Question

我需要执行一项相当复杂的任务，所以请耐心等待。我猜它是可能的，但如果没有，请告诉我。

假设我有以下数据

set.seed(123)
date1 <- c(seq(as.Date("2011-11-1"),as.Date("2012-1-1"),by = "months"),seq(as.Date("2011-12-1"),as.Date("2012-3-1"),by = "months"))
date2 <- c(seq(as.Date("2011-12-1"),as.Date("2012-1-1"),by = "months"),seq(as.Date("2011-11-1"),as.Date("2012-1-1"),by = "months"))
variables <- c(rep("Number of Coins",3),rep("Number of Shoes",4),rep("Number of Coins",2),rep("Number of Shoes",3))
date <- c(date1,date2)
names <- c(rep("Jim",7),rep("Arnold",5))
value <- rnorm(12)
df <- data.frame(names, date, variables, value)

    names       date       variables       value
1     Jim 2011-11-01 Number of Coins -0.56047565
2     Jim 2011-12-01 Number of Coins -0.23017749
3     Jim 2012-01-01 Number of Coins  1.55870831
4     Jim 2011-12-01 Number of Shoes  0.07050839
5     Jim 2012-01-01 Number of Shoes  0.12928774
6     Jim 2012-02-01 Number of Shoes  1.71506499
7     Jim 2012-03-01 Number of Shoes  0.46091621
8  Arnold 2011-12-01 Number of Coins -1.26506123
9  Arnold 2012-01-01 Number of Coins -0.68685285
10 Arnold 2011-11-01 Number of Shoes -0.44566197
11 Arnold 2011-12-01 Number of Shoes  1.22408180
12 Arnold 2012-01-01 Number of Shoes  0.35981383

此数据的问题在于变量名称占用了一列。我想为Number of Shoes和Number of Coins创建两列，但我想确保日期保持不变。理想情况下，我想将此数据框转换为此

    names    date Number.of.Coins Number.of.Shoes
1     Jim 11/1/11      -0.5604756              NA
2     Jim 12/1/11      -0.2301775      0.07050839
3     Jim  1/1/12       1.5587083      0.12928773
4     Jim  2/1/12              NA      1.71506499
5     Jim  3/1/12              NA      0.46091621
6 Arnold  11/1/11              NA     -0.44566197
7  Arnold 12/1/11      -1.2650612      1.22408180
8  Arnold  1/1/12      -0.6868529      0.35981383

因此，日期范围将是每个变量的最小日期，即每个变量的最大日期。这将产生对NA的需求。我想在每个name内执行此操作。希望有意义！

Answer 1

正如@ Ajinkya Kale建议的那样，您可以使用reshape2包处理此任务。

dcast(df, names + date ~ variables, value.var = "value")

如果您想确保日期顺序符合时间顺序，可以在arrange()包中使用dplyr。

arrange(dcast(df, names + date ~ variables, value.var = "value"), names, date)

#   names       date Number of Coins Number of Shoes
#1 Arnold 2011-11-01              NA     -0.44566197
#2 Arnold 2011-12-01      -1.2650612      1.22408180
#3 Arnold 2012-01-01      -0.6868529      0.35981383
#4    Jim 2011-11-01      -0.5604756              NA
#5    Jim 2011-12-01      -0.2301775      0.07050839
#6    Jim 2012-01-01       1.5587083      0.12928774
#7    Jim 2012-02-01              NA      1.71506499
#8    Jim 2012-03-01              NA      0.46091621

Answer 2

另一种选择是使用spread

中的tidyr

library(tidyr)
spread(df, variables, value)
#   names       date Number of Coins Number of Shoes
#1 Arnold 2011-11-01              NA     -0.44566197
#2 Arnold 2011-12-01      -1.2650612      1.22408180
#3 Arnold 2012-01-01      -0.6868529      0.35981383
#4    Jim 2011-11-01      -0.5604756              NA
#5    Jim 2011-12-01      -0.2301775      0.07050839
#6    Jim 2012-01-01       1.5587083      0.12928774
#7    Jim 2012-02-01              NA      1.71506499
#8    Jim 2012-03-01              NA      0.46091621

在R中格式化数据帧

2 个答案: