如何使用R对数据资产进行统一存储

Question

这是用于创建虚拟对象的libraryI

install.packages("fastDummies")
library(fastDummies)

这是数据集

winners <- data.frame(
          city = c("SaoPaulito", "NewAmsterdam", "BeatifulCow"),
          year = c(1990, 2000, 1990),
          crime = 1:3)

让他们在这些城市中创建超级假人：

dummy_cols(winners, select_columns = c("city"))

结果是

city year crime city_SaoPaulito city_NewAmsterdam city_BeatifulCow
1   SaoPaulito 1990     1               1                 0                0
2 NewAmsterdam 2000     2               0                 1                0
3  BeatifulCow 1990     3               0                 0                1

那么问题是，如果我想返回到之前的数据集，有什么想法吗？

提前感谢！

Answer 1

我们可以使用dcast

library(data.table)
dcast(setDT(winners), crime ~ city, length)

如果我们需要获取输入，它将是

subset(df1, select = 1:3)
#         city year crime
#1   SaoPaulito 1990     1
#2 NewAmsterdam 2000     2
#3  BeatifulCow 1990     3

或使用melt

melt(setDT(df1), measure = patterns("_"))[value == 1, .(city, year, crime)]
#          city year crime
#1:   SaoPaulito 1990     1
#2: NewAmsterdam 2000     2
#3:  BeatifulCow 1990     3

数据

df1 <- structure(list(city = c("SaoPaulito", "NewAmsterdam", "BeatifulCow"
), year = c(1990L, 2000L, 1990L), crime = 1:3, city_SaoPaulito = c(1L, 
0L, 0L), city_NewAmsterdam = c(0L, 1L, 0L), city_BeatifulCow = c(0L, 
0L, 1L)), class = "data.frame", row.names = c("1", "2", "3"))

Answer 2

如果每行中只有一个city作为1，则可以跳过虚拟列

df[, 1:3]


#           city year crime
#1   SaoPaulito 1990     1
#2 NewAmsterdam 2000     2
#3  BeatifulCow 1990     3

如果您可以拥有多个城市，那么使用dplyr和tidyr::gather的一种方式是

library(dplyr)

df %>%
  tidyr::gather(key, value, starts_with("city_")) %>%
  filter(value == 1) %>%
  select(-value, -key)