问题描述
我正在尝试对R中的df列的子集进行热编码，
一种热编码是将分类变量转换成可以提供给ML算法的形式的过程，通过将字符串列转换为二进制列中的每个字符串，可以更好地进行预测。
假设我们有一个如下所示的df：
mes work_location birth_place
01/01/2000 China Chile
01/02/2000 Mexico Japan
01/03/2000 China Chile
01/04/2000 China Argentina
01/05/2000 USA Poland
01/06/2000 Mexico Poland
01/07/2000 USA Finland
01/08/2000 USA Finland
01/09/2000 Japan norway
01/10/2000 Japan Kenia
01/11/2000 Japan Mali
01/12/2000 India Mali
这是热编码的代码：
## function to hot-encode ##
columna_dummy <- function(df,columna) {
df %>%
mutate_at(columna,~paste(columna,eval(as.symbol(columna)),sep = "_")) %>%
mutate(valor = 1) %>%
spread(key = columna,value = valor,fill = 0)
}
## selecting columns ##
columnas <- c("work_location","birth_place")
## applying loop to repeat columna_dummy function for each df column ##
for(i in 1:length(columnas)){
new_dataset <- columna_dummy(df,i)
}
控制台输出：
Error: Problem with `mutate()` input `mes`.
x objeto '1' no enconTrado
i Input `mes` is `(structure(function (...,.x = ..1,.y = ..2,. = ..1) ...`.
Run `rlang::last_error()` to see where the error occurred.
Called from: signal_abort(cnd)
列mes是日期类列，但是不包含在原子向量列中
仍然会引发上述错误，
对于所选字符串df列中的每个字符串，预期输出应类似于以下内容：
（我无法添加每列，但work_location_China是一个示例 列的外观）
mes work_location birth_place work_location_China
01/01/2000 China Chile 1
01/02/2000 Mexico Japan 0
01/03/2000 China Chile 1
01/04/2000 China Argentina 1
01/05/2000 USA Poland 0
01/06/2000 Mexico Poland 0
01/07/2000 USA Finland 0
01/08/2000 USA Finland 0
01/09/2000 Japan norway 0
01/10/2000 Japan Kenia 0
01/11/2000 Japan Mali 0
01/12/2000 India Mali 0
还有其他方法可以应用此循环吗？
解决方法
我们在传递字符串时，可以选择select列（select可以带引号/不带引号），创建1s列（“ valor”）和行号列（ 'rn'），然后从'long'改成'wide'（pivot_wider）
library(dplyr)
library(tidyr)
library(purrr)
library(stringr)
columna_dummy <- function(df,columna) {
df %>%
select(columna) %>%
mutate(valor = 1,rn = row_number()) %>%
pivot_wider(names_from = all_of(columna),values_from = valor,values_fill = 0) %>%
select(-rn)
}
-测试
对于多个列，一个选项是用map遍历感兴趣的列名，应用该函数并用_dfc绑定它们，然后绑定到原始数据集（{{1} }）
bind_cols-输出
out <- imap_dfc(setNames(c("work_location","birth_place"),c("work_location","birth_place")),~ {
nm1 <- as.character(.y)
columna_dummy(df = df,columna = .x) %>%
rename_all(~ str_c(nm1,.,sep="_"))
}) %>%
bind_cols(df,.)
数据
head(out,2)
# mes work_location birth_place work_location_China work_location_Mexico work_location_USA work_location_Japan
#1 01/01/2000 China Chile 1 0 0 0
#2 01/02/2000 Mexico Japan 0 1 0 0
# work_location_India birth_place_Chile birth_place_Japan birth_place_Argentina birth_place_Poland birth_place_Finland
#1 0 1 0 0 0 0
#2 0 0 1 0 0 0
# birth_place_Norway birth_place_Kenia birth_place_Mali
#1 0 0 0
#2 0 0 0
通过使用purrr库，我解决了这个问题：
## data ##
df <- structure(list(mes = c("01/01/2000","01/02/2000","01/03/2000","01/04/2000","01/05/2000","01/06/2000","01/07/2000","01/08/2000","01/09/2000","01/10/2000","01/11/2000","01/12/2000"),work_location = c("China","Mexico","China","USA","Japan","India"),birth_place = c("Chile","Chile","Argentina","Poland","Finland","Norway","Kenia","Mali","Mali")),class = "data.frame",row.names = c(NA,-12L))
## function to hot-encode ##
columna_dummy <- function(df,columna) {
df %>%
mutate_at(columna,~paste(columna,eval(as.symbol(columna)),sep = "_")) %>%
mutate(valor = 1) %>%
spread(key = columna,value = valor,fill = 0)
}
## vector of columns ##
columnas <- c("work_location","birth_place")
## hot_encoded_dataset ##
library(purrr)
hot_encoded_dataset <- purrr :: map(columnas,columna_dummy,df = df) %>%
reduce(inner_join)