按組連續的天數-突變函數Dplyr中的錯誤

Question

這是以下問題的延續： 在R中按組記錄連續天數

答案適用於我發布的示例中的數據集，但我意識到我的實際數據集有問題，並且出現了一個錯誤，指出： Error: incompatible size (0), expecting 1 (the group size) or 1

下面是出現錯誤的數據集和可復制示例。 有人知道為什么會這樣嗎？

DATE <- as.Date(c('2016-10-26', '2016-10-30', '2016-10-26', '2016-10-20', '2016-10-21', '2016-10-17', '2016-10-26', '2016-10-17', '2016-10-18', '2016-10-20', '2016-10-17', '2016-10-18', '2016-10-17', '2016-10-18', '2016-10-19','2016-10-18', '2016-10-19','2016-10-17','2016-10-17','2016-10-19','2016-10-19','2016-10-20','2016-10-19','2016-10-20','2016-10-30'))
`Parent` <- c('A','A','A','A','A','A','A','B', 'B', 'B', 'C', 'C', 'D', 'D', 'D', 'D', 'D', 'E', 'E', 'F', 'G', 'G', 'G', 'G', 'G')
Child <- c('ab', 'ac', 'ad', 'ae', 'ae','af', 'af','ba', 'ba', 'ba', 'ca', 'cb', 'da', 'da', 'da', 'db', 'db', 'ea', 'eb', 'fa', 'ga', 'ga', 'gb', 'gb', 'gb')
salary <- c(290.45, 0.00, 336.51, 2238.56, 2256.75, 725.73, 319.69, 46.48, 42.13, 43.22, 0.41, 865.20, 1889.80, 2691.97, 3016.80, 8636.18, 8540.24, 1587.21, 1416.63, 79.62,1967.95,1947.35,34925.58,31158.51,6973.54)
avg_child_salary <- c(500.29, 526.27, 492.00, 1197.25, 1197.25, 474.10, 474.10, 21.68, 21.68, 21.68, 0.05, 199.90, 575.31, 575.31, 575.31, 1701.82, 1701.82, 495.48, 316.93, 26.16, 582.66, 582.66, 18089.83, 18089.83, 18089.83)
Callout <- c('LOW', 'LOW', 'LOW', 'HIGH', 'HIGH', 'HIGH', 'LOW', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'LOW')
employ.data <- data.frame(DATE, Parent, Child, avg_child_salary, salary, Callout)

employ.data

         DATE Parent Child avg_child_salary   salary Callout
1  2016-10-26      A    ab           500.29   290.45     LOW
2  2016-10-30      A    ac           526.27     0.00     LOW
3  2016-10-26      A    ad           492.00   336.51     LOW
4  2016-10-20      A    ae          1197.25  2238.56    HIGH
5  2016-10-21      A    ae          1197.25  2256.75    HIGH
6  2016-10-17      A    af           474.10   725.73    HIGH
7  2016-10-26      A    af           474.10   319.69     LOW
8  2016-10-17      B    ba            21.68    46.48    HIGH
9  2016-10-18      B    ba            21.68    42.13    HIGH
10 2016-10-20      B    ba            21.68    43.22    HIGH
11 2016-10-17      C    ca             0.05     0.41    HIGH
12 2016-10-18      C    cb           199.90   865.20    HIGH
13 2016-10-17      D    da           575.31  1889.80    HIGH
14 2016-10-18      D    da           575.31  2691.97    HIGH
15 2016-10-19      D    da           575.31  3016.80    HIGH
16 2016-10-18      D    db          1701.82  8636.18    HIGH
17 2016-10-19      D    db          1701.82  8540.24    HIGH
18 2016-10-17      E    ea           495.48  1587.21    HIGH
19 2016-10-17      E    eb           316.93  1416.63    HIGH
20 2016-10-19      F    fa            26.16    79.62    HIGH
21 2016-10-19      G    ga           582.66  1967.95    HIGH
22 2016-10-20      G    ga           582.66  1947.35    HIGH
23 2016-10-19      G    gb         18089.83 34925.58    HIGH
24 2016-10-20      G    gb         18089.83 31158.51    HIGH
25 2016-10-30      G    gb         18089.83  6973.54     LOW

然后，我要從該數據集中收集包含2016-10-30所有行，然后在單獨的列中，根據employee.data數據幀，用LOW或HIGH標注計數連續的天數。 連續天數必須在“標注”旁邊的新列中。 這是在應用錯誤的腳本之前：

yesterday <- as.Date(Sys.Date()-37)
df2<-filter(employ.data, DATE == yesterday)
df2 

         DATE Parent Child avg_child_salary   salary Callout  
2  2016-10-30      A    ac           526.27     0.00     LOW                          
25 2016-10-30      G    gb         18089.83  6973.54     LOW                          

嘗試的代碼如下：

library(dplyr)
yesterday <- as.Date(Sys.Date()-37) ##because today is 12/6/16
df2 <- employ.data %>% group_by(Child) %>%
  mutate(`Consec. Days with Callout`=cumsum(rev(cumprod(rev((yesterday-DATE)==(which(DATE == yesterday)-row_number()) & Callout==Callout[DATE == yesterday]))))) %>% filter(DATE == yesterday)

最后，對於此特定示例，它需要看起來像這樣：

         DATE Parent Child avg_child_salary   salary Callout  Consec. Days with Callout
2  2016-10-30      A    ac           526.27     0.00     LOW                          1
25 2016-10-30      G    gb         18089.83  6973.54     LOW                          1

然后出現錯誤：

Error: incompatible size (0), expecting 1 (the group size) or 1

Answer 1

問題是，對於某些組，找不到yesterday的行。 可以通過定義一個檢查該功能的函數來解決此問題，而不是在mutate中內聯該函數：

library(dplyr)
compute.consec.days <- function(date, callout, yesterday, rown) {
  j <- which(date == yesterday)
  if (length(j)==0) NA else cumsum(rev(cumprod(rev((yesterday-date)==(j-rown) & callout==callout[date == yesterday]))))
}

該函數檢查which DATE是yesterday 。 如果找不到該組，則將返回integer(0) 。 我們通過返回值j的length進行檢查。 如果為TRUE ，則連續兩天返回NA ，這沒有關系，因為以下filter將刪除該組（即找不到yesterday ）； 否則，我們將像以前一樣計算連續的天數。 這樣可以避免錯誤。 現在，使用此功能和您新發布的數據：

yesterday <- as.Date("2016-10-30")
out <- employ.data %>% group_by(Child) %>%
  mutate(`Consec. Days with Callout`=compute.consec.days(DATE,Callout,yesterday,row_number())) %>%
  filter(DATE == yesterday)
##Source: local data frame [2 x 7]
##Groups: Child [2]
##
##        DATE Parent  Child avg_child_salary  salary Callout Consec. Days with Callout
##      <date> <fctr> <fctr>            <dbl>   <dbl>  <fctr>                     <dbl>
##1 2016-10-30      A     ac           526.27    0.00     LOW                         1
##2 2016-10-30      G     gb         18089.83 6973.54     LOW                         1

---

更新以支持昨天不是組中最后日期的情況

如果查詢yesterday不是任何的最后一天， Child組，那么我們需要修改我們的compute.consec.days功能，例如：

compute.consec.days <- function(date, callout, yesterday, rown) {
  j <- which(date == yesterday)
  if (length(j)==0) NA else {
    ## first compute the condition
    cond <- (yesterday-date)==(j-rown) & callout==callout[date == yesterday]
    ## then evaluate consecutive days only with this vector up to
    ## the row corresponding to yesterday. Then add the result with NAs
    ## because mutate is a windowing function
    c(cumsum(rev(cumprod(rev(cond[1:j[1]])))),rep(NA,length(date)-j[1]))
  }
}

例如，如果給定新發布的數據，昨天的查詢為"2016-10-20" ，則結果為：

yesterday <- as.Date("2016-10-20")
out <- employ.data %>% group_by(Child) %>%
  mutate(`Consec. Days with Callout`=compute.consec.days(DATE,Callout,yesterday,row_number())) %>%
  filter(DATE == yesterday)
##Source: local data frame [4 x 7]
##Groups: Child [4]
##
##        DATE Parent  Child avg_child_salary   salary Callout Consec. Days with Callout
##      <date> <fctr> <fctr>            <dbl>    <dbl>  <fctr>                     <dbl>
##1 2016-10-20      A     ae          1197.25  2238.56    HIGH                         1
##2 2016-10-20      B     ba            21.68    43.22    HIGH                         1
##3 2016-10-20      G     ga           582.66  1947.35    HIGH                         2
##4 2016-10-20      G     gb         18089.83 31158.51    HIGH                         2

原始查詢為"2016-10-30" ，我們仍然得到原始結果：

yesterday <- as.Date("2016-10-30")
out <- employ.data %>% group_by(Child) %>%
  mutate(`Consec. Days with Callout`=compute.consec.days(DATE,Callout,yesterday,row_number())) %>%
  filter(DATE == yesterday)
##Source: local data frame [2 x 7]
##Groups: Child [2]
##
##        DATE Parent  Child avg_child_salary  salary Callout Consec. Days with Callout
##      <date> <fctr> <fctr>            <dbl>   <dbl>  <fctr>                     <dbl>
##1 2016-10-30      A     ac           526.27    0.00     LOW                         1
##2 2016-10-30      G     gb         18089.83 6973.54     LOW                         1