Question

我总是冒出很多错误。我仍然在获得价值观，但我不确定它们的准确程度。在修复这些错误时，我尝试从顶部开始并使用我所拥有的变量定义，但是出现了空白。

_.chain(values) . groupBy(name) . map(key) . join('&') . value()

model.frame.default中的错误（formula = insampley~insamplex，data = insample，：变量的无效类型（列表）＆＃39; insampley＆＃39;

contdepdata <- read.delim("/Users/Jkels/Documents/Introduction to 
Computational Statistics/winequality-
red.csv",sep=",",header=TRUE,stringsAsFactors=FALSE)
contdepdata <- na.omit(contdepdata)
numericvars <- c(1:11)
dim(contdepdata)
head(contdepdata)

insample <- contdepdata[1:500,]
outsample <- contdepdata[501:1000,]
insamplex <- insample[x,]
outsamplex <- outsample[x,]
insampley <- insample[y,]
outsampley <- outsample[y,]

x <- insample[,numericvars]
y <- insample$quality

lambdalevels <- 10^seq(7,-2,length=100)

lmout <- lm(insampley~insamplex, data=insample)

cbind（1，outsamplex）中的错误％*％lmout $ coefficients：需要数字/复杂矩阵/向量参数

Answer 1

我认为这与你想要的很接近。 lm使用的公式/数据框策略与glmnet使用的矩阵策略之间存在巨大差异，我认为这是绊倒你的一部分。坦率地说，我发现glmnet对矩阵的使用令人失望（想想也许这是必要的，我以前从未使用过它。）

library(dplyr)
library(magrittr)
library(glmnet)

data = 
  data_frame(a = rnorm(10),
             b = rnorm(10),
             c = "delete",
             quality = rnorm(10)) %>%
  extract(c(1:2, which(names(.) == "quality")))

data.training = 
  data %>%
  slice(1:5)

data.test = 
  data %>%
  slice(6:10)

quality.predict = 
  ( quality ~ . ) %>%
  lm(data.training) %>%
  predict(data.test) %>%
  unname

MSE.standard = 
  data.test %>%
  summarize(MSE = sum( (quality - quality.predict) ^ 2 / n() ) ) %>%
  use_series(MSE)

lambda = 10^seq(7, -2, length = 10)

regression.lasso = cv.glmnet(data.training %>% 
                               select(., -quality) %>% 
                               as.matrix,
                             data.training$quality,
                             alpha = 1,
                             lambda = lambda )

quality.lasso = 
  predict(regression.lasso$glmnet.fit,
          regression.lasso$lambda.min,
          newx = data.test %>%
            select(-quality) %>%
            as.matrix
          )[,1]

MSE.lasso = 
  data.training %>% 
  summarize(MSE = sum( (quality - quality.lasso) ^ 2 / n() ) ) %>%
  use_series(MSE)

R中有很多错误

1 个答案: