英文:
Fastest way to convert 29th February into 1st March in R?
问题
我有一个在R中的日期向量（"YYYY-MM-DD"），其中包含一些2月29日。为了简化我的分析，我想将这些日期转换为3月1日。我正在使用一个for循环来做这个，但我想知道是否有一种更快的方法，因为我将把它应用到成千上万个日期。
这是一个可重现的示例：
library(lubridate)dates0 <- as.POSIXlt(as.Date(c("1945-04-09 UTC", "1957-06-10 UTC", "1924-02-28 UTC","1921-06-22 UTC","1926-05-16 UTC", "1920-02-29 UTC")))dates0[6]for(i in 1:6){if(day(dates0[i])==29 & month(dates0[i])==2) day(dates0[i]) <- 1; month(dates0[i]) <- 3}dates0[6]
希望这对你有帮助。
英文:
I have a vector of dates ("YYYY-MM-DD") in R, which contain some 29th February. For simplicity of my analysis, I want to convert these dates into 1st March. I am doing this using a for loop, but I wondered if there is a faster way as I will apply it to tens of thousands of dates.
Here is a reproducible example:
library(lubridate)dates0 <- as.POSIXlt(as.Date(c("1945-04-09 UTC", "1957-06-10 UTC", "1924-02-28 UTC","1921-06-22 UTC","1926-05-16 UTC", "1920-02-29 UTC")))dates0[6]for(i in 1:6){if(day(dates0[i])==29 & month(dates0[i])==2) day(dates0[i]) <- 1; month(dates0[i]) <- 3}dates0[6]
答案1
得分: 3
尝试
tmp = 月份(dates0) == 2 & 天数(dates0) == 29dates0[tmp] = dates0[tmp] + 天数(1)
英文:
Try
tmp=month(dates0)==2 & day(dates0)==29dates0[tmp]=dates0[tmp]+days(1)
答案2
得分: 3
以下是您要翻译的内容：
A more onerous method than user2974951 offered, but more easily generalised if the date change is more than a single day step:tochange < - day(dates0) == 29 & month(dates0) == 2month(dates0[tochange]) < - 3day(dates0[tochange]) < - 1In terms of fastest:dates1 < - rep(dates0, times = 1000)OP < - \(x) {for(i in 1:length(x)){if(day(x[i])==29 & month(x[i])==2) day(x[i]) < - 1; month(x[i]) < - 3}}setBoth < - \(x) {tochange < - day(x) == 29 & month(x) == 2month(x[tochange]) < - 3day(x[tochange]) < - 1}incrementDay < - \(x) {tmp=month(x)==2 & day(x)==29x[tmp]=x[tmp]+days(1)}microbenchmark(OP = OP(dates1),setBoth = setBoth(dates1),incrementDay = incrementDay(dates1))Unit: millisecondsexpr min lq mean median uq max neval cldOP 1.1611 1.38955 1.839490 1.67015 1.8911 8.3134 100 asetBoth 1.2863 1.44650 3.456468 1.63335 1.7768 158.1885 100 aincrementDay 1.1924 1.41305 1.765843 1.53635 1.7587 7.8440 100 aThere doesn't seem to be a lot in it, my way likely slowest.
请注意，代码部分没有被翻译，只翻译了文本内容。
英文:
A more onerous method than user2974951 offered, but more easily generalised if the date change is more than a single day step:
tochange <- day(dates0) == 29 & month(dates0) == 2month(dates0[tochange]) <- 3day(dates0[tochange]) <- 1
In terms of fastest:
dates1 <- rep(dates0, times = 1000)OP <- \(x) {for(i in 1:length(x)){if(day(x[i])==29 & month(x[i])==2) day(x[i]) <- 1; month(x[i]) <- 3}}setBoth <- \(x) {tochange <- day(x) == 29 & month(x) == 2month(x[tochange]) <- 3day(x[tochange]) <- 1}incrementDay <- \(x) {tmp=month(x)==2 & day(x)==29x[tmp]=x[tmp]+days(1)}microbenchmark(OP = OP(dates1),setBoth = setBoth(dates1),incrementDay = incrementDay(dates1))Unit: millisecondsexpr min lq mean median uq max neval cldOP 1.1611 1.38955 1.839490 1.67015 1.8911 8.3134 100 asetBoth 1.2863 1.44650 3.456468 1.63335 1.7768 158.1885 100 aincrementDay 1.1924 1.41305 1.765843 1.53635 1.7587 7.8440 100 a
There doesn't seem to be a lot in it, my way likely slowest.
答案3
得分: 2
如果不需要使用POSIXlt，可以在使用Date时提高速度，当找到2月29日时，只需添加1，这在base中可以正常工作。
i <- which(format(dates1, "%m%d") == "0229")`[<-`(dates1, i, value = dates1[i] + 1)
数据
dates1 <- as.Date(c("1945-04-09 UTC", "1957-06-10 UTC", "1924-02-28 UTC","1921-06-22 UTC", "1926-05-16 UTC", "1920-02-29 UTC"))dates0 <- as.POSIXlt(dates1)
基准测试
incrementDay <- function(x) { #@user2974951tmp <- month(x) == 2 & day(x) == 29x[tmp] <- x[tmp] + days(1)x}idLub <- function(dates1) {i <- which(month(dates1) == 2 & day(dates1) == 29)`[<-`(dates1, i, value = dates1[i] + 1)}idBase <- function(dates1) {i <- which(format(dates1, "%m%d") == "0229")`[<-`(dates1, i, value = dates1[i] + 1)}bench::mark(incrementDay = incrementDay(dates1),idLub = idLub(dates1),idBase = idBase(dates1))
结果
expression min median `itr/sec` mem_alloc `gc/sec` n_itr n_gc total_time<bch:expr> <bch:t> <bch:t> <dbl> <bch:byt> <dbl> <int> <dbl> <bch:tm>1 increment… 441.3µs 454.7µs 2161. 328B 14.4 1048 7 485ms2 idLub 46.2µs 48.8µs 20060. 21.8KB 21.0 9531 10 475ms3 idBase 28.8µs 30.9µs 31793. 0B 22.3 9993 7 314ms
在这种情况下，使用Date而不是POSIXlt可以将时间减少10倍。
英文:
In case there is no need to use POSIXlt it would gain much speed when using Date, where, when found the 29 Feb., simply 1 needs to be added and this works in base.
i <- which(format(dates1, "%m%d") == "0229")`[<-`(dates1, i, value = dates1[i] + 1) }
Data
dates1 <- as.Date(c("1945-04-09 UTC", "1957-06-10 UTC", "1924-02-28 UTC","1921-06-22 UTC","1926-05-16 UTC", "1920-02-29 UTC"))dates0 <- as.POSIXlt(dates1)
Benchmark
incrementDay <- \(x) { #@user2974951tmp <- month(x)==2 & day(x)==29x[tmp] <- x[tmp]+days(1)x }idLub <- \(dates1) {i <- which(month(dates1)==2 & day(dates1)==29)`[<-`(dates1, i, value = dates1[i] + 1) }idBase <- \(dates1) {i <- which(format(dates1, "%m%d") == "0229")`[<-`(dates1, i, value = dates1[i] + 1) }bench::mark(incrementDay = incrementDay(dates1),idLub = idLub(dates1),idBase = idBase(dates1) )
Result
expression min median `itr/sec` mem_alloc `gc/sec` n_itr n_gc total_time<bch:expr> <bch:t> <bch:t> <dbl> <bch:byt> <dbl> <int> <dbl> <bch:tm>1 increment… 441.3µs 454.7µs 2161. 328B 14.4 1048 7 485ms2 idLub 46.2µs 48.8µs 20060. 21.8KB 21.0 9531 10 475ms3 idBase 28.8µs 30.9µs 31793. 0B 22.3 9993 7 314ms
Using Date instead of POSIXlt decreases the time in this case by a factor of 10.
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