7

在 中实现的数据科学 (TM) 的整洁模型中modelr，重新采样的数据使用list-columns进行组织：

library(modelr)
library(tidyverse)

# create the k-folds
df_heights_resampled = heights %>% 
  crossv_kfold(k = 10, id = "Resample ID")

可以对maplist-column 中的每个训练数据集建立模型，并通过ping list-columntrain来计算性能指标。maptest

如果需要对多个模型执行此操作，则需要对每个模型重复此操作。

# create a list of formulas 
formulas_heights = formulas(
  .response = ~ income, 
  model1  = ~ height + weight + marital + sex,
  model2 = ~ height + weight + marital + sex + education
) 

# fit each of the models in the list of formulas
df_heights_resampled = df_heights_resampled %>% 
  mutate(
    model1 = map(train, function(train_data) {
      lm(formulas_heights[[1]], data = train_data)
    }),
    model2 = map(train, function(train_data) {
      lm(formulas_heights[[2]], data = train_data)
    })
  )

# score the models on the test sets
df_heights_resampled = df_heights_resampled %>% 
  mutate(
    rmse1 = map2_dbl(.x = model1, .y = test, .f = rmse),
    rmse2 = map2_dbl(.x = model2, .y = test, .f = rmse)
  )

这使：

> df_heights_resampled
# A tibble: 10 × 7
            train           test `Resample ID`   model1   model2    rmse1    rmse2
           <list>         <list>         <chr>   <list>   <list>    <dbl>    <dbl>
1  <S3: resample> <S3: resample>            01 <S3: lm> <S3: lm> 58018.35 53903.99
2  <S3: resample> <S3: resample>            02 <S3: lm> <S3: lm> 55117.37 50279.38
3  <S3: resample> <S3: resample>            03 <S3: lm> <S3: lm> 49005.82 44613.93
4  <S3: resample> <S3: resample>            04 <S3: lm> <S3: lm> 55437.07 51068.90
5  <S3: resample> <S3: resample>            05 <S3: lm> <S3: lm> 48845.35 44673.88
6  <S3: resample> <S3: resample>            06 <S3: lm> <S3: lm> 58226.69 54010.50
7  <S3: resample> <S3: resample>            07 <S3: lm> <S3: lm> 56571.93 53322.41
8  <S3: resample> <S3: resample>            08 <S3: lm> <S3: lm> 46084.82 42294.50
9  <S3: resample> <S3: resample>            09 <S3: lm> <S3: lm> 59762.22 54814.55
10 <S3: resample> <S3: resample>            10 <S3: lm> <S3: lm> 45328.48 41882.79

问题：

如果要探索的模型数量很大，这会很快变得很麻烦。modelr提供fit_with允许迭代多个模型（以多个公式为特征）但似乎不允许像train上面的模型中那样的列表列的函数。我假设其中一个*map*函数系列可以实现这一点（invoke_map？），但无法弄清楚如何。

4

2 回答 2

3

map您可以使用和以编程方式构建调用lazyeval::interp。我很好奇是否有一个纯粹的purrr解决方案，但问题是您想要创建多个列，并且您需要多次调用。也许一个purrr解决方案会创建另一个包含所有模型的列表列。

library(lazyeval)
model_calls <- map(formulas_heights, 
                   ~interp(~map(train, ~lm(form, data = .x)), form = .x))
score_calls <- map(names(model_calls), 
                   ~interp(~map2_dbl(.x = m, .y = test, .f = rmse), m = as.name(.x)))
names(score_calls) <- paste0("rmse", seq_along(score_calls))

df_heights_resampled %>% mutate_(.dots = c(model_calls, score_calls))

# A tibble: 10 × 7
            train           test `Resample ID`   model1   model2    rmse1    rmse2
           <list>         <list>         <chr>   <list>   <list>    <dbl>    <dbl>
1  <S3: resample> <S3: resample>            01 <S3: lm> <S3: lm> 44720.86 41452.07
2  <S3: resample> <S3: resample>            02 <S3: lm> <S3: lm> 54174.38 48823.03
3  <S3: resample> <S3: resample>            03 <S3: lm> <S3: lm> 56854.21 52725.62
4  <S3: resample> <S3: resample>            04 <S3: lm> <S3: lm> 53312.38 48797.48
5  <S3: resample> <S3: resample>            05 <S3: lm> <S3: lm> 61883.90 57469.17
6  <S3: resample> <S3: resample>            06 <S3: lm> <S3: lm> 55709.83 50867.26
7  <S3: resample> <S3: resample>            07 <S3: lm> <S3: lm> 53036.06 48698.07
8  <S3: resample> <S3: resample>            08 <S3: lm> <S3: lm> 55986.83 52717.94
9  <S3: resample> <S3: resample>            09 <S3: lm> <S3: lm> 51738.60 48006.74
10 <S3: resample> <S3: resample>            10 <S3: lm> <S3: lm> 45061.22 41480.35
于 2016-10-23T20:20:00.700 回答
0

受我自己的问题的启发，我认为这个问题有类似的方法。

首先，定义一个函数，该函数可以在列表列结构中获取数据和公式的参数，并使用输入估计模型：

est_model <- function(data, formula, ...) {
  map(list(data), formula, ...)[[1]]
}

然后用每个 CV 折叠和公式对估计多个模型会很简单：

library(gapminder)
library(tidyverse)
library(modelr)

cv_gm <- gapminder %>% 
  crossv_kfold(k = 10, id = "Resample ID")

# Assume 4 different formulae
formulae_tbl <- tibble::frame_data(
  ~model, ~fmla,
  1, ~lm(lifeExp ~ year, data = .),
  2, ~lm(lifeExp ~ year + pop, data = .),
  3, ~lm(lifeExp ~ year + gdpPercap, data = .),
  4, ~lm(lifeExp ~ year + pop + gdpPercap, data = .)
)

cv_gm_results <- cv_gm %>% 
  tidyr::crossing(formulae_tbl)

cv_gm_results <- cv_gm_results %>% 
  mutate(fit=map2(train, fmla, est_model),
         rmse=map2_dbl(fit, test, .f = rmse))

可以说，根据整洁的数据哲学，最好继续使用cv_gm_results，但如果你想要它在原始问题中的形状（h/t this question）：

cv_gm_results %>% 
  select(`Resample ID`, model, fit, rmse) %>% 
  gather(variable, value, fit, rmse) %>%
  unite(temp, variable, model, sep="") %>%
  spread(temp, value) %>% 
  mutate_at(.cols=vars(starts_with("rmse")), .funs=flatten_dbl)

# A tibble: 10 × 9
   `Resample ID`     fit1     fit2     fit3     fit4    rmse1    rmse2
           <chr>   <list>   <list>   <list>   <list>    <dbl>    <dbl>
1             01 <S3: lm> <S3: lm> <S3: lm> <S3: lm> 11.32344 11.32201
2             02 <S3: lm> <S3: lm> <S3: lm> <S3: lm> 11.34626 11.33175
3             03 <S3: lm> <S3: lm> <S3: lm> <S3: lm> 11.62480 11.60221
4             04 <S3: lm> <S3: lm> <S3: lm> <S3: lm> 10.80946 10.81421
5             05 <S3: lm> <S3: lm> <S3: lm> <S3: lm> 11.52413 11.52384
6             06 <S3: lm> <S3: lm> <S3: lm> <S3: lm> 12.10914 12.08134
7             07 <S3: lm> <S3: lm> <S3: lm> <S3: lm> 11.97641 12.00809
8             08 <S3: lm> <S3: lm> <S3: lm> <S3: lm> 12.30191 12.31489
9             09 <S3: lm> <S3: lm> <S3: lm> <S3: lm> 11.96970 11.95617
10            10 <S3: lm> <S3: lm> <S3: lm> <S3: lm> 11.30289 11.30294
# ... with 2 more variables: rmse3 <dbl>, rmse4 <dbl>

更新

事实证明 est_model() 不是必需的，purrr只要at_depth()符合我们的目的：

cv_gm_results <- cv_gm_results %>% 
  mutate(fit=map2(train, fmla, ~at_depth(.x, 0, .y)),
         rmse=map2_dbl(fit, test, .f = rmse))
于 2017-05-09T00:15:54.883 回答