让我们说我想对不同样本的mtcars数据集运行几次线性回归模型。 这个想法是，对于for循环中的每次迭代，每次运行线性回归时都要存储predict（）方法的结果 对于其他样品。一个小例子如下：
## Perform model once on a Sample and use model on full dataset:
Sample_Size <- 10
Sample <- mtcars[sample(nrow(mtcars), Sample_Size), ]
Model <- lm(formula = mpg ~ wt, data = Sample)
Predictions <- predict(Model,newdata=mtcars)
## Gets us a list with predicted wt for each car:
Predictions <- t(Predictions)
这产生
> Predictions
Mazda RX4 Mazda RX4 Wag Datsun 710 Hornet 4 Drive Hornet Sportabout
[1,] 25.80494 23.89161 28.05592 21.34051 19.65228
Valiant Duster 360 Merc 240D Merc 230 Merc 280 Merc 280C Merc 450SE
[1,] 19.50221 18.67685 21.52809 21.82822 19.65228 19.65228 14.92523
Merc 450SL Merc 450SLC Cadillac Fleetwood Lincoln Continental
[1,] 17.47633 17.10117 6.071394 4.765828
.... and so on for other cars
我想每次在for循环中多次执行此过程 选择其他样本并获得相应的Predictions（）列表， 并将所有Predictions（）结果按行存储在数据框中。
假设我为两个不同的样本运行模型。结果数据框的每一行都应该是该样本的上面的结果，例如：
Mazda RX4 Mazda RX4 Wag Datsun 710 Hornet 4 Drive Hornet Sportabout
[1,] 25.80494 23.89161 28.05592 21.34051 19.65228
[2,] 22.80492 22.89147 28.05532 21.34231 20.65290
Valiant Duster 360 Merc 240D Merc 230 Merc 280 Merc 280C Merc 450SE
[1,] 19.50221 18.67685 21.52809 21.82822 19.65228 19.65228 14.92523
[2,] 21.83492 23.84147 29.02532 21.34231 20.35290 18.45228 13.92523
... and so on for other cars.
关于如何执行此操作的任何想法？我已经开发出一些东西，但要么 抛出错误或仅存储最后的结果...我在这里想念什么？
这是我到目前为止所拥有的：
### Inside a for loop, to get a dataframe of Predictions:
Bootstrap_times <- 2
Sample_Size <- 10
Predictions <- list()
Results <-vector ("list",Bootstrap_times)## Stores the Predictions for each run
for(i in 1:Bootstrap_times){
### Take a sample
Sample[[i]] <- mtcars[sample(nrow(mtcars), Sample_Size), ]
### Do the regression on the sample
Model[[i]] <- lm(formula = mpg ~ wt, data = Sample[[i]])
### Perform the predict() on the sample
Predictions[[i]] <- predict(Model[[i]],newdata=mtcars)
### put the result as a line on the dataframe Results
Predictions[[i]] <- t(Predictions[[i]])
return(Predictions)
}
但是，我不断得到：
[[<-.data.frame（*tmp*中的错误，i，值=列表（mpg = c（13.3， 10.4 ，：替换有10行，数据有0
答案 0 :(得分：2)
我更喜欢使用magic_for()，但是您也可以很容易地使用R为基数。
这是一个例子：
Bootstrap_times <- 2
Sample_Size <- 10
Sample <- mtcars[sample(nrow(mtcars), Sample_Size), ]
Model <- lm(formula = mpg ~ wt, data = Sample)
Predictions <- predict(Model,newdata=mtcars)
## You like how I line up arrows, right?
Predictions <- t(Predictions)
Predictions <- list()
Results <-vector ("list",Bootstrap_times)## Stores the Predictions for each run
magicfor::magic_for()
for(i in 1:Bootstrap_times){
### Take a sample
Sample[[i]] <- mtcars[sample(nrow(mtcars), Sample_Size), ]
### Do the regression on the sample
Model[[i]] <- lm(formula = mpg ~ wt, data = Sample[[i]])
### Perform the predict() on the sample
put(predict(Model[[i]],newdata=mtcars))
}
tmp<-magicfor::magic_result_as_dataframe()
tmp
i predict(Model[[i]],newdata=mtcars) 1 1 22.858806 2 2 20.922763 3 1 25.136504 4 2 18.341372 5 1 16.633098 6 2 16.481252 7 1 15.646096 8 2 18.531180 9 1 18.834873 10 2 16.633098 11 1 16.633098 12 2 11.849933 13 1 14.431324 14 2 14.051708 15 1 2.890988 16 2 1.569924 17 1 2.169717 18 2 26.047583 19 1 30.489093 20 2 28.818782 21 1 24.035616 22 2 16.025712 23 1 16.671060 24 2 13.596168 25 1 13.558206 26 2 28.059549 27 1 26.503122 28 2 31.263511 29 1 18.683026 30 2 21.719957 31 1 15.646096 32 2 21.644034 33 1 22.978374 34 2 21.584264 35 1 24.618503 36 2 19.725450 37 1 18.495353 38 2 18.386011 39 1 17.784630 40 2 19.862128 41 1 20.080812 42 2 18.495353 43 1 18.495353 44 2 15.051081 45 1 16.909894 46 2 16.636540 47 1 8.599905 48 2 7.648629 49 1 8.080530 50 2 25.274555 51 1 28.472808 52 2 27.270046 53 1 23.825774 54 2 18.057985 55 1 18.522689 56 2 16.308514 57 1 16.281178 58 2 26.723336 59 1 25.602581 60 2 29.030452 61 1 19.971470 62 2 22.158309 63 1 17.784630 64 2 22.103638
答案 1 :(得分：1)
我的版本：
# load data
data(mtcars)
N <- nrow(mtcars)
# bootstrap parameters
sample_size <- 10
bootstrap_times <- 20
# create empty storage matrix of results
# one row per bootstrap sample, one column per predicted weight
res_mat <- matrix(NA, nrow=bootstrap_times, ncol=N)
colnames(res_mat) <- rownames(mtcars)
# do bootstrap
for (i in seq(bootstrap_times)) {
this_sample <- sample(N, sample_size, replace=FALSE)
reg_result <- lm(mpg ~ wt, data=mtcars[this_sample,])
res_mat[i,] <- predict(reg_result, mtcars)
}
答案 2 :(得分：1)
这是使用嵌套data.frames的tidyverse方法：
library(tidyverse)
Bootstrap_times <- 2
Sample_Size <- 10
Predictions <- data.frame(SampleID = 1:Bootstrap_times) %>%
group_by(SampleID) %>%
nest() %>%
mutate(data = data %>% map(~mtcars[sample(nrow(mtcars), Sample_Size), ]),
Model = data %>% map(~lm(formula = mpg ~ wt, data = .)),
Predictions = map2(Model, data, ~predict(.x, newdata = .y))) %>%
select(SampleID, Predictions) %>%
unnest()
结果：
# A tibble: 20 x 2
SampleID Predictions
<int> <dbl>
1 1 22.7
2 1 16.2
3 1 19.7
4 1 21.5
5 1 18.7
6 1 17.4
7 1 23.3
8 1 10.7
9 1 18.8
10 1 19.8
11 2 11.4
12 2 19.6
13 2 11.7
14 2 18.1
15 2 21.1
16 2 18.6
17 2 16.2
18 2 23.5
19 2 19.7
20 2 20.7
此方法的优点是非常容易从模型中提取其他信息（使用broom并合并为一个数据。frame输出：
library(broom)
data.frame(SampleID = 1:Bootstrap_times) %>%
group_by(SampleID) %>%
nest() %>%
mutate(data = data %>% map(~mtcars[sample(nrow(mtcars), Sample_Size), ]),
Model = data %>% map(~lm(formula = mpg ~ wt, data = .) %>% augment())) %>%
select(-data) %>%
unnest()
结果：
# A tibble: 20 x 11
SampleID .rownames mpg wt .fitted .se.fit .resid .hat .sigma .cooksd .std.resid
<int> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 Dodge Challenger 15.5 3.52 17.2 0.689 -1.72 0.106 2.15 0.0442 -0.862
2 1 Datsun 710 22.8 2.32 23.5 0.940 -0.655 0.198 2.24 0.0148 -0.346
3 1 Cadillac Fleetwood 10.4 5.25 8.24 1.52 2.16 0.515 1.93 1.15 1.47
4 1 Merc 450SE 16.4 4.07 14.4 0.863 2.04 0.167 2.10 0.112 1.06
5 1 Ford Pantera L 15.8 3.17 19.0 0.672 -3.24 0.101 1.85 0.147 -1.62
6 1 Lotus Europa 30.4 1.51 27.6 1.39 2.75 0.432 1.79 1.14 1.73
7 1 Volvo 142E 21.4 2.78 21.1 0.751 0.334 0.126 2.26 0.00207 0.169
8 1 Merc 280C 17.8 3.44 17.6 0.678 0.163 0.103 2.26 0.000378 0.0812
9 1 Mazda RX4 Wag 21 2.88 20.6 0.724 0.428 0.117 2.25 0.00308 0.215
10 1 Camaro Z28 13.3 3.84 15.6 0.773 -2.26 0.134 2.06 0.102 -1.15
11 2 Merc 280 19.2 3.44 19.7 1.09 -0.470 0.108 3.53 0.00138 -0.151
12 2 Toyota Corolla 33.9 1.84 28.2 1.65 5.66 0.251 2.52 0.658 1.98
13 2 Hornet Sportabout 18.7 3.44 19.7 1.09 -0.970 0.108 3.51 0.00588 -0.311
14 2 Mazda RX4 Wag 21 2.88 22.7 1.07 -1.69 0.106 3.47 0.0173 -0.540
15 2 Chrysler Imperial 14.7 5.34 9.50 2.42 5.20 0.539 2.02 3.15 2.32
16 2 Camaro Z28 13.3 3.84 17.5 1.26 -4.23 0.145 3.08 0.163 -1.39
17 2 Valiant 18.1 3.46 19.6 1.09 -1.46 0.110 3.48 0.0136 -0.469
18 2 Porsche 914-2 26 2.14 26.6 1.43 -0.611 0.188 3.52 0.00490 -0.205
19 2 Merc 280C 17.8 3.44 19.7 1.09 -1.87 0.108 3.45 0.0219 -0.600
20 2 Lotus Europa 30.4 1.51 30.0 1.91 0.441 0.335 3.52 0.00677 0.164
注意：
使用此方法，您甚至不需要预测步骤（除非您正在使用新数据），因为您拥有.fitted中的augment个值。
由于未设置种子，因此第一和第二输出之间的预测有所不同。