我想比较使用相同预测变量但不同模型参数的一组模型的模型性能。这似乎是使用broom
来创建整洁输出的地方,但我无法弄明白。
这里有一些非工作代码可以帮助我们思考:
seq(1:10) %>%
do(fit = knn(train_Market, test_Market, train_Direction, k=.), score = mean(fit==test_Direction)) %>%
tidy()
对于更多背景信息,这是我们试图整合的ISLR实验室的一部分。您可以在此处查看整个实验室:https://github.com/AmeliaMN/tidy-islr/blob/master/lab3/lab3.Rmd
[更新:可重现的例子]这里很难做出一个最小的例子,因为在模型拟合之前需要进行数据争论,但这应该是可重现的:
library(ISLR)
library(dplyr)
train = Smarket %>%
filter(Year < 2005)
test = Smarket %>%
filter(Year >= 2005)
train_Market = train %>%
select(Lag1, Lag2)
test_Market = test %>%
select(Lag1, Lag2)
train_Direction = train %>%
select(Direction) %>%
.$Direction
set.seed(1)
knn_pred = knn(train_Market, test_Market, train_Direction, k=1)
mean(knn_pred==test_Direction)
knn_pred = knn(train_Market, test_Market, train_Direction, k=3)
mean(knn_pred==test_Direction)
knn_pred = knn(train_Market, test_Market, train_Direction, k=4)
mean(knn_pred==test_Direction)
等
答案 0 :(得分:3)
由于你的每个knn(和oracle)的输出都是一个向量,这对于tidyr的obj()
(与purrr的unnest
和map
结合是一个很好的例子:
rep_along
然后将library(class)
library(purrr)
library(tidyr)
set.seed(1)
predictions <- data_frame(k = 1:5) %>%
unnest(prediction = map(k, ~ knn(train_Market, test_Market, train_Direction, k = .))) %>%
mutate(oracle = rep_along(prediction, test_Direction))
变量组织为:
predictions
可以很容易地总结出来:
# A tibble: 1,260 x 3
k prediction oracle
<int> <fctr> <fctr>
1 1 Up Up
2 1 Down Up
3 1 Up Down
4 1 Up Up
5 1 Up Up
6 1 Down Up
7 1 Down Down
8 1 Down Up
9 1 Down Up
10 1 Up Up
# ... with 1,250 more rows
同样,你不需要扫帚,因为每个输出都是一个因素,但如果它是一个模型,你可以使用扫帚的predictions %>%
group_by(k) %>%
summarize(accuracy = mean(prediction == oracle))
或tidy
,然后以类似的方式取消它。 / p>
这种方法的一个重要方面是它对许多参数组合很灵活,通过将它们与tidyr的augment
(或crossing
)相结合,并使用expand.grid
将函数应用于每个参数行。例如,您可以在invoke_rows
旁边尝试l
的变体:
k
返回:
crossing(k = 2:5, l = 0:1) %>%
invoke_rows(knn, ., train = train_Market, test = test_Market, cl = train_Direction) %>%
unnest(prediction = .out) %>%
mutate(oracle = rep_along(prediction, test_Direction)) %>%
group_by(k, l) %>%
summarize(accuracy = mean(prediction == oracle))