Question

我在日期系列和一个组中汇总数据时出现问题，其中一个日期在一个但不是所有组中都缺失。

dates <- seq.Date(as.Date("2010-01-01"), by=7, length.out=5)
dates.2 <- dates[-2]
all.dates <- c(dates, dates, dates.2, dates.2)
subgroups <- c(rep("a", 5), rep("b", 5), rep("c", 4), rep("d", 4))
groups <- c(rep("X", 10), rep("Y", 8))
set.seed(2)

  df.1 <- data.frame(Date = all.dates, 
  Group = groups, 
  Subgrp = subgroups,
  Cost = runif(18,100,200)
)
df.1

         Date Group Subgrp     Cost
1  2010-01-01     X      a 118.4882
2  2010-01-08     X      a 170.2374
3  2010-01-15     X      a 157.3326
4  2010-01-22     X      a 116.8052
5  2010-01-29     X      a 194.3839
6  2010-01-01     X      b 194.3475
7  2010-01-08     X      b 112.9159
8  2010-01-15     X      b 183.3449
9  2010-01-22     X      b 146.8019
10 2010-01-29     X      b 154.9984
11 2010-01-01     Y      c 155.2674
12 2010-01-15     Y      c 123.8895
13 2010-01-22     Y      c 176.0513
14 2010-01-29     Y      c 118.0820
15 2010-01-01     Y      d 140.5282
16 2010-01-15     Y      d 185.3548
17 2010-01-22     Y      d 197.6398
18 2010-01-29     Y      d 122.5825

> ag.1 <- aggregate(Cost ~ Group + Date, FUN=sum,  data=df.1)
> ag.1
  Group       Date     Cost
1     X 2010-01-01 312.8357
2     Y 2010-01-01 295.7956
3     X 2010-01-08 283.1533
4     X 2010-01-15 340.6775
5     Y 2010-01-15 309.2443
6     X 2010-01-22 263.6070
7     Y 2010-01-22 373.6912
8     X 2010-01-29 349.3823
9     Y 2010-01-29 240.6646

小组Y未在2010-01-08上付款，但ag.1对象在此日期对群组Y保持沉默。我希望ag.1有一行反映这一点：

> ag.1
      Group       Date     Cost
    1     X 2010-01-01 312.8357
    2     Y 2010-01-01 295.7956
    3     X 2010-01-08 283.1533
    3a    Y 2010-01-08   0.0000
    4     X 2010-01-15 340.6775
    5     Y 2010-01-15 309.2443

我在na.omit=na.pass函数中尝试aggregate但是（1）我真的不知道这是做什么的，（2）它没有改变输出。

欢迎不使用aggregate的建议，但更愿意使用基础包。

Answer 1

expand.grid可用于填写缺失的条目。

df.2 <- expand.grid(Date = unique(dates),Group = unique(groups))
df <- merge(df.1,df.2,all=TRUE)

aggregate(Cost ~ Group + Date, FUN=sum,  data=df, na.action=na.pass)

编辑：根据OP的提示，我找到了对aggregate电话的适当调整。

   Group       Date     Cost
1      X 2010-01-01 312.8357
2      Y 2010-01-01 295.7956
3      X 2010-01-08 283.1533
4      Y 2010-01-08       NA
5      X 2010-01-15 340.6775
6      Y 2010-01-15 309.2443
7      X 2010-01-22 263.6070
8      Y 2010-01-22 373.6912
9      X 2010-01-29 349.3823
10     Y 2010-01-29 240.6646

Answer 2

1）只要任何日期至少有一个具有该日期的组，那么就这样做：

> as.data.frame(xtabs(Cost ~ Date + Group, df.1), responseName = "Cost")
         Date Group     Cost
1  2010-01-01     X 312.8357
2  2010-01-08     X 283.1533
3  2010-01-15     X 340.6775
4  2010-01-22     X 263.6070
5  2010-01-29     X 349.3823
6  2010-01-01     Y 295.7956
7  2010-01-08     Y   0.0000
8  2010-01-15     Y 309.2443
9  2010-01-22     Y 373.6912
10 2010-01-29     Y 240.6646

事实上，如果这个布局合适，上面的xtabs部分可能就是你所需要的：

> xtabs(Cost ~ Date + Group, df.1)
            Group
Date                X        Y
  2010-01-01 312.8357 295.7956
  2010-01-08 283.1533   0.0000
  2010-01-15 340.6775 309.2443
  2010-01-22 263.6070 373.6912
  2010-01-29 349.3823 240.6646

2）如果有没有任何组有条目的日期，则将日期转换为包含在级别中的非出现日期的因子：

> # define levels to be all weeks between minimum date and 2010-02-05
> levs <- as.character(seq(min(df.1$Date), as.Date("2010-02-05"), by = 7))
> df.2 <- transform(df.1, Date = factor(Date, sort(unique(levs))))
>
> # now repeat using df.2
> as.data.frame(xtabs(Cost ~ Date + Group, df.2), responseName = "Cost")
         Date Group     Cost
1  2010-01-01     X 312.8357
2  2010-01-08     X 283.1533
3  2010-01-15     X 340.6775
4  2010-01-22     X 263.6070
5  2010-01-29     X 349.3823
6  2010-02-05     X   0.0000
7  2010-01-01     Y 295.7956
8  2010-01-08     Y   0.0000
9  2010-01-15     Y 309.2443
10 2010-01-22     Y 373.6912
11 2010-01-29     Y 240.6646
12 2010-02-05     Y   0.0000

> xtabs(Cost ~ Date + Group, df.2)
            Group
Date                X        Y
  2010-01-01 312.8357 295.7956
  2010-01-08 283.1533   0.0000
  2010-01-15 340.6775 309.2443
  2010-01-22 263.6070 373.6912
  2010-01-29 349.3823 240.6646
  2010-02-05   0.0000   0.0000

R按时间序列聚合，缺少某些组的日期

2 个答案: