问题

我需要创建一些代码，让我的团队能够持续获取每年的季度信息。我需要一个能够输出例如2020年第一季度或2500年第四季度的函数。

到目前为止我想出来的代码和经理的评论如下：

答案1

得分: 1

这会生成一个包含yearqtr、开始日期和结束日期列的三列数据框。输入可以是Date、yearqtr或字符对象。在最后一种情况下，它应该采用类似于示例中的yyyy-q的形式。

yearqtr对象直接表示年份和季度，没有包含日期，并在内部表示为year + fraction，其中fraction为0、1/4、2/4、3/4，对应一年的4个季度，因此添加1/4会得到下一个季度，添加1会得到下一年的同一个季度。

    library(zoo)
    
    gen_qtrs <- function(from = Sys.Date(), to) {
      qtrs <- seq(as.yearqtr(from), as.yearqtr(to), by = 1/4)
      data.frame(qtrs, from = as.Date(qtrs), to = as.Date(qtrs, frac = 1))
    }
    gen_qtrs(to = "2024-4")
    ##      qtrs       from         to
    ## 1 2023 Q1 2023-01-01 2023-03-31
    ## 2 2023 Q2 2023-04-01 2023-06-30
    ## 3 2023 Q3 2023-07-01 2023-09-30
    ## 4 2023 Q4 2023-10-01 2023-12-31
    ## 5 2024 Q1 2024-01-01 2024-03-31
    ## 6 2024 Q2 2024-04-01 2024-06-30
    ## 7 2024 Q3 2024-07-01 2024-09-30
    ## 8 2024 Q4 2024-10-01 2024-12-31

这些也可以工作，并且会给出相同的输出：

    gen_qtrs(to = as.Date("2024-12-31"))
    gen_qtrs(to = as.yearqtr("2024-4"))

library(zoo)

gen_qtrs <- function(from = Sys.Date(), to) {
  qtrs <- seq(as.yearqtr(from), as.yearqtr(to), by = 1/4)
  data.frame(qtrs, from = as.Date(qtrs), to = as.Date(qtrs, frac = 1))
}
gen_qtrs(to = "2024-4")
##      qtrs       from         to
## 1 2023 Q1 2023-01-01 2023-03-31
## 2 2023 Q2 2023-04-01 2023-06-30
## 3 2023 Q3 2023-07-01 2023-09-30
## 4 2023 Q4 2023-10-01 2023-12-31
## 5 2024 Q1 2024-01-01 2024-03-31
## 6 2024 Q2 2024-04-01 2024-06-30
## 7 2024 Q3 2024-07-01 2024-09-30
## 8 2024 Q4 2024-10-01 2024-12-31

gen_qtrs(to = as.Date("2024-12-31"))
gen_qtrs(to = as.yearqtr("2024-4"))

答案2

得分: 0

这段代码是一个用于获取指定季度起止日期的R语言函数。它可以容忍输入的格式不同，比如季度和年份的顺序、大小写、以及额外的空格。以下是一些测试示例：

对于简单的情况，比如输入 'q3 2020'，将返回起止日期：

```r
getStartEndDate('q3 2020')
#>        start        end
#> 1 2020-07-01 2020-09-30

这个函数也可以处理未来几个世纪的日期：

getStartEndDate('q4 2500')
#>        start        end
#> 1 2500-10-01 2500-12-31

它支持处理多个输入，可以将输入以向量形式传递：

getStartEndDate(c('q1 2023', 'q3 2024 ', 'q2 2025'))
#>        start        end
#> 1 2023-01-01 2023-03-31
#> 2 2024-07-01 2024-09-30
#> 3 2025-04-01 2025-06-30

另外，它对输入的顺序、大小写和额外的空格都具有容忍性：

getStartEndDate(c('2023 q1', 'Q3 2024 ', ' q2 2025 '))
#>        start        end
#> 1 2023-01-01 2023-03-31
#> 2 2024-07-01 2024-09-30
#> 3 2025-04-01 2025-06-30

如果输入不符合要求，它会给出有用的错误信息：

getStartEndDate('Q3')
#> Error in FUN(X[[i]], ...) : 
#>   "quarter" must be a string in format "q1 2020"

<details>
<summary>英文:</summary>

It&#39;s not clear what format you want the input to be, but I assume you want it to be a string in the format &quot;q3 2020&quot;. We can write a function that will be tolerant of whether the quarter comes before or after the year, and whether the &quot;q&quot; is uppercase or lowercase, and also allow vectorized input:
``` r
getStartEndDate &lt;- function(quarter) {
  # Convert input to lowercase letters and remove any whitespace around text
  quarter &lt;- trimws(tolower(quarter))

  # Split the input strings by the whitespace in the middle to get
  # a list of length-2 strings, (the year and the quarter)
  quarter &lt;- strsplit(quarter, &#39;\\s+&#39;)

  # Only the year portion will convert from string to numeric without NAs
  # so we find this part for each pair of strings and get a vector of all 
  # the years. We throw an error if there are not two strings for each
  # item on our list, and if we fail to extract a year, sapply will return
  # a list instead of a vector so we also throw an error at this stage:
  years &lt;- sapply(quarter, function(x) {
    if(length(x) != 2) stop(&#39;&quot;quarter&quot; must be a string in format &quot;q1 2020&quot;&#39;)
    x &lt;- suppressWarnings(as.numeric(x))
    x[!is.na(x)]})
  if(is.list(years)) stop(&#39;&quot;quarter&quot; must be a string in format &quot;q1 2020&quot;&#39;)
  
  # Go through our list and pick out the q1, q2, q3, or q4, converting it to
  # the number 1:4 by removing the q and calling as.numeric
  quarters &lt;- sapply(quarter, function(x) {
    x &lt;- grep(&#39;q(1|2|3|4)&#39;, x, value = TRUE)
    if(length(x) != 1) stop(&#39;&quot;quarter&quot; must be a string in format &quot;q1 2020&quot;&#39;)
    as.numeric(sub(&#39;q&#39;, &#39;&#39;, x))})
  
  # Now we use Map to get a vector of quarters starting from each year, and 
  # subset the correct one using our quarters. These are made into a one-row
  # data frame for each year-quarter then all bound into a single data frame.
  do.call(rbind, Map(function(y, q) {
    dates &lt;- seq(as.Date(paste0(y, &#39;-01-01&#39;)), by = &#39;quarter&#39;, length = 5)
    data.frame(start = dates[q], end = dates[q + 1] - 1)}, y = years, q = quarters))
}

Testing on a simple case, we have:

getStartEndDate('q3 2020')
#>        start        end
#> 1 2020-07-01 2020-09-30

This will work for many centuries to come:

getStartEndDate('q4 2500')
#>        start        end
#> 1 2500-10-01 2500-12-31

It can take multiple inputs in vetorised form:

getStartEndDate(c('q1 2023', 'q3 2024 ', 'q2 2025'))
#>        start        end
#> 1 2023-01-01 2023-03-31
#> 2 2024-07-01 2024-09-30
#> 3 2025-04-01 2025-06-30

It is tolerant of ordering, capitalization and accidental whitespaces:

getStartEndDate(c('2023 q1', 'Q3 2024 ', ' q2 2025 '))
#>        start        end
#> 1 2023-01-01 2023-03-31
#> 2 2024-07-01 2024-09-30
#> 3 2025-04-01 2025-06-30

And throws a helpful error message if the wrong input is received:

getStartEndDate('Q3')
#> Error in FUN(X[[i]], ...) : 
#>   "quarter" must be a string in format "q1 2020" 

^{Created on 2023-02-16 with reprex v2.0.2}