按组连续几天 - 变异函数Dplyr中的错误

时间：2016-12-06 16:04:34

标签： r join dplyr

这是问题的延续：Record Consecutive Days by Group in R

答案适用于我发布的示例中的数据集，但我意识到我的实际数据集存在问题，并且出现错误：Error: incompatible size (0), expecting 1 (the group size) or 1

以下是出现错误的数据集和可重现的示例。有人知道为什么会这样吗？

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，我希望从此数据集中收集包含2016-10-30的所有行，然后在单独的列中，根据标注LOW或HIGH计算连续天数employ.data数据框。连续天数需要在Callout旁边的新列中。这是在应用错误脚本之前：

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

尝试的代码如下：

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，对于这个特定的例子，它需要看起来像这样：

         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

然后出现错误：

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

1 个答案:

答案 0 :(得分：2)

问题是，对于某些群组，找不到yesterday的行。这可以通过定义一个检查它的函数来修复，而不是在mutate：

中内联函数 
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]))))
}

此功能检查which DATE是yesterday。如果找不到该组，则会返回integer(0)。我们通过返回值length的{​​{1}}进行检查。如果这是j，我们会连续几天返回TRUE，这无关紧要，因为以下NA会删除该组（即找不到filter）;否则，我们像以前一样计算连续的天数。这可以避免错误。现在，使用此功能和您新发布的数据：

yesterday

更新以支持昨天不是最后日期的情况

如果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 的查询不是任何yesterday群组的最后一天，那么我们需要修改我们的Child功能：

compute.consec.days

例如，如果给定新发布的数据，昨天的查询为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])) } } ，则会导致：

"2016-10-20"

使用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 的原始查询，我们仍然可以获得原始结果：

"2016-10-30"
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