Filling missing dates in a grouped time series - a tidyverse-way?

问题

Given a data.frame that contains a time series and one or ore grouping fields. So we have several time series - one for each grouping combination. But some dates are missing. So, what's the easiest (in terms of the most "tidyverse way") of adding these dates with the right grouping values?

Normally I would say I generate a data.frame with all dates and do a full_join with my time series. But now we have to do it for each combination of grouping values -- and fill in the grouping values.

Let's look at an example:

First I create a data.frame with missing values:

library(dplyr)
library(lubridate)

set.seed(1234)
# Time series should run vom 2017-01-01 til 2017-01-10
date <- data.frame(date = seq.Date(from=ymd("2017-01-01"), to=ymd("2017-01-10"), by="days"), v = 1)
# Two grouping dimensions
d1   <- data.frame(d1 = c("A", "B", "C", "D"), v = 1)
d2   <- data.frame(d2 = c(1, 2, 3, 4, 5), v = 1)

# Generate the data.frame
df <- full_join(date, full_join(d1, d2)) %>%
  select(date, d1, d2) 
# and ad to value columns
df$v1 <- runif(200)
df$v2 <- runif(200)

# group by the dimension columns
df <- df %>% 
  group_by(d1, d2)

# create missing dates
df.missing <- df %>%
  filter(v1 <= 0.8)

# So now  2017-01-01 and 2017-01-10, A, 5 are missing now
df.missing %>%
  filter(d1 == "A" & d2 == 5)

# A tibble: 8 x 5
# Groups:   d1, d2 [1]
        date     d1    d2         v1        v2
      <date> <fctr> <dbl>      <dbl>     <dbl>
1 2017-01-02      A     5 0.21879954 0.1335497
2 2017-01-03      A     5 0.32977018 0.9802127
3 2017-01-04      A     5 0.23902573 0.1206089
4 2017-01-05      A     5 0.19617465 0.7378315
5 2017-01-06      A     5 0.13373890 0.9493668
6 2017-01-07      A     5 0.48613541 0.3392834
7 2017-01-08      A     5 0.35698708 0.3696965
8 2017-01-09      A     5 0.08498474 0.8354756

So to add the missing dates I generate a data.frame with all dates:

start <- min(df.missing$date)
end   <- max(df.missing$date)

all.dates <- data.frame(date=seq.Date(start, end, by="day"))

No I want to do something like (remember: df.missing is group_by(d1, d2))

df.missing %>%
  do(my_join())

So let's define my_join():

my_join <- function(data) {
  # get value of both dimensions
  d1.set <- data$d1[[1]]
  d2.set <- data$d2[[1]]

  tmp <- full_join(data, all.dates) %>%
    # First we need to ungroup.  Otherwise we can't change d1 and d2 because they are grouping variables
    ungroup() %>%
    mutate(
      d1 = d1.set,
      d2 = d2.set 
    ) %>%
    group_by(d1, d2)

  return(tmp)
}

Now we can call my_join() for each combination and have a look at "A/5"

df.missing %>%
  do(my_join(.)) %>%
  filter(d1 == "A" & d2 == 5)

# A tibble: 10 x 5
# Groups:   d1, d2 [1]
         date     d1    d2         v1        v2
       <date> <fctr> <dbl>      <dbl>     <dbl>
 1 2017-01-02      A     5 0.21879954 0.1335497
 2 2017-01-03      A     5 0.32977018 0.9802127
 3 2017-01-04      A     5 0.23902573 0.1206089
 4 2017-01-05      A     5 0.19617465 0.7378315
 5 2017-01-06      A     5 0.13373890 0.9493668
 6 2017-01-07      A     5 0.48613541 0.3392834
 7 2017-01-08      A     5 0.35698708 0.3696965
 8 2017-01-09      A     5 0.08498474 0.8354756
 9 2017-01-01      A     5         NA        NA
10 2017-01-10      A     5         NA        NA

Great! That's what we were looking for. But we need to define d1 and d2 in my_join and it feels a little bit clumsy.

So, is there any tidyverse-way of this solution?

P.S.: I've put the code into a gist: https://gist.github.com/JerryWho/1bf919ef73792569eb38f6462c6d7a8e

回答1:

tidyr has some great tools for these sorts of problems. Take a look at complete.

library(dplyr)
library(tidyr)
library(lubridate)

want <- df.missing %>% 
  ungroup() %>%
  complete(nesting(d1, d2), date = seq(min(date), max(date), by = "day"))

want %>% filter(d1 == "A" & d2 == 5) 

#> # A tibble: 10 x 5
#> # Groups:   d1 [1]
#>        d1    d2       date         v1        v2
#>    <fctr> <dbl>     <date>      <dbl>     <dbl>
#>  1      A     5 2017-01-01         NA        NA
#>  2      A     5 2017-01-02 0.21879954 0.1335497
#>  3      A     5 2017-01-03 0.32977018 0.9802127
#>  4      A     5 2017-01-04 0.23902573 0.1206089
#>  5      A     5 2017-01-05 0.19617465 0.7378315
#>  6      A     5 2017-01-06 0.13373890 0.9493668
#>  7      A     5 2017-01-07 0.48613541 0.3392834
#>  8      A     5 2017-01-08 0.35698708 0.3696965
#>  9      A     5 2017-01-09 0.08498474 0.8354756
#> 10      A     5 2017-01-10         NA        NA

回答2:

Here's a tidyverse way starting with df.missing

library(tidyverse)
ans <- df.missing %>% 
          nest(date) %>% 
          mutate(data = map(data, ~seq.Date(start, end, by="day"))) %>% 
          unnest(data) %>%
          rename(date = data) %>%
          left_join(., df.missing, by=c("date","d1","d2"))

ans %>% filter(d1 == "A" & d2 == 5)

Output

      d1    d2       date         v1        v2
   <fctr> <dbl>     <date>      <dbl>     <dbl>
 1      A     5 2017-01-01         NA        NA
 2      A     5 2017-01-02 0.21879954 0.1335497
 3      A     5 2017-01-03 0.32977018 0.9802127
 4      A     5 2017-01-04 0.23902573 0.1206089
 5      A     5 2017-01-05 0.19617465 0.7378315
 6      A     5 2017-01-06 0.13373890 0.9493668
 7      A     5 2017-01-07 0.48613541 0.3392834
 8      A     5 2017-01-08 0.35698708 0.3696965
 9      A     5 2017-01-09 0.08498474 0.8354756
10      A     5 2017-01-10         NA        NA

-------------------------------------------------------------------------------------------------
Here's an alternative approach that uses expand.grid and dplyr verbs

with(df.missing, expand.grid(unique(date), unique(d1), unique(d2))) %>%
  setNames(c("date", "d1", "d2")) %>%
  left_join(., df.missing, by=c("date","d1","d2"))

output (head)

          date d1 d2          v1          v2
1   2017-01-01  A  1 0.113703411 0.660754634
2   2017-01-02  A  1 0.316612455 0.422330675
3   2017-01-03  A  1 0.553333591 0.424109178
4   2017-01-04  A  1          NA          NA
5   2017-01-05  A  1          NA          NA
6   2017-01-06  A  1 0.035456727 0.352998502

回答3:

Here read.zoo creates a wide form zoo object and to that we merge the dates. Then we convert that back to a long data frame using fortify.zoo and spread out out v1 and v2 using spread.

Note that:

if we can assume that each date appears in at least one combination of the split variables, i.e. sort(unique(df.missing$date)) contains all the dates, then we could omit the merge line and no joins would have to be done at all. The test data df.missing shown in the question does have this property:
```
all(all.dates$date %in% df.missing$date)
## [1] TRUE
```
we could stop after the merge (or after read.zoo if each date is present at least once as in prior point) if a wide form zoo object can be used as that already has all the dates.

In the code below the line marked ### can be omitted with the development version of zoo (1.8.1):

library(dplyr)
library(tidyr)
library(zoo)

split.vars <- c("d1", "d2")
df.missing %>%
   as.data.frame %>%     ###
   read.zoo(split = split.vars) %>%
   merge(zoo(, seq(start(.), end(.), "day"))) %>%
   fortify.zoo(melt = TRUE) %>%
   separate(Series, c("v", split.vars)) %>%
   spread(v, Value)

Update: Note simplification in zoo 1.8.1 .

回答4:

package tsibble function fill_gaps should do the job easily.

library(tsibble)
df.missing %>% 
  # tsibble format
  as_tsibble(key = c(d1, d2), index = date) %>% 
  # fill gaps
  fill_gaps(.full = TRUE)

来源：https://stackoverflow.com/questions/46130246/filling-missing-dates-in-a-grouped-time-series-a-tidyverse-way

标签

dplyr

time-series

tidyverse