glm`predict（）`错误：没有适用于'；预测'；适用于“类”的对象；列表“；
我在控制输入预测函数的对象类型时遇到问题。下面是我的简化函数，它生成glm`predict（）`错误：没有适用于'；预测'；适用于“类”的对象；列表“；,r,regression,glm,predict,R,Regression,Glm,Predict,我在控制输入预测函数的对象类型时遇到问题。下面是我的简化函数，它生成glm对象 fitOneSample <- function(x,data,sampleSet) { #how big of a set are we going to analyze? Pick a number between 5,000 & 30,000, then select that many rows to study sampleIndices <- 1:5000 #now ra
glmfitOneSample <- function(x,data,sampleSet)
{
#how big of a set are we going to analyze? Pick a number between 5,000 & 30,000, then select that many rows to study
sampleIndices <- 1:5000
#now randomly pick which columns to study
colIndices <- 1:10
xnames <- paste(names(data[,colIndices]),sep = "")
formula <- as.formula(paste("target ~ ", paste(xnames,collapse = "+")))
glm(formula,family=binomial(link=logit),data[sampleIndices,])
}
myFit <- fitOneSample(1,data,sampleSet)
fits <- sapply(1:2,fitOneSample,data,sampleSet)
all.equal(myFit,fits[,1]) #different object types
#this works
probability <- predict(myFit,newdata = data)
#this doesn't
probability2 <- predict(fits[,1],newdata = data)
# Error in UseMethod("predict") :
# no applicable method for 'predict' applied to an object of class "list"
fitOneSample我想我现在可以恢复你的状况了
fits <- sapply(names(trees),
function (y) do.call(lm, list(formula = paste0(y, " ~ ."), data = trees)))
由于我们使用了sapply
，每次迭代都返回相同的lm
对象或长度为12的列表，因此结果将简化为12*3
矩阵：
class(fits)
# "matrix"
dim(fits)
# 12 3
矩阵索引fits[，1]
有效
如果您选中str（fits[，1]）
，它看起来几乎像一个正常的lm
对象。但如果您进一步检查：
class(fits[, 1])
# "list"
Em？它没有“lm”类因此，S3
分派方法在调用通用函数predict
时将失败：
predict(x)
#Error in UseMethod("predict") :
# no applicable method for 'predict' applied to an object of class "list"
这可以看作是一个很好的例子，说明sapply
具有破坏性。我们想要lappy
，或者至少，sapply（…，simplify=FALSE）
：
您可以通过以下方式修复代码：
fits <- lapply(1:2,fitOneSample,data,sampleSet)
probability2 <-predict(fits[[1]],newdata = data)
fits试试fits谢谢@cryo111。很好，谢谢你。我的主要问题是在循环期间保留结构。你的评论让我意识到sapply破坏了结构-很高兴知道。我修改了先前的函数以使用lappy。
fits <- lapply(names(trees),
function (y) do.call(lm, list(formula = paste0(y, " ~ ."), data = trees)))
class(fits[[1]])
# "lm"
predict(fits[[1]])
# 1 2 3 4 5 6 7 8
# 9.642878 9.870295 9.941744 10.742507 10.801587 10.886282 10.859264 10.957380
# 9 10 11 12 13 14 15 16
#11.588754 11.289186 11.946525 11.458400 11.536472 11.835338 11.133042 11.783583
# 17 18 19 20 21 22 23 24
#13.547349 12.252715 12.603162 12.765403 14.002360 13.364889 14.535617 15.016944
# 25 26 27 28 29 30 31
#15.628799 17.945166 17.958236 18.556671 17.229448 17.131858 21.888147
fits <- lapply(1:2,fitOneSample,data,sampleSet)
probability2 <-predict(fits[[1]],newdata = data)