Question

我想知道如何为每个参数组合创建未来的时间戳（BranchId，Hour，weekdays）

BranchId      Hour   weekdays     ActivityDate     Total   
    1          11      3      2018-02-06T00:00:00   18
    1          11      3      2018-02-13T00:00:00   23
    1          12      3      2018-02-06T00:00:00   15
    1          12      3      2018-02-13T00:00:00   13
    1          13      3      2018-02-06T00:00:00   24
    1          13      3      2018-02-13T00:00:00   22

目前我只能为一个组合创建未来的时间戳，如下所示：

BranchId    Hour    weekdays    ActivityDate       Total
    1        11        3     2018-02-06T00:00:00Z   18
    1        11        3     2018-02-13T00:00:00Z   23
    1        11        3     2018-02-20T00:00:00Z   
    1        11        3     2018-02-27T00:00:00Z   
    1        11        3     2018-03-06T00:00:00Z   
    1        11        3     2018-03-13T00:00:00Z

代码是：

min.date <- min(data$ActivityDate)
max.date <- max(data$ActivityDate)


unique.time <- seq(from = min.date,  to = max.date, by = "week")

forecast.time <- seq(from = max.date, by = observation.freq, length.out = 4 + 1)[-1]
all.time <- c(unique.time, forecast.time)
all.time <- data.frame(BranchId = data$BranchId[1], Hour = data$Hour[1], weekdays = data$weekdays[1],ActivityDate = all.time)

# Join the combination with original data
data <- join(all.time, data, by = c("BranchId","Hour", "weekdays", "ActivityDate"), type = "left")

当我在其上应用此代码时，结果出错了，它不会为每个组合创建未来的时间戳：

BranchId    Hour    weekdays    ActivityDate      Total
   1         11        3    2018-02-06T00:00:00Z    18
   1         11        3    2018-02-13T00:00:00Z    23
   1         12        3    2018-02-20T00:00:00Z    
   1         12        3    2018-02-27T00:00:00Z    
   1         13        3    2018-03-06T00:00:00Z    
   1         13        3    2018-03-13T00:00:00Z

我是否需要生成多个函数或for循环来接近它？

Answer 1

使用R，您可以使用以下代码创建每周间隔的额外日期。使用包pad中的函数padr，您可以定义向data.frame添加日期的时间间隔。使用group选项告诉函数应该使用哪些变量来创建新的时间轴或填写缺少的日期。您可以指定开始日期和结束日期，以便在这些日期之间进行所有操作，否则它将是data.frame中可用的min和max日期。

min.date <- min(df$ActivityDate)
max.date <- as.Date("2018-03-13T00:00:00Z")

library(padr)

df <- pad(df, interval = "week" , start_val = min.date, end_val = max.date, group = c("BranchId", "Hour", "weekdays"))

# this step can be skipped if you want to keep NA's instead of 0 
df <- fill_by_value(df, value = 0)
df

   BranchId Hour weekdays ActivityDate Total
1         1   11        3   2018-02-06    18
2         1   11        3   2018-02-13    23
3         1   11        3   2018-02-20     0
4         1   11        3   2018-02-27     0
5         1   11        3   2018-03-06     0
6         1   11        3   2018-03-13     0
7         1   12        3   2018-02-06    15
8         1   12        3   2018-02-13    13
9         1   12        3   2018-02-20     0
10        1   12        3   2018-02-27     0
11        1   12        3   2018-03-06     0
12        1   12        3   2018-03-13     0
13        1   13        3   2018-02-06    24
14        1   13        3   2018-02-13    22
15        1   13        3   2018-02-20     0
16        1   13        3   2018-02-27     0
17        1   13        3   2018-03-06     0
18        1   13        3   2018-03-13     0

数据：

df <- structure(list(BranchId = c(1L, 1L, 1L, 1L, 1L, 1L), 
                     Hour = c(11L, 11L, 12L, 12L, 13L, 13L), 
                     weekdays = c(3L, 3L, 3L, 3L, 3L, 3L),
                     ActivityDate = as.Date(c("2018-02-06T00:00:00", "2018-02-13T00:00:00","2018-02-06T00:00:00", 
                                      "2018-02-13T00:00:00", "2018-02-06T00:00:00", "2018-02-13T00:00:00")), 
                     Total = c(18L, 23L, 15L, 13L, 24L, 22L)), 
                .Names = c("BranchId", "Hour", "weekdays", "ActivityDate", "Total"),
                class = "data.frame", row.names = c(NA, -6L))

Answer 2

在R中完成

df%>%group_by(BranchId,Hour,weekdays)%>%complete(ActivityDate = seq.Date(min(ActivityDate), min(ActivityDate)+42, by="week"),fill=list(Total=0))
# A tibble: 21 x 5
# Groups:   BranchId, Hour, weekdays [3]
   BranchId  Hour weekdays ActivityDate Total
      <int> <int>    <int>       <date> <dbl>
 1        1    11        3   2018-02-06    18
 2        1    11        3   2018-02-13    23
 3        1    11        3   2018-02-20     0
 4        1    11        3   2018-02-27     0
 5        1    11        3   2018-03-06     0
 6        1    11        3   2018-03-13     0
 7        1    11        3   2018-03-20     0
 8        1    12        3   2018-02-06    15
 9        1    12        3   2018-02-13    13
10        1    12        3   2018-02-20     0

为每个参数组合创建未来时间戳

2 个答案: