r - 用 1 或 0 填充列，如果该列名与在同一数据帧内的其他命名列中找到的变量匹配

Question

view(fastcars)

   day    car1   car2   car3
1 day1   red  silver   blue
2 day2  blue    red   green
3 day3  blue  white   green
4 day4 green  black     red
5 day5 black    red  silver

将所有颜色的汽车合并到一个具有唯一名称的列表中。

cars <- stack(fastcars[, c(2:4)])
cars <- t(unique(cars[,1]))

将颜色作为列名添加到数据帧的末尾

fastcars[c(cars)] <- NA

   day  car1   car2   car3  red blue green black silver white
1 day1   red  silver  blue  NA   NA    NA    NA     NA    NA
2 day2  blue    red  green  NA   NA    NA    NA     NA    NA
3 day3  blue  white  green  NA   NA    NA    NA     NA    NA
4 day4 green  black    red  NA   NA    NA    NA     NA    NA
5 day5 black    red silver  NA   NA    NA    NA     NA    NA

如果列名与 car1、car2 和/或 car3 列中的变量匹配，则希望用 1 或 0 填充 NA。

day car1     car2    car3   red  blue green black silver white
day1     red   silver    blue      1    1     0     0      1     0
day2    blue      red   green      1    1     1     0      0     0
day3    blue    white   green      0    1     1     0      0     1
day4   green    black     red      1    0     1     1      0     0
day5   black      red   silver     1    0     0     1      1     0`

我相信这里的链接与我想要做的很接近，但无法弄清楚如何在我现有的数据框中创建它。 https://stackoverflow.com/a/30274596/3837899

#Generate example dataframe with character column

example <- as.data.frame(c("A", "A", "B", "F", "C", "G", "C", "D", "E", "F"))
names(example) <- "strcol"

#For every unique value in the string column, create a new 1/0 column
#This is what Factors do "under-the-hood" automatically when passed to    function requiring numeric data

for(level in unique(example$strcol)){
  example[paste("dummy", level, sep = "_")] <- ifelse(example$strcol == level, 1, 0)
}

Answer 1

这里有几个选项。

选项 1:我们可以在重新转换熔化的数据后使用 data.table 合并。

library(data.table) # v1.9.6
## make 'df' a data.table
setDT(df)
## melt, cast, and merge on 'day'
df[dcast(melt(df, "day"), day ~ value, fun.aggregate = length), on = "day"]
#     day  car1   car2   car3 black blue green red silver white
# 1: day1   red silver   blue     0    1     0   1      1     0
# 2: day2  blue    red  green     0    1     1   1      0     0
# 3: day3  blue  white  green     0    1     1   0      0     1
# 4: day4 green  black    red     1    0     1   1      0     0
# 5: day5 black    red silver     1    0     0   1      1     0

选项 2:这是一个不太吸引人但很可靠的 R 方法。

## make sure car columns are character (may not be necessary)
df[-1] <- lapply(df[-1], as.character)
## get unique values of car columns
u <- unique(unlist(df[-1]))
## match 'u' with each row in 'df'
l <- lapply(seq_len(nrow(df)), function(i) as.numeric(u %in% df[i, -1]))
## bring the data together
cbind(df, setNames(do.call(rbind.data.frame, l), u))
#    day  car1   car2   car3 red blue green black silver white
# 1 day1   red silver   blue   1    1     0     0      1     0
# 2 day2  blue    red  green   1    1     1     0      0     0
# 3 day3  blue  white  green   0    1     1     0      0     1
# 4 day4 green  black    red   1    0     1     1      0     0
# 5 day5 black    red silver   1    0     0     1      1     0

数据:

df <-structure(list(day = structure(1:5, .Label = c("day1", "day2", 
"day3", "day4", "day5"), class = "factor"), car1 = structure(c(4L, 
2L, 2L, 3L, 1L), .Label = c("black", "blue", "green", "red"), class = "factor"), 
    car2 = structure(c(3L, 2L, 4L, 1L, 2L), .Label = c("black", 
    "red", "silver", "white"), class = "factor"), car3 = structure(c(1L, 
    2L, 2L, 3L, 4L), .Label = c("blue", "green", "red", "silver"
    ), class = "factor")), .Names = c("day", "car1", "car2", 
"car3"), class = "data.frame", row.names = c("1", "2", "3", "4", 
"5"))