将不同符号的持续时间与 lubridate 进行比较

发布于 2025-01-14 07:59:11 字数 1370 浏览 2 评论 0原文

我正在使用带有日期的数据集：

a <- structure(list(ID1 = c(1, 2, 3), ID2 = c(4,
5, 6), dob.x = structure(c(-9954, -8960, -1219), class = "Date"),
    dob.y = structure(c(-1057, -1157, -9752), class = "Date")), row.names = c(NA,                                                                            
3L), class = "data.frame")

  ID1 ID2      dob.x      dob.y
1   1   4 1942-10-01 1967-02-09
2   2   5 1945-06-21 1966-11-01
3   3   6 1966-08-31 1943-04-21

我需要获取这两个日期之间的差异，然后使用 filter() 函数将其与固定的年数进行比较，例如

a %>%
filter(diff > 21)

我尝试使用lubridate 来完成此任务，但如果我计算日期之间的差异，我无法比较它们，因为它们有不同的符号：

> a %>% mutate(diff = as.duration(interval(dob.x, dob.y)))                                                                                                   
  ID1 ID2      dob.x      dob.y                        diff
1   1   4 1942-10-01 1967-02-09   768700800s (~24.36 years)
2   2   5 1945-06-21 1966-11-01   674179200s (~21.36 years)
3   3   6 1966-08-31 1943-04-21 -737251200s (~-23.36 years)

> a %>% mutate(diff = as.duration(interval(dob.x, dob.y))) %>% filter(diff > years(22))
  ID1 ID2      dob.x      dob.y                      diff
1   1   4 1942-10-01 1967-02-09 768700800s (~24.36 years)

当我有负值时，如何保留间隔超过例如 22 年的日期对？

原文

I am working with a dataset with dates:

a <- structure(list(ID1 = c(1, 2, 3), ID2 = c(4,
5, 6), dob.x = structure(c(-9954, -8960, -1219), class = "Date"),
    dob.y = structure(c(-1057, -1157, -9752), class = "Date")), row.names = c(NA,                                                                            
3L), class = "data.frame")

  ID1 ID2      dob.x      dob.y
1   1   4 1942-10-01 1967-02-09
2   2   5 1945-06-21 1966-11-01
3   3   6 1966-08-31 1943-04-21

I need to get the difference between those two dates, and then use a filter() function to compare it with a fixed number of years, e.g.

a %>%
filter(diff > 21)

I tried to use lubridate for this task, but if I compute the difference between dates I cannot compare them, as they have different signs:

> a %>% mutate(diff = as.duration(interval(dob.x, dob.y)))                                                                                                   
  ID1 ID2      dob.x      dob.y                        diff
1   1   4 1942-10-01 1967-02-09   768700800s (~24.36 years)
2   2   5 1945-06-21 1966-11-01   674179200s (~21.36 years)
3   3   6 1966-08-31 1943-04-21 -737251200s (~-23.36 years)

> a %>% mutate(diff = as.duration(interval(dob.x, dob.y))) %>% filter(diff > years(22))
  ID1 ID2      dob.x      dob.y                      diff
1   1   4 1942-10-01 1967-02-09 768700800s (~24.36 years)

How do I keep pairs of dates with intervals over e.g. 22 years when I have negative values?

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念三年u 2025-01-21 07:59:11

abs() 适用于持续时间，因为它们存储为带有符号的向量元素（以秒为单位）：

> a %>% mutate(diff = abs(as.duration(interval(dob.x, dob.y)))) %>% filter(diff > years(22))
  ID1 ID2      dob.x      dob.y                      diff
1   1   4 1942-10-01 1967-02-09 768700800s (~24.36 years)
2   3   6 1966-08-31 1943-04-21 737251200s (~23.36 years)

abs() works on durations, as they're stored as elements of a vector with a sign (in seconds):

> a %>% mutate(diff = abs(as.duration(interval(dob.x, dob.y)))) %>% filter(diff > years(22))
  ID1 ID2      dob.x      dob.y                      diff
1   1   4 1942-10-01 1967-02-09 768700800s (~24.36 years)
2   3   6 1966-08-31 1943-04-21 737251200s (~23.36 years)

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