累积日期范围内的总和

huangapple
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英文:

Cumulative sum for specific range of dates

问题

I'm trying to calculate the rowise cumulative sum of Rates from DATE to DATE_following.

我试图计算从DATE到DATE_following的逐行累积总和Rates。

For example:

例如：

  1. library(bizdays)
  2. library(lubridate)
  4. set.seed(1)
  5. dat <- seq.Date(from = as.Date(as.Date("2023-04-06")- days(10)),
  6.                 to = as.Date(as.Date("2023-04-06")),
  7.                 by = "day") %>% 
  8.   data.frame(DATE = .) %>% 
  9.   mutate(Rates = sample(seq(from=1,to=10,by=1), size = length(DATE),replace=TRUE),
  10.          DATE_following = modified.following(DATE %m+% days(3)))
  12. dat
  13.         DATE Rates DATE_following
  14. 1  2023-03-27     9     2023-03-30
  15. 2  2023-03-28     4     2023-03-31
  16. 3  2023-03-29     7     2023-04-01
  17. 4  2023-03-30     1     2023-04-02
  18. 5  2023-03-31     2     2023-04-03
  19. 6  2023-04-01     7     2023-04-04
  20. 7  2023-04-02     2     2023-04-05
  21. 8  2023-04-03     3     2023-04-06
  22. 9  2023-04-04     1     2023-04-07
  23. 10 2023-04-05     5     2023-04-08
  24. 11 2023-04-06     5     2023-04-09

The output i'm trying to get is:

我想要的输出是：

  1. Result: 9+4+7+1 = 21 (the sum of Rates from 2023-03-27 to 2023-03-30 )

  2. Result: 4+7+1+2 = 14 ...

  3. 结果：9+4+7+1 = 21（从2023-03-27到2023-03-30的Rates总和）

  4. 结果：4+7+1+2 = 14...

  1.          DATE Rates DATE_following Results
  2. 1  2023-03-27     9     2023-03-30      21
  3. 2  2023-03-28     4     2023-03-31      14
  4. 3  2023-03-29     7     2023-04-01      17
  5. 4  2023-03-30     1     2023-04-02      12
  6. 5  2023-03-31     2     2023-04-03      14
  7. 6  2023-04-01     7     2023-04-04      13
  8. 7  2023-04-02     2     2023-04-05      11
  9. 8  2023-04-03     3     2023-04-06      14
  10. 9  2023-04-04     1     2023-04-07      NA
  11. 10 2023-04-05     5     2023-04-08      NA
  12. 11 2023-04-06     5     2023-04-09      NA

Is it possible to get this result using dplyr functions like rowwise() and cumsum()? My main problem is that I don't know how to define this condition within these functions.

是否可以使用dplyr函数如rowwise()和cumsum()来获得这个结果？我的主要问题是不知道如何在这些函数内定义这个条件。

英文:

I'm trying to calculate the rowise cumulative sum of Rates from DATE to DATE_following.

For example:

  1. library(tidyverse)
  2. library(bizdays)
  3. library(lubridate)
  5. set.seed(1)
  6. dat &lt;- seq.Date(from = as.Date(as.Date(&quot;2023-04-06&quot;)- days(10)),
  7.                 to = as.Date(as.Date(&quot;2023-04-06&quot;)),
  8.                 by = &quot;day&quot;)  %&gt;% 
  9.   data.frame(DATE = .) %&gt;% 
  10.   mutate(Rates = sample(seq(from=1,to=10,by=1), size = length(DATE),replace=TRUE),
  11.          DATE_following = modified.following(DATE %m+% days(3)))
  13. dat
  14.         DATE Rates DATE_following
  15. 1  2023-03-27     9     2023-03-30
  16. 2  2023-03-28     4     2023-03-31
  17. 3  2023-03-29     7     2023-04-01
  18. 4  2023-03-30     1     2023-04-02
  19. 5  2023-03-31     2     2023-04-03
  20. 6  2023-04-01     7     2023-04-04
  21. 7  2023-04-02     2     2023-04-05
  22. 8  2023-04-03     3     2023-04-06
  23. 9  2023-04-04     1     2023-04-07
  24. 10 2023-04-05     5     2023-04-08
  25. 11 2023-04-06     5     2023-04-09

The output i'm trying to get is:

  1. Result: 9+4+7+1 = 21 (the sum of Rates from 2023-03-27 to 2023-03-30 )
  2. Result: 4+7+1+2 = 14 ...
  1.          DATE Rates DATE_following Results
  2. 1  2023-03-27     9     2023-03-30      21
  3. 2  2023-03-28     4     2023-03-31      14
  4. 3  2023-03-29     7     2023-04-01      17
  5. 4  2023-03-30     1     2023-04-02      12
  6. 5  2023-03-31     2     2023-04-03      14
  7. 6  2023-04-01     7     2023-04-04      13
  8. 7  2023-04-02     2     2023-04-05      11
  9. 8  2023-04-03     3     2023-04-06      14
  10. 9  2023-04-04     1     2023-04-07      NA
  11. 10 2023-04-05     5     2023-04-08      NA
  12. 11 2023-04-06     5     2023-04-09      NA

Is it possible to get this result using dplyr functions like rowwise() and cumsum()? My main problem is that I don't know how to define this condition within these functions.

答案1

得分: 4

如果您想要对连续的四个Rates进行滚动求和，您可以使用zoo的rollsum()函数：

  1. library(dplyr)
  2. library(zoo)
  4. dat %>% 
  5.   mutate(Result = rollsum(Rates, k = 4, fill = NA_real_, align = "left"))

这将返回：

  1. # A tibble: 11 × 5
  2.       no DATE       Rates DATE_following Result
  3.    <dbl> <date>     <dbl> <date>          <dbl>
  4.  1     1 2023-03-27     9 2023-03-30         21
  5.  2     2 2023-03-28     4 2023-03-31         14
  6.  3     3 2023-03-29     7 2023-04-01         17
  7.  4     4 2023-03-30     1 2023-04-02         12
  8.  5     5 2023-03-31     2 2023-04-03         14
  9.  6     6 2023-04-01     7 2023-04-04         13
  10.  7     7 2023-04-02     2 2023-04-05         11
  11.  8     8 2023-04-03     3 2023-04-06         14
  12.  9     9 2023-04-04     1 2023-04-07         NA
  13. 10    10 2023-04-05     5 2023-04-08         NA
  14. 11    11 2023-04-06     5 2023-04-09         NA

基于LeMarque的评论，还有一个稍微通用的答案：

  1. dat2 %>% 
  2.   mutate(days = as.integer(DATE_following - DATE) + 1,
  3.          res = rollapply(data = Rates, width = days, FUN = sum, align = "left", fill = NA_real_))

这将返回：

  1. # A tibble: 11 × 6
  2.       no DATE       Rates DATE_following  days   res
  3.    <dbl> <date>     <dbl> <date>         <dbl> <dbl>
  4.  1     1 2023-03-27     9 2023-03-30         4    21
  5.  2     2 2023-03-28     4 2023-03-31         4    14
  6.  3     3 2023-03-29     7 2023-04-01         4    17
  7.  4     4 2023-03-30     1 2023-04-02         4    12
  8.  5     5 2023-03-31     2 2023-04-10        11    NA
  9.  6     6 2023-04-01     7 2023-04-04         4    13
  10.  7     7 2023-04-02     2 2023-04-05         4    11
  11.  8     8 2023-04-03     3 2023-04-06         4    14
  12.  9     9 2023-04-04     1 2023-04-07         4    NA
  13. 10    10 2023-04-05     5 2023-04-08         4    NA
  14. 11    11 2023-04-06     5 2023-04-09         4    NA

由于第5行的DATE_following在数据中不存在，此版本返回NA。此外，这个版本不是对连续的四天进行求和，而是计算了DATE和DATE_following之间的天数，并将它们应用于滚动求和。

英文:

If you want a rolling sum for four consecutive Rates, you could use zoos rollsum() function:

  1. library(dplyr)
  2. library(zoo)
  4. dat %&gt;% 
  5.   mutate(Result = rollsum(Rates, k = 4, fill = NA_real_, align = &quot;left&quot;))

This returns

  1. # A tibble: 11 &#215; 5
  2.       no DATE       Rates DATE_following Result
  3.    &lt;dbl&gt; &lt;date&gt;     &lt;dbl&gt; &lt;date&gt;          &lt;dbl&gt;
  4.  1     1 2023-03-27     9 2023-03-30         21
  5.  2     2 2023-03-28     4 2023-03-31         14
  6.  3     3 2023-03-29     7 2023-04-01         17
  7.  4     4 2023-03-30     1 2023-04-02         12
  8.  5     5 2023-03-31     2 2023-04-03         14
  9.  6     6 2023-04-01     7 2023-04-04         13
  10.  7     7 2023-04-02     2 2023-04-05         11
  11.  8     8 2023-04-03     3 2023-04-06         14
  12.  9     9 2023-04-04     1 2023-04-07         NA
  13. 10    10 2023-04-05     5 2023-04-08         NA
  14. 11    11 2023-04-06     5 2023-04-09         NA

A slightly more general answer based on LeMarque's comment:

  1. dat2 %&gt;% 
  2.   mutate(days = as.integer(DATE_following - DATE) + 1,
  3.          res = rollapply(data = Rates, width = days, FUN = sum, align = &quot;left&quot;, fill = NA_real_))

This returns

  1. # A tibble: 11 &#215; 6
  2.       no DATE       Rates DATE_following  days   res
  3.    &lt;dbl&gt; &lt;date&gt;     &lt;dbl&gt; &lt;date&gt;         &lt;dbl&gt; &lt;dbl&gt;
  4.  1     1 2023-03-27     9 2023-03-30         4    21
  5.  2     2 2023-03-28     4 2023-03-31         4    14
  6.  3     3 2023-03-29     7 2023-04-01         4    17
  7.  4     4 2023-03-30     1 2023-04-02         4    12
  8.  5     5 2023-03-31     2 2023-04-10        11    NA
  9.  6     6 2023-04-01     7 2023-04-04         4    13
  10.  7     7 2023-04-02     2 2023-04-05         4    11
  11.  8     8 2023-04-03     3 2023-04-06         4    14
  12.  9     9 2023-04-04     1 2023-04-07         4    NA
  13. 10    10 2023-04-05     5 2023-04-08         4    NA
  14. 11    11 2023-04-06     5 2023-04-09         4    NA

Since the DATE_following in row 5 isn't present in the data, this version returns NA. Furthermore this version doesn't sum four consecutive days but calculates the days between DATE and DATE_following and applies them to the rolling sum.

Data

  1. dat &lt;- structure(list(no = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11), DATE = structure(c(19443, 
  2. 19444, 19445, 19446, 19447, 19448, 19449, 19450, 19451, 19452, 
  3. 19453), class = &quot;Date&quot;), Rates = c(9, 4, 7, 1, 2, 7, 2, 3, 1, 
  4. 5, 5), DATE_following = structure(c(19446, 19447, 19448, 19449, 
  5. 19450, 19451, 19452, 19453, 19454, 19455, 19456), class = &quot;Date&quot;)), class = c(&quot;spec_tbl_df&quot;, 
  6. &quot;tbl_df&quot;, &quot;tbl&quot;, &quot;data.frame&quot;), row.names = c(NA, -11L), spec = structure(list(
  7.     cols = list(no = structure(list(), class = c(&quot;collector_double&quot;, 
  8.     &quot;collector&quot;)), DATE = structure(list(format = &quot;&quot;), class = c(&quot;collector_date&quot;, 
  9.     &quot;collector&quot;)), Rates = structure(list(), class = c(&quot;collector_double&quot;, 
  10.     &quot;collector&quot;)), DATE_following = structure(list(format = &quot;&quot;), class = c(&quot;collector_date&quot;, 
  11.     &quot;collector&quot;))), default = structure(list(), class = c(&quot;collector_guess&quot;, 
  12.     &quot;collector&quot;)), skip = 1L), class = &quot;col_spec&quot;))
  14. dat2 &lt;- structure(list(no = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11), DATE = structure(c(19443, 
  15. 19444, 19445, 19446, 19447, 19448, 19449, 19450, 19451, 19452, 
  16. 19453), class = &quot;Date&quot;), Rates = c(9, 4, 7, 1, 2, 7, 2, 3, 1, 
  17. 5, 5), DATE_following = structure(c(19446, 19447, 19448, 19449, 
  18. 19457, 19451, 19452, 19453, 19454, 19455, 19456), class = &quot;Date&quot;)), class = c(&quot;spec_tbl_df&quot;, 
  19. &quot;tbl_df&quot;, &quot;tbl&quot;, &quot;data.frame&quot;), row.names = c(NA, -11L), spec = structure(list(
  20.     cols = list(no = structure(list(), class = c(&quot;collector_double&quot;, 
  21.     &quot;collector&quot;)), DATE = structure(list(format = &quot;&quot;), class = c(&quot;collector_date&quot;, 
  22.     &quot;collector&quot;)), Rates = structure(list(), class = c(&quot;collector_double&quot;, 
  23.     &quot;collector&quot;)), DATE_following = structure(list(format = &quot;&quot;), class = c(&quot;collector_date&quot;, 
  24.     &quot;collector&quot;))), default = structure(list(), class = c(&quot;collector_guess&quot;, 
  25.     &quot;collector&quot;)), skip = 1L), class = &quot;col_spec&quot;))

答案2

得分: 2

这是您需要的内容：

  1. library(zoo)
  2. library(tidyverse)
  4. set.seed(1)
  5. dat <- seq.Date(from = as.Date(today()- days(10)), to = as.Date(today()), by = "day") %>%
  6.   data.frame(DATE = .) %>%
  7.   mutate(Rates = sample(seq(from=1,to=10,by=1), size = length(DATE),,replace=TRUE),
  8.          DATE_following = DATE %m+% days(3),
  9.          Results = rollapply(data = Rates, width = 4, FUN = sum, align = "left", fill = NA, partial = TRUE)) %>%
  10.   mutate( Results = ifelse(DATE_following %in% DATE, Results, NA))
  12. dat

希望这对您有帮助。

英文:

Did you need something like this:

  1. library(zoo)
  2. library(tidyverse)
  3. set.seed(1)
  4. dat &lt;- seq.Date(from = as.Date(today()- days(10)), to = as.Date(today()), by = &quot;day&quot;)  %&gt;% 
  5. data.frame(DATE = .) %&gt;% 
  6. mutate(Rates = sample(seq(from=1,to=10,by=1), size = length(DATE),,replace=TRUE),
  7. DATE_following = DATE %m+% days(3),
  8. Results = rollapply(data = Rates, width = 4, FUN = sum, align = &quot;left&quot;, fill = NA, partial = TRUE)) %&gt;%
  9. mutate( Results = ifelse(DATE_following %in% DATE, Results, NA))

dat

which results in:

  1.           DATE Rates DATE_following Results
  2. 1  2023-03-27     3     2023-03-30      20
  3. 2  2023-03-28     3     2023-03-31      24
  4. 3  2023-03-29     8     2023-04-01      27
  5. 4  2023-03-30     6     2023-04-02      27
  6. 5  2023-03-31     7     2023-04-03      28
  7. 6  2023-04-01     6     2023-04-04      22
  8. 7  2023-04-02     8     2023-04-05      20
  9. 8  2023-04-03     7     2023-04-06      20
  10. 9  2023-04-04     1     2023-04-07      NA
  11. 10 2023-04-05     4     2023-04-08      NA
  12. 11 2023-04-06     8     2023-04-09      NA

Please check and let me know...

答案3

得分: 2

Sure, here are the translated code parts:

如果你想要一个滚动总和，你可以使用 filter。

  1. rev(filter(rev(dat$Rates), rep(1,4), side=1))
  2. #rev(stats::filter(rev(dat$Rates), rep(1,4), side=1)) #如果使用了遮盖了 stats::filter 的 dplyr
  3. # [1] 21 14 17 12 14 13 11 14 NA NA NA

如果需要匹配日期：

  1. mapply(\(a,b) if(is.na(b)) NA else sum(dat$Rates[a:b]),
  2.   seq_len(nrow(dat)), match(dat$DATE_following, dat$DATE))
  3. # [1] 21 14 17 12 14 13 11 14 NA NA NA

或者在日期没有排序且不需要匹配所有日期或无需匹配的情况下：

  1. mapply(\(a,b) sum(dat$Rates[dat$DATE >= a & dat$DATE <= b]),
  2.   dat$DATE, dat$DATE_following)
  3. # [1] 21 14 17 12 14 13 11 14 11 10  5

感谢 @Martin Gal 提供的数据！

英文:

If you want a rolling sum you can use filter.

  1. rev(filter(rev(dat$Rates), rep(1,4), side=1))
  2. #rev(stats::filter(rev(dat$Rates), rep(1,4), side=1)) #In case using dplyr which is masking stats::filter
  3. # [1] 21 14 17 12 14 13 11 14 NA NA NA

In case the dates should be matched:

  1. mapply(\(a,b) if(is.na(b)) NA else sum(dat$Rates[a:b]),
  2. seq_len(nrow(dat)), match(dat$DATE_following, dat$DATE))
  3. # [1] 21 14 17 12 14 13 11 14 NA NA NA

Or in case it is not sorted and there is no need that all dates are present nor need to match:

  1. mapply(\(a,b) sum(dat$Rates[dat$DATE &gt;= a &amp; dat$DATE &lt;= b]),
  2. +   dat$DATE, dat$DATE_following)
  3. # [1] 21 14 17 12 14 13 11 14 11 10  5

Thanks to @Martin Gal for providing the data!

答案4

得分: 1

以下是 data.table 中的两种方法：

  1. frollsum 方法
  1. library(data.table)
  2. setDT(dat)
  3. dat[,
  4.   Results := frollsum(Rates, DATE_following - DATE + 1, adaptive = TRUE)
  5. ][,
  6.   Results := Results[order(is.na(Results))]
  7. ][]
  1. non-equi join 方法
  1. library(data.table)
  2. setDT(dat)
  3. dat[
  4.   dat[
  5.     dat,
  6.     on = .(DATE &lt;= DATE_following)
  7.   ][
  8.     DATE_following &gt;= DATE,
  9.     .(Results = sum(Rates)),
  10.     i.DATE
  11.   ],
  12.   Results := Results * (match(DATE_following, DATE) &gt; 0),
  13.   on = .(DATE = i.DATE)
  14. ][]

输出

  1.           DATE Rates DATE_following Results
  2.  1: 2023-03-27     9     2023-03-30      21
  3.  2: 2023-03-28     4     2023-03-31      14
  4.  3: 2023-03-29     7     2023-04-01      17
  5.  4: 2023-03-30     1     2023-04-02      12
  6.  5: 2023-03-31     2     2023-04-03      14
  7.  6: 2023-04-01     7     2023-04-04      13
  8.  7: 2023-04-02     2     2023-04-05      11
  9.  8: 2023-04-03     3     2023-04-06      14
  10.  9: 2023-04-04     1     2023-04-07      NA
  11. 10: 2023-04-05     5     2023-04-08      NA
  12. 11: 2023-04-06     5     2023-04-09      NA
英文:

Here are some data.table option

  1. frollsum approach
  1. library(data.table)
  2. setDT(dat)
  3. dat[
  4. ,
  5. Results := frollsum(Rates, DATE_following - DATE + 1, adaptive = TRUE)
  6. ][
  7. ,
  8. Results := Results[order(is.na(Results))]
  9. ][]
  1. non-equi join approach
  1. library(data.table)
  2. setDT(dat)
  3. dat[
  4. dat[
  5. dat,
  6. on = .(DATE &lt;= DATE_following)
  7. ][
  8. DATE_following &gt;= DATE,
  9. .(Results = sum(Rates)),
  10. i.DATE
  11. ],
  12. Results := Results * (match(DATE_following, DATE) &gt; 0),
  13. on = .(DATE = i.DATE)
  14. ][]

Output

  1.           DATE Rates DATE_following Results
  2. 1: 2023-03-27     9     2023-03-30      21
  3. 2: 2023-03-28     4     2023-03-31      14
  4. 3: 2023-03-29     7     2023-04-01      17
  5. 4: 2023-03-30     1     2023-04-02      12
  6. 5: 2023-03-31     2     2023-04-03      14
  7. 6: 2023-04-01     7     2023-04-04      13
  8. 7: 2023-04-02     2     2023-04-05      11
  9. 8: 2023-04-03     3     2023-04-06      14
  10. 9: 2023-04-04     1     2023-04-07      NA
  11. 10: 2023-04-05     5     2023-04-08      NA
  12. 11: 2023-04-06     5     2023-04-09      NA

答案5

得分: 1

以下是代码的翻译部分：

  1. # remotes::install_github(&quot;NicChr/timeplyr&quot;)
  2. library(timeplyr)
  3. library(dplyr)
  4. library(lubridate)
  5. set.seed(1)
  6. dat &lt;- seq.Date(from = as.Date(Sys.Date()- days(10)),
  7.                 to = as.Date(Sys.Date()),
  8.                 by = &quot;day&quot;)  %&gt;% 
  9.   data.frame(DATE = .) %&gt;% 
  10.   mutate(Rates = sample(seq(from=1,to=10,by=1), size = length(DATE),,replace=TRUE),
  11.          DATE_following =DATE %m+% days(3))
  12. # Vectorised seq function
  13. x1 &lt;- time_seq_v(dat$DATE, dat$DATE_following, by = &quot;days&quot;)
  14. # Label these based on the relevant rows
  15. x2 &lt;- rep(seq_len(nrow(dat)), time_seq_len(dat$DATE, dat$DATE_following, by = &quot;days&quot;))
  17. # Add these sequences to the data
  18. dat &lt;- dat %&gt;%
  19. mutate(dates = split(x1, x2))
  21. # Sum
  22. my_sum &lt;- numeric(nrow(dat))
  23. for (i in seq_len(nrow(dat))){
  24.   my_sum[[i]] &lt;- sum(dat$Rates[dat$DATE %in% dat$dates[[i]]])
  25. }
  26. dat$Result &lt;- my_sum
  28. dat

希望这对您有所帮助。

英文:

A rolling sum would indeed be more efficient, but this could be a potential solution if your start and end points are dynamic.

  1. # remotes::install_github(&quot;NicChr/timeplyr&quot;)
  2. library(timeplyr)
  3. library(dplyr)
  4. library(lubridate)
  5. set.seed(1)
  6. dat &lt;- seq.Date(from = as.Date(Sys.Date()- days(10)),
  7.                 to = as.Date(Sys.Date()),
  8.                 by = &quot;day&quot;)  %&gt;% 
  9.   data.frame(DATE = .) %&gt;% 
  10.   mutate(Rates = sample(seq(from=1,to=10,by=1), size = length(DATE),,replace=TRUE),
  11.          DATE_following =DATE %m+% days(3))
  12. # Vectorised seq function
  13. x1 &lt;- time_seq_v(dat$DATE, dat$DATE_following, by = &quot;days&quot;)
  14. # Label these based on the relevant rows
  15. x2 &lt;- rep(seq_len(nrow(dat)), time_seq_len(dat$DATE, dat$DATE_following, by = &quot;days&quot;))
  17. # Add these sequences to the data
  18. dat &lt;- dat %&gt;%
  19.   as_tibble() %&gt;%
  20.   mutate(dates = split(x1, x2))
  24. # Sum
  25. my_sum &lt;- numeric(nrow(dat))
  26. for (i in seq_len(nrow(dat))){
  27.   my_sum[[i]] &lt;- sum(dat$Rates[dat$DATE %in% dat$dates[[i]]])
  28. }
  29. dat$Result &lt;- my_sum
  31. dat
  32. #&gt; # A tibble: 11 x 5
  33. #&gt;    DATE       Rates DATE_following dates        Result
  34. #&gt;    &lt;date&gt;     &lt;dbl&gt; &lt;date&gt;         &lt;named list&gt;  &lt;dbl&gt;
  35. #&gt;  1 2023-03-27     9 2023-03-30     &lt;date [4]&gt;       21
  36. #&gt;  2 2023-03-28     4 2023-03-31     &lt;date [4]&gt;       14
  37. #&gt;  3 2023-03-29     7 2023-04-01     &lt;date [4]&gt;       17
  38. #&gt;  4 2023-03-30     1 2023-04-02     &lt;date [4]&gt;       12
  39. #&gt;  5 2023-03-31     2 2023-04-03     &lt;date [4]&gt;       14
  40. #&gt;  6 2023-04-01     7 2023-04-04     &lt;date [4]&gt;       13
  41. #&gt;  7 2023-04-02     2 2023-04-05     &lt;date [4]&gt;       11
  42. #&gt;  8 2023-04-03     3 2023-04-06     &lt;date [4]&gt;       14
  43. #&gt;  9 2023-04-04     1 2023-04-07     &lt;date [4]&gt;       11
  44. #&gt; 10 2023-04-05     5 2023-04-08     &lt;date [4]&gt;       10
  45. #&gt; 11 2023-04-06     5 2023-04-09     &lt;date [4]&gt;        5

<sup>Created on 2023-04-06 with reprex v2.0.2</sup>

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  • 本文由 发表于 2023年4月6日 19:15:49
  • 转载请务必保留本文链接：https://go.coder-hub.com/75948891.html
  • cumsum
  • cumulative-sum
  • dataframe
  • r
  • rowwise
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