在R中的两个月之间插入月日期
在R中有两个日期,我想在它们之间创建一个日期向量,间隔一个月。然而,R似乎使用了31天的间隔(有时?)。例如:在R中的两个月之间插入月日期,r,date,R,Date,在R中有两个日期,我想在它们之间创建一个日期向量,间隔一个月。然而,R似乎使用了31天的间隔(有时?)。例如: x <- as.Date("31-01-1900", "%d-%m-%Y") y <- as.Date("31-01-1901", "%d-%m-%Y") seq(x, y, by="month") [1] "1900-01-31" "1900-03-03" "1900-03-31" "1900-05-01" "1900-05-31" [6] "1900-07-01"
x <- as.Date("31-01-1900", "%d-%m-%Y")
y <- as.Date("31-01-1901", "%d-%m-%Y")
seq(x, y, by="month")
[1] "1900-01-31" "1900-03-03" "1900-03-31" "1900-05-01" "1900-05-31"
[6] "1900-07-01" "1900-07-31" "1900-08-31" "1900-10-01" "1900-10-31"
[11] "1900-12-01" "1900-12-31" "1901-01-31"
R知道月份的长度吗,还是我必须手工完成?我们可以使用
lubridate
library(lubridate)
x <- as.Date("31-01-1900", "%d-%m-%Y")
y <- as.Date("31-01-1901", "%d-%m-%Y")
c(x, x %m+% months(1:11), y)
[1] "1900-01-31" "1900-02-28" "1900-03-31" "1900-04-30"
[5] "1900-05-31" "1900-06-30" "1900-07-31" "1900-08-31"
[9] "1900-09-30" "1900-10-31" "1900-11-30" "1900-12-31"
[13] "1901-01-31"
您可以生成每个月第一个的序列,然后减去1
seq(as.Date("1900-02-01"), length = 12, by="1 month") - 1
# [1] "1900-01-31" "1900-02-28" "1900-03-31" "1900-04-30" "1900-05-31" "1900-06-30"
# [7] "1900-07-31" "1900-08-31" "1900-09-30" "1900-10-31" "1900-11-30" "1900-12-31"
## the same can be achieved with `seq.Date()`
# seq.Date(as.Date("1900-02-01"), by = "1 month", length.out = 12) - 1
您可以查看lubridate的
package,不过这是?seq.POSIXt的一段摘录:“使用“month”首先提前月份,而不更改日期:如果这导致一个月中的某一天无效,则将其向前计算到下一个月。”因此,这些函数中的这一点看起来是固定的。这两个答案都非常好,但我更喜欢这个,因为它不需要包装;我通常会先尝试一个基本的解决方案。
lubridate::add_with_rollback(x, months(0:12))
[1] "1900-01-31" "1900-02-28" "1900-03-31" "1900-04-30"
[5] "1900-05-31" "1900-06-30" "1900-07-31" "1900-08-31"
[9] "1900-09-30" "1900-10-31" "1900-11-30" "1900-12-31"
[13] "1901-01-31"
seq(as.Date("1900-02-01"), length = 12, by="1 month") - 1
# [1] "1900-01-31" "1900-02-28" "1900-03-31" "1900-04-30" "1900-05-31" "1900-06-30"
# [7] "1900-07-31" "1900-08-31" "1900-09-30" "1900-10-31" "1900-11-30" "1900-12-31"
## the same can be achieved with `seq.Date()`
# seq.Date(as.Date("1900-02-01"), by = "1 month", length.out = 12) - 1