在tidy model of data science (TM)中实施的modelr中，重新抽样数据使用list-columns进行整理：
bhathiya@bhathiya-x1:/$ curl -v -k -X OPTIONS https://localhost:8243/userinfo
* Trying 127.0.0.1...
* Connected to localhost (127.0.0.1) port 8243 (#0)
* found 173 certificates in /etc/ssl/certs/ca-certificates.crt
* found 697 certificates in /etc/ssl/certs
* ALPN, offering http/1.1
* SSL connection using TLS1.2 / ECDHE_RSA_AES_128_GCM_SHA256
* server certificate verification SKIPPED
* server certificate status verification SKIPPED
* common name: localhost (matched)
* server certificate expiration date OK
* server certificate activation date OK
* certificate public key: RSA
* certificate version: #3
* subject: C=US,ST=CA,L=Mountain View,O=WSO2,CN=localhost
* start date: Fri, 19 Feb 2010 07:02:26 GMT
* expire date: Tue, 13 Feb 2035 07:02:26 GMT
* issuer: C=US,ST=CA,L=Mountain View,O=WSO2,CN=localhost
* compression: NULL
* ALPN, server did not agree to a protocol
> OPTIONS /userinfo HTTP/1.1
> Host: localhost:8243
> User-Agent: curl/7.47.0
> Accept: */*
>
< HTTP/1.1 200 OK
< Accept: */*
< Access-Control-Allow-Origin: *
< Access-Control-Allow-Methods: GET
< Host: localhost:8243
< Access-Control-Allow-Headers: authorization,Access-Control-Allow-Origin,Content-Type,SOAPAction
< Date: Sun, 23 Oct 2016 14:43:27 GMT
< Transfer-Encoding: chunked
<
可以为列表列library(modelr)
library(tidyverse)
# create the k-folds
df_heights_resampled = heights %>%
crossv_kfold(k = 10, id = "Resample ID")
中的每个训练数据集map建模，并通过train ping列表列{{}来计算性能指标1}}。
如果需要对多个模型进行此操作，则需要对每个模型重复此操作。
map给出：
test如果要探索的模型数量很大，这可能会非常麻烦。 # 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)
)
提供> 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
函数，允许迭代多个模型（由多个公式表征），但似乎不允许模型中的modelr列表列以上。我假设fit_with函数族中的一个函数可以实现这一点（train？），但是无法弄清楚如何。
答案 0 :(得分：3)
您可以使用map和lazyeval::interp以编程方式构建呼叫。我很好奇是否有纯purrr解决方案，但问题是您要创建多个列，并且需要多次调用。也许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
答案 1 :(得分：0)
受到my own question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]]
}
然后直接估计每个CV折叠和公式对的多个模型：
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，最好与cv_gm_results一起使用，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>
<强>更新
事实证明，est_model（）不是必需的，purrr提供符合我们目的的at_depth()：
cv_gm_results <- cv_gm_results %>%
mutate(fit=map2(train, fmla, ~at_depth(.x, 0, .y)),
rmse=map2_dbl(fit, test, .f = rmse))