当我使用as.Date(as.yearqtr(test[,1],format ="%qQ%Y"),frac =1)时,我有一个问题,但它返回一个错误,并且四年没有更改到日期。错误是:error in as.yearqtr(as.numeric(x)) (list) object cannot be coerced to type 'double'
这是我在R中的数据帧。
TIME VALUE
1Q2019 1
2Q2019 2
3Q2019 3
4Q2019 4
理想的输出是
TIME VALUE
2019-03-31 1
2019-06-30 2
2019-09-30 3
2019-12-31 4
发布于 2020-04-25 02:12:51
我们可以使用zoo转换为日期,并使用frac获取本季度的最后日期。我们使用一些RegEx来重新排列成zoo的合适格式:
df$TIME=as.Date(as.yearqtr(gsub("(\\d)(Q)(\\d{1,})","\\3 Q\\1",df$TIME)),frac = 1)
df
TIME VALUE
1 2019-03-31 1
2 2019-06-30 2
3 2019-09-30 3
4 2019-12-31 4数据:
df <-structure(list(TIME = structure(1:4, .Label = c("1Q2019", "2Q2019",
"3Q2019", "4Q2019"), class = "factor"), VALUE = 1:4), class = "data.frame", row.names = c(NA,
-4L))发布于 2020-04-25 02:22:30
下面是一个函数,它将返回一个日期向量,假设输入向量的形式为2019年第一季度...
dateStrings <- c("1Q2019","2Q2019","3Q2019","4Q2019","1Q2020")
lastDayOfQuarter <- function(x){
require(lubridate)
result <- NULL
months <-c(3,6,9,12)
days <- c(31,30,30,31)
for(i in 1:length(x)) {
qtr <- as.numeric(substr(x[i],1,1))
result[i] <- mdy(paste(months[qtr],days[qtr],(substr(x[i],3,6)),sep="-"))
}
as.Date(result)
}
lastDayOfQuarter(dateStrings)和输出:
>lastDayOfQuarter(dateStrings)
[1] "2019-03-31" "2019-06-30" "2019-09-30" "2019-12-31" "2020-03-31"
> https://stackoverflow.com/questions/61414089
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