Find the minimum date between two maximum dates based off unique values in a column

Question

Data example.

date1 = seq(as.Date("2019/01/01"), by = "month", length.out = 29)
date2= seq(as.Date("2019/05/01"), by = "month", length.out = 29)

subproducts1=rep("1",29)
subproducts2=rep("2",29)

b1 <- c(rnorm(29,5))
b2 <- c(rnorm(29,5))

dfone <- data.frame("date"= c(date1,date2),
                "subproduct"= 
                  c(subproducts1,subproducts2),
                "actuals"= c(b1,b2))

Max Date for Subproduct 1 is May 2021 and max date for Subproduct 2 is Sept 2021.

Question: Is there a way to:

1. Find the max date for both unique subproduct and
2. Find the minimum date from the two max dates all in one step?

在这种情况下，最终结果应该是 2021 年 5 月，并且能够处理多个子产品。

Answer 1

我们可以使用slice_max由“子产品”分组后，pull将date与获得min，将其分配到一个新的对象

library(dplyr)
dfone %>%
    group_by(subproduct) %>%
    slice_max(n = 1, order_by = date) %>%
    ungroup %>%
    pull(date) %>%
    min -> Min_date

-输出

 Min_date
[1] "2021-05-01"

另一种选择是arrange行并filter使用duplicated

dfone %>%
    arrange(subproduct, desc(date)) %>% 
    filter(!duplicated(subproduct)) %>% 
    pull(date) %>%
    min

Answer 2

对于您的第一个目标，您可以尝试subset+ave如下所示

out1 <- subset(
    dfone,
    ave(date, subproduct, FUN = max) == date
)

这使

         date subproduct  actuals
29 2021-05-01          1 5.728420
58 2021-09-01          2 3.455491

对于您的第二个目标，基于out1，您可以尝试

out2 <- subset(
    out1,
    date == min(date)
)

这使

         date subproduct  actuals
29 2021-05-01          1 5.083229

Answer 3

这也可以在基础 R 中完成。最后我使用了Reduce这样的解决方案，以便可以将解决方案推广到任意数量的subproducts 和dates 而不仅仅是 2 个值，就像这里的情况一样。

Reduce(function(x, y) min(x, y),
       lapply(unique(dfone$subproduct), \(x){
         max(dfone$date[dfone$subproduct == x])
       }))

[1] "2021-05-01"