首页 > 解决方案 > R按最接近的日期合并两个数据框

问题描述

我有两个大数据框,dfA并且dfB,我在这里生成了简单的示例

dfA = data.frame(id=c("Apple", "Banana", "Carrot", "Dates", "Egg"),
                    Answer_Date=as.Date(c("2013-12-07", "2014-12-07", "2015-12-07", "2016-12-07", "2017-12-07" )),
                    x1 = c(1,  2,  3,  4,  5),
                    x2 = c(10, 20, 30, 40, 50))

    Browse[2]> dfA
      id Answer_Date x1 x2
1  Apple  2013-12-07  1 10
2 Banana  2014-12-07  2 20
3 Carrot  2015-12-07  3 30
4  Dates  2016-12-07  4 40
5    Egg  2017-12-07  5 50

dfB = data.frame(id=c("Apple", "Apple", "Banana", "Banana", "Banana"),
                    Answer_Date=as.Date(c("2013-12-05", "2014-12-07", "2015-12-10", "2018-11-07", "2019-11-07" )),
                    x3 = c(5,  4,  3,  2,  1),
                    x4 = c(50, 40, 30, 20, 10))
Browse[2]> dfB
      id Answer_Date x3 x4
1  Apple  2013-12-05  5 50
2  Apple  2014-12-07  4 40
3 Banana  2014-12-10  3 30
4 Banana  2018-11-07  2 20
5 Banana  2019-11-07  1 10

我想按最接近的日期合并它们,以便我得到 dfA 和 dfB 中存在的项目,它们与id完全匹配,并且与 Answer_Date尽可能接近(即两个日期之间日期差的最小绝对值)。在这种情况下,我想得到

dfC
      id Answer_Date.x Answer_Date.y x1 x2 x3 x4
1  Apple    2013-12-07    2013-12-05  1 10  5 50
2 Banana    2014-12-07    2014-12-10  2 20  3 30

不幸的是,与 merge() 苦苦挣扎并尝试了我在 StackOverflow 上找到的各种解决方案并没有解决我的问题,只会让我感到困惑。有人会向我指出正确的解决方案吗,最好简单解释一下它为什么起作用?

真诚的,并提前非常感谢

托马斯飞利浦

标签: rdataframedplyrmerge

解决方案


左连接dfBdfA,取每行日期之间的差异并选择每个 id 的最小差异。

left_join(dfA, dfB, by = "id") %>%
  mutate(date_diff = abs(Answer_Date.x - Answer_Date.y)) %>%
  group_by(id) %>%
  filter(date_diff == min(date_diff)) %>%
  select(id, Answer_Date.x, Answer_Date.y, starts_with("x"), date_diff)

然后输出是:

# A tibble: 2 x 8
# Groups:   id [2]
  id     Answer_Date.x Answer_Date.y    x1    x2    x3    x4 date_diff
  <fct>  <date>        <date>        <dbl> <dbl> <dbl> <dbl> <drtn>   
1 Apple  2013-12-07    2013-12-05        1    10     5    50 2 days   
2 Banana 2014-12-07    2014-12-10        2    20     3    30 3 days 

顺便说一句,在您的示例代码Answer_Date中,定义中的第三个dfB应该是"2014-12-10"而不是"2015-12-10".


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