Question

我与R合作，我有这些数据：

   data <- structure(list(Col1 = 1:9, Col2 = structure(c(2L, 2L, 2L, 1L, 
3L, 3L, 3L, 3L, 3L), .Label = c("Administrative ", "National", 
"Regional"), class = "factor"), Col3 = structure(c(NA, 3L, 4L, 
NA, 2L, 3L, 1L, 4L, 3L), .Label = c("bike", "boat", "car", "truck"
), class = "factor"), Col4 = c(56L, 65L, 58L, 62L, 24L, 25L, 
120L, 89L, 468L), X = c(NA, NA, NA, NA, NA, NA, NA, NA, NA), 
    X.1 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA)), .Names = c("Col1", 
"Col2", "Col3", "Col4", "X", "X.1"), class = "data.frame", row.names = c(NA, 
-9L))

我想重新安排它以查看可用或不可用的内容。输出看起来像这样：

    result <- structure(list(Col1 = c(1L, 4L, 5L), Col2 = structure(c(2L, 1L, 
3L), .Label = c("Administrative ", "National", "Regional"), class = "factor"), 
    car = c(1L, 0L, 1L), truck = c(1L, 0L, 1L), boat = c(0L, 
    0L, 1L), bike = c(0L, 0L, 1L)), .Names = c("Col1", "Col2", 
"car", "truck", "boat", "bike"), class = "data.frame", row.names = c(NA, 
-3L))

我尝试使用聚合但我仍远未达到结果。帮助将是

t <- aggregate(data$Col2, by=list(data$Col3), c)

欢迎提供帮助！

Answer 1

我们可以使用dcast中的data.table length作为fun.aggregate

library(data.table)
dcast(setDT(data), Col2~ Col3, length)[, 1:5, with = FALSE]

Answer 2

这是使用基础R的想法，

#convert to character
data[2:3] <- lapply(data[2:3], as.character)

#get unique elements to tabulate
i1 <- unique(data$Col3)
i1 <- i1[!is.na(i1)]


setNames(data.frame(do.call(rbind, lapply(split(data$Col3, data$Col2), function(i) 
                                          as.integer(match(i1, i, nomatch = 0) > 0)))), i1)

给出，

               car truck boat bike
Administrative    0     0    0    0
National          1     1    0    0
Regional          1     1    1    1

Answer 3

如果你感兴趣，这是一个dplyr解决方案，虽然akrun的解决方案看起来更简洁：

library(tidyverse)

result <- data %>%
  group_by(Col2, Col3) %>%
  summarise(tot = sum(Col4)) %>%
  mutate(bool = if_else(tot > 0, 1, 0)) %>%
  select(Col2, Col3, bool) %>%
  spread(key = Col3, value = bool, fill = 0) %>%
  select(-`<NA>`)

Answer 4

这是使用table和一些强制的另一种基本R方法。

(table(data$Col2, data$Col3) > 0) + 0L

                  bike boat car truck
  Administrative     0    0   0     0
  National           0    0   1     1
  Regional           1    1   1     1

table计算实例，为NA返回0。然后，我们强制使用> 0来逻辑删除大于1的值，然后返回到+ 0L的整数。

每组重新安排一张桌子

4 个答案: