首页 > 解决方案 > 映射列表并创建训练和测试拆分

问题描述

我有一个数据框列表,我想将每个列表拆分为一个训练和测试集。

我正在尝试的是以下内容:

library(furrr)
library(purrr)
library(dplyr)
train <- df %>% 
  future_map(., ~as_tibble(.) %>% 
               sample_frac(0.70))

test <- df %>% 
  future_map2(., ~anti_join(df, train, by = "ID"))

我在test代码部分遇到错误。

错误告诉我:

缺少参数“.f”,没有默认值

.f我要应用的功能是功能anti_join

数据:

df <- list(structure(list(ID = c(32854L, 87264L, 10046L, 19734L, 46776L, 
69104L), Var1 = c(0.422291041272086, 0.421857549178741, 0.833626138132112, 
0.0548106066414176, 0.624276309555977, 0.71510623528032), Var2 = c(12.1523922938606, 
11.8078760576117, 15.0796093186172, 14.6689943885814, 16.2489940803754, 
12.6618473977339)), row.names = c(NA, -6L), class = c("tbl_df", 
"tbl", "data.frame")), structure(list(ID = c(64813L, 28351L, 
98206L, 11662L, 39650L, 48688L), Var1 = c(0.7736786, 0.9184715, 
0.6961003, 0.904335, 0.2900523, 0.5886406), Var2 = c(14.9093, 
14.37623, 13.50509, 13.84473, 12.35784, 14.02979)), row.names = c(NA, 
-6L), class = c("tbl_df", "tbl", "data.frame")), structure(list(
    ID = c(34745L, 20879L, 16576L, 23177L, 64122L, 23805L), Var1 = c(0.4781384, 
    0.89258, 1.993566, 0.291347, 0.1067107, 0.866149), Var2 = c(12.96561, 
    12.33563, 11.8721, 14.79747, 13.08436, 14.70748)), row.names = c(NA, 
-6L), class = c("tbl_df", "tbl", "data.frame")), structure(list(
    ID = c(42108L, 11726L, 78484L, 36898L, 60224L, 61154L), Var1 = c(0.877778213319744, 
    1.33963228732897, 0.806292976614067, 0.972259512214242, NA, 
    0.756496381825957), Var2 = c(11.5246035772182, 11.6156242530741, 
    12.9551613288682, 13.1341296830114, NA, 13.9791350710985)), row.names = c(NA, 
-6L), class = c("tbl_df", "tbl", "data.frame")))

标签: rpurrrfurrr

解决方案


由于您使用的是future_map2(),因此您需要将 a.x.y参数输入到anti_join()函数中。

您的示例丢失了train,因此future_map2()认为这anti_join()是您的.y论点,因此它认为.f丢失了,因此出现了错误。

test <- df %>% 
  future_map2(.x = ., 
              .y = train, 
              .f = ~anti_join(.x, .y, by = "ID"))

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