我正在使用按group_by
包dplyr
函数分组的数据集。
每个组都有自己的时间索引,即假定由12个月的序列组成。
这意味着它可以从1月开始到12月结束,或者在其他情况下可以从前一年的6月开始到明年5月结束。
以下是数据集示例:
ID DATE
8 2017-01-31
8 2017-02-28
8 2017-03-31
8 2017-04-30
8 2017-05-31
8 2017-06-30
8 2017-07-31
8 2017-08-31
8 2017-09-30
8 2017-10-31
8 2017-11-30
8 2017-12-31
32 2017-01-31
32 2017-02-28
32 2017-03-31
32 2017-04-30
32 2017-05-31
32 2017-06-30
32 2017-07-31
32 2017-08-31
32 2017-09-30
32 2017-10-31
32 2017-11-30
32 2017-12-31
45 2016-09-30
45 2016-10-31
45 2016-11-30
45 2016-12-31
45 2017-01-31
45 2017-02-28
45 2017-03-31
45 2017-04-30
45 2017-05-31
45 2017-06-30
45 2017-07-31
45 2017-08-31
问题在于,如果存在所谓的"跳跃",换句话说,如果日期一致,我就无法确认或验证数据集维度。 r中是否有任何简单的方法可以执行此操作,也许是来自tibbletime
包的一些修改/组合功能。
任何帮助都将受到赞赏。
提前谢谢。
答案 0 :(得分:0)
以下是我通常使用data.table
解决此问题的方法 - 来自base的cut.Date()
和seq.Date()
函数是逻辑的核心,所以你使用相同的如果需要,可以使用dplyr
。
library(data.table)
## Convert to data.table
setDT(df)
## Convert DATE to a date in case it wasn't already
df[,DATE := as.Date(DATE)]
## Order by ID and Date
setkey(df,ID,DATE)
## Create a column with the month of each date
df[,Month := as.Date(cut.Date(DATE, breaks = "months"))]
## Generate a sequence of Dates by month for the number of observations
## in each group -- .N
df[,ExpectedMonth := seq.Date(from = min(Month),
by = "months",
length.out = .N), by = .(ID)]
## Create a summary table to test whether an ID had 12 observations where
## the actual month was equal to the expected month
Test <- df[Month == ExpectedMonth, .(Valid = ifelse(.N == 12L,TRUE,FALSE)), by = .(ID)]
print(Test)
# ID Valid
# 1: 8 TRUE
# 2: 32 TRUE
# 3: 45 TRUE
## Do a no-copy join of Test to df based on ID
## and create a column in df based on the 'Valid' column in Test
df[Test, Valid := i.Valid, on = "ID"]
## The final output:
head(df)
# ID DATE Month ExpectedMonth Valid
# 1: 8 2017-01-31 2017-01-01 2017-01-01 TRUE
# 2: 8 2017-02-28 2017-02-01 2017-02-01 TRUE
# 3: 8 2017-03-31 2017-03-01 2017-03-01 TRUE
# 4: 8 2017-04-30 2017-04-01 2017-04-01 TRUE
# 5: 8 2017-05-31 2017-05-01 2017-05-01 TRUE
# 6: 8 2017-06-30 2017-06-01 2017-06-01 TRUE
如果你真的想要使用自我加入并跳过创建Test
setDT(df)
df[,DATE := as.Date(DATE)]
setkey(df,ID,DATE)
df[,Month := as.Date(cut.Date(DATE, breaks = "months"))]
df[,ExpectedMonth := seq.Date(from = min(Month), by = "months", length.out = .N), keyby = .(ID)]
df[df[Month == ExpectedMonth,.(Valid = ifelse(.N == 12L,TRUE,FALSE)),keyby = .(ID)], Valid := i.Valid]