r - 使用 gbm() 进行分类 - 错误

Question

cancer <- read.csv('breast-cancer-wisconsin.data', header = FALSE, na.strings="?")
cancer <- cancer[complete.cases(cancer),]
names(cancer)[11] <- "class"
cancer[, 11] <- factor(cancer[, 11], labels = c("benign", "malignant"))
library(gbm)

• 数据
• 资料说明

首先，我使用 complete.cases 删除“NA”值，并将第十一列“类”作为因子。我想使用“类”作为响应变量和其他列，除了第一个，作为预测变量。

在我第一次尝试时，我输入了：

boost.cancer <- gbm(class ~ .-V1, data = cancer, distribution = "bernoulli") 

Error in gbm.fit(x, y, offset = offset, distribution = distribution, w = w,  : 
Bernoulli requires the response to be in {0,1}

然后，我使用类的对比而不是类。

boost.cancer <- gbm(contrasts(class) ~ .-V1, distribution = "bernoulli", data = cancer)

Error in model.frame.default(formula = contrasts(class) ~ . - V1, data = cancer,  : 
variable lengths differ (found for 'V1')

我该如何纠正这些错误？我确定我的方法有问题。

Answer 1

正如错误所说，您的响应不在 [0,1] 中。您可以这样做而不是创建因子：

> cancer$class <- (cancer$class -2)/2

> boost.cancer <- gbm(class ~ .-V1, data = cancer, distribution = "bernoulli")
> boost.cancer
gbm(formula = class ~ . - V1, distribution = "bernoulli", data = cancer)
A gradient boosted model with bernoulli loss function.
100 iterations were performed.
There were 9 predictors of which 4 had non-zero influence.

Answer 2

您还可以使用：

boost.cancer <- gbm((unclass(class)-1) ~ .-V1, data = cancer, distribution = "bernoulli") 摘要(boost.cancer)

在“预测”功能并准确确定混淆矩阵时做类似的事情。