使用
tidyverse
library(tidyverse)
f1 <- function(data, wordToCompare, colsToCompare) {
wordToCompare <- enquo(wordToCompare)
data %>%
select(colsToCompare) %>%
mutate(!! wordToCompare := map(., ~
.x == as_label(wordToCompare)) %>%
reduce(`|`) %>%
as.integer)
}
f1(df1, Carol, c("first", 'middle', 'last'))
f1(df1, Sarah, c("first", 'middle', 'last'))
或者也可以使用pmap
来完成这个任务
df1 %>%
mutate(Carol = pmap_int(.[c('first', 'middle', 'last')],
~ +('Carol' %in% c(...))))
可以封装成一个函数
f2 <- function(data, wordToCompare, colsToCompare) {
wordToCompare <- enquo(wordToCompare)
data %>%
mutate(!! wordToCompare := pmap_int(.[colsToCompare],
~ +(as_label(wordToCompare) %in% c(...))))
}
f2(df1, Carol, c("first", 'middle', 'last'))
注意:tidyverse的两种方法都不需要进行任何重塑
使用base R
,我们可以循环遍历“第一列”、“中间列”和“最后一列”,并使用==
进行比较,以获取逻辑vector
的list
,然后使用Reduce
将其缩减为单个逻辑vector
,并使用|
强制转换为二进制+
df1$Carol <- +(Reduce(`|`, lapply(df1[2:4], `==`, 'Carol')))
df1
注意:此帖子有重复内容。例如:这里
数据
df1 <- structure(list(id = 1:3, first = c("Carol", "Sarah", "Josh"),
middle = c("Jenny", "Carol", "David"), last = c("Smith",
"Roberts", "Richardson"), Age = c(15L, 20L, 22L)),
class = "data.frame", row.names = c(NA,
-3L))