Efficient way to Fill Time-Series per group

问题

I was looking for a way to fill a time series data set by time, per group. The very very inefficient way I was using was to split the data set per group and apply a custom time-series fill function (create sequence between max and min, and merge) in all elements of that list. Needless to say, this operations would not go pass the splitting.

My dataset looks like,

    source                 grp cnt
 1:     83 2017-06-06 13:00:00   1
 2:     83 2017-06-06 23:00:00   1
 3:     83 2017-06-07 03:00:00   1
 4:     83 2017-06-07 07:00:00   2
 5:     83 2017-06-07 13:00:00   1
 6:     83 2017-06-07 19:00:00   1
 7:     83 2017-06-08 00:00:00   1
 8:     83 2017-06-08 14:00:00   1
 9:     83 2017-06-08 15:00:00   1
10:     83 2017-06-08 20:00:00   1
11:    137 2017-06-04 02:00:00   1
12:    137 2017-06-04 05:00:00   1
13:    137 2017-06-04 23:00:00   1
...

My attempt was to use tidyverse methods by utilising the complete function, i.e.

library(tidyverse)

d1 %>% 
 group_by(source) %>% 
 complete(source, grp = seq(min(grp), max(grp), by = 'hour'))

However, after about 40-45 seconds, a progress bar appeared (apparently a neat feature in some tidyverse functions - I suspect complete in this case) which estimated 9 hours to completion. My dataset is very very big and this is not the lightest operation, so something really efficient is what I am looking for.

DATA

#dput(d1)
structure(list(source = c("83", "83", "83", "83", "83", "83", 
"83", "83", "83", "83", "137", "137", "137", "137", "137", "137", 
"137", "137", "137", "137", "137", "137", "137", "137"), grp = structure(c(1496743200, 
1496779200, 1496793600, 1496808000, 1496829600, 1496851200, 1496869200, 
1496919600, 1496923200, 1496941200, 1496530800, 1496541600, 1496606400, 
1496617200, 1496649600, 1496696400, 1496808000, 1496844000, 1496876400, 
1496962800, 1497880800, 1497888000, 1497978000, 1497996000), class = c("POSIXct", 
"POSIXt"), tzone = ""), cnt = c(1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L
)), .Names = c("source", "grp", "cnt"), row.names = c(NA, -24L
), class = "data.frame")

回答1:

It appears that data.table is really much faster than the tidyverse option. So merely translating the above into data.table(compliments of @Frank) completed the operation in little under 3 minutes.

library(data.table)

mDT = setDT(d1)[, .(grp = seq(min(grp), max(grp), by = "hour")), by = source]
new_D <- d1[mDT, on = names(mDT)]

new_D <- new_D[, cnt := replace(cnt, is.na(cnt), 0)] #If needed

回答2:

This can be done using zoo as well. This is an order of magnitude faster than the code and data in the question but not as fast as the data.table solution although there exists the possibility of speeding it iup further if the last line of code shown below is not needed.

We read d1 into a zoo object z splitting it to give a multivariate time series having a column for each source. We then merge that with a zero width series having all the times and fortify that back to a data frame using the melt=TRUE argument to get a long form data.frame. If a wide form multivariate zoo series can be used then you could skip the last line in which case it would then be even faster.

library(zoo)

z <- read.zoo(d1, split = 1, index = 2) # wide form
zz <- merge(z, zoo(, seq(start(z), end(z), "hour"))) # expand
fortify(zz, melt = TRUE) # convert to long form data.frame

来源：https://stackoverflow.com/questions/46712379/efficient-way-to-fill-time-series-per-group

标签

data.table

tidyverse