Question

我要生成列：“ PriorityCountInLast7Days”。对于给定的员工A，此列计算最近7天的案例数，其中PRIORITY与当前案例相同。我该如何在R中使用前4列？

data <- data.frame(
    Date = c("2018-06-01", "2018-06-03", "2018-06-03", "2018-06-03",  "2018-06-04", "2018-06-01", "2018-06-02", "2018-06-03"),
Emp1 = c("A","A","A","A","A","A","B","B","B"),
Case = c("A1", "A2", "A3", "A4", "A5", "A6", "B1", "B2", "B3"),
Priority = c(0,0,0,1,2,0,0,0,0),
PriorityCountinLast7days = c(0,1,2,1,1,3,1,2,3))

+------------+------+------+----------+--------------------------+
|    Date    | Emp1 | Case | Priority | PriorityCountinLast7days |
+------------+------+------+----------+--------------------------+
| 2018-06-01 | A    | A1   |        0 |                        0 |
| 2018-06-03 | A    | A2   |        0 |                        1 |
| 2018-06-03 | A    | A3   |        0 |                        2 |
| 2018-06-03 | A    | A4   |        1 |                        1 |
| 2018-06-03 | A    | A5   |        2 |                        1 |
| 2018-06-04 | A    | A6   |        0 |                        3 |
| 2018-06-01 | B    | B1   |        0 |                        1 |
| 2018-06-02 | B    | B2   |        0 |                        2 |
| 2018-06-03 | B    | B3   |        0 |                        3 |
+------------+------+------+----------+--------------------------+

Answer 1

您可以在整个数据集上使用迭代条件总和来完成此滚动窗口。这是什么意思？在for循环中，您可以检查当前日期> =要包含的日期，以及要包含> =的日期到7天前的日期，并且要包含的个案是==当前的个案。循环中的这种逻辑组合将为您创建此滚动过滤器。这是一个函数：

rollPriority <- function(data, window = 7){
  stopifnot(all(c("Date","Case","Priority") %in% colnames(data))) # string error check
  data$Date <- as.Date(data$Date)
  for(i in 1:nrow(data)){
    #priorxdays <= dates we want <= current date
    datecheck <- (data$Date[i] - (window-1)) <= data$Date & data$Date <= data$Date[i]
    casecheck <- data$Case == data$Case[i]
    data$PriorityCountinLastXdays[i] = sum(data$Priority[which(datecheck & casecheck)])
  }
  Xdays <- which(colnames(data) == "PriorityCountinLastXdays")
  colnames(data)[Xdays] <- paste0("PriorityCountinLast", window, "days")
  return(data)
}

将来，请提供可重复输出的示例数据。您会注意到，我们仅看到4天的信息就无法满足您预期的7天滚动输出。一种快速的方法是使用expand.grid()生成组合，并使用set.seed()保留采样输出：

# Reproducible Example Data
dat <- expand.grid(Date = seq.Date(as.Date("2018-06-01"),
                                   as.Date("2018-06-4"), 
                                   by = "day"), 
                   Case = as.factor(sort(apply(expand.grid(c("A","B"),1:2), 
                                               1, 
                                               paste0, 
                                               collapse = ""))))
# Ensures random sampling is identical each time
set.seed(42); 
dat$Priority <- sample(0:1, nrow(dat), replace = T)

# The function
rollPriority(dat, 2)
#         Date Case Priority PriorityCountinLast2days
#1  2018-06-01   A1        1                        1
#2  2018-06-02   A1        1                        2
#3  2018-06-03   A1        0                        1
#4  2018-06-04   A1        1                        1
#5  2018-06-01   A2        1                        1
#6  2018-06-02   A2        1                        2
#7  2018-06-03   A2        1                        2
#8  2018-06-04   A2        0                        1
#9  2018-06-01   B1        1                        1
#10 2018-06-02   B1        1                        2
#11 2018-06-03   B1        0                        1
#12 2018-06-04   B1        1                        1
#13 2018-06-01   B2        1                        1
#14 2018-06-02   B2        0                        1
#15 2018-06-03   B2        0                        0
#16 2018-06-04   B2        1                        1

这样，某人更容易准确地为您提供帮助。

计算因素在滚动窗口中出现的次数

1 个答案: