填写data.table缺失日期的最快方法

Question

我正在data.table从包含日期,订单,金额等字段的CSV文件中加载.

输入文件偶尔没有所有日期的数据.例如,如下所示:

> NADayWiseOrders
           date orders  amount guests
  1: 2013-01-01     50 2272.55    149
  2: 2013-01-02      3   64.04      4
  3: 2013-01-04      1   18.81      0
  4: 2013-01-05      2   77.62      0
  5: 2013-01-07      2   35.82      2

在上面的03年1月和6月6日没有任何条目.

想要用缺省值填充缺失的条目(例如,订单为零,金额等),或者最后一个vaue(例如,03-Jan将重用02-Jan值,06-Jan将重用05-Jan价值观等.)

使用此类默认值填充缺失日期数据缺口的最佳/最佳方法是什么？

这里的答案建议使用allow.cartesian = TRUE,并且expand.grid对于缺少工作日 - 它可能适用于工作日(因为它们只是7个工作日) - 但不确定这是否也是正确的约会方式,特别是如果我们处理多个年度数据.

Answer 1

惯用的data.table方式(使用滚动连接)是这样的:

setkey(NADayWiseOrders, date)
all_dates <- seq(from = as.Date("2013-01-01"), 
                   to = as.Date("2013-01-07"), 
                   by = "days")

NADayWiseOrders[J(all_dates), roll=Inf]
         date orders  amount guests
1: 2013-01-01     50 2272.55    149
2: 2013-01-02      3   64.04      4
3: 2013-01-03      3   64.04      4
4: 2013-01-04      1   18.81      0
5: 2013-01-05      2   77.62      0
6: 2013-01-06      2   77.62      0
7: 2013-01-07      2   35.82      2

Answer 2

不确定它是否是最快的,但如果NA数据中没有s,它将起作用:

# just in case these aren't Dates. 
NADayWiseOrders$date <- as.Date(NADayWiseOrders$date)
# all desired dates.
alldates <- data.table(date=seq.Date(min(NADayWiseOrders$date), max(NADayWiseOrders$date), by="day"))
# merge
dt <- merge(NADayWiseOrders, alldates, by="date", all=TRUE)
# now carry forward last observation (alternatively, set NA's to 0)
require(xts)
na.locf(dt)

Answer 3

以下是您如何填补小组内的空白

# a toy dataset with gaps in the time series
dt <- as.data.table(read.csv(textConnection('"group","date","x"
"a","2017-01-01",1
"a","2017-02-01",2
"a","2017-05-01",3
"b","2017-02-01",4
"b","2017-04-01",5')))
dt[,date := as.Date(date)]

# the desired dates by group
indx <- dt[,.(date=seq(min(date),max(date),"months")),group]

# key the tables and join them using a rolling join
setkey(dt,group,date)
setkey(indx,group,date)
dt[indx,roll=TRUE]

#>    group       date x
#> 1:     a 2017-01-01 1
#> 2:     a 2017-02-01 2
#> 3:     a 2017-03-01 2
#> 4:     a 2017-04-01 2
#> 5:     a 2017-05-01 3
#> 6:     b 2017-02-01 4
#> 7:     b 2017-03-01 4
#> 8:     b 2017-04-01 5