12000,500),
f = c(1,1,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1))
Train <- createDataPartition(f, p=0.6, list=FALSE) #60% of data set is used as training.
training <- A[ Train, ]
testing <- A[ -Train, ]
mod_fit <- train(f ~ a + b + c + d + e, data=training, method="glm", family="binomial")
答案 2 :(得分：0)
我尝试运行逻辑回归模型。我写了这段代码：
vtk_volume_mapper_->RemoveAllClippingPlanes();
vtk_volume_mapper_->RemoveAllInputs();
vtk_volume_mapper_->ReleaseGraphicsResources(vtk_render_window_);
vtk_volume_->ReleaseGraphicsResources(vtk_render_window_);
vtk_renderer_->ReleaseGraphicsResources(vtk_render_window_);我收到了最后一条命令的消息：
install.packages('caret')
library(caret)
setwd('C:\\Users\\BAHOZ\\Documents\\')
D<-read.csv(file = "D.csv",header = T)
D<-read.csv(file = 'DataSet.csv',header=T)
names(D)
set.seed(111134)
Train<-createDataPartition(D$X, p=0.7,list = FALSE)
training<-D[Train,]
length(training$age)
testing<-D[-Train,]
length(testing$age)
mod_fit<-train(X~age + gender + total.Bilirubin + direct.Bilirubin + total.proteins + albumin + A.G.ratio+SGPT + SGOT + Alkphos,data=training,method="glm", family="binomial")
summary(mod_fit)
exp(coef(mod_fit$finalModel))
通过运行此命令，我可以预测我的数据，
(Intercept) age gender total.Bilirubin direct.Bilirubin total.proteins albumin A.G.ratio
0.01475027 1.01596886 1.03857883 1.00022899 1.78188072 1.00065332 1.01380334 1.00115742
SGPT SGOT Alkphos
3.93498241 0.05616662 38.29760014
但是如果我设置了predict(mod_fit , newdata=testing)
或type="prob"
type="raw"它出错了：
dimnames（out）中的错误<-
predict(mod_fit , newdata=testing, type = "prob")：'dimnames'[2]的长度不等于数组范围