Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
684 views
in Technique[技术] by (71.8m points)

dataframe - efficiently locf by groups in a single R data.table

I have a large, wide data.table (20m rows) keyed by a person ID but with lots of columns (~150) that have lots of null values. Each column is a recorded state / attribute that I wish to carry forward for each person. Each person may have anywhere from 10 to 10,000 observations and there are about 500,000 people in the set. Values from one person can not 'bleed' into the following person, so my solution must respect the person ID column and group appropriately.

For demonstration purposes - here's a very small sample input:

DT = data.table(
  id=c(1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3),
  aa=c("A", NA, "B", "C", NA, NA, "D", "E", "F", NA, NA, NA),
  bb=c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA),
  cc=c(1, NA, NA, NA, NA, 4, NA, 5, 6, NA, 7, NA)
)

It looks like this:

    id aa bb cc
 1:  1  A NA  1
 2:  1 NA NA NA
 3:  1  B NA NA
 4:  1  C NA NA
 5:  2 NA NA NA
 6:  2 NA NA  4
 7:  2  D NA NA
 8:  2  E NA  5
 9:  3  F NA  6
10:  3 NA NA NA
11:  3 NA NA  7
12:  3 NA NA NA

My expected output looks like this:

    id aa bb cc
 1:  1  A NA  1
 2:  1  A NA  1
 3:  1  B NA  1
 4:  1  C NA  1
 5:  2 NA NA NA
 6:  2 NA NA  4
 7:  2  D NA  4
 8:  2  E NA  5
 9:  3  F NA  6
10:  3  F NA  6
11:  3  F NA  7
12:  3  F NA  7

I've found a data.table solution that works, but it's terribly slow on my large data sets:

DT[, na.locf(.SD, na.rm=FALSE), by=id]

I've found equivalent solutions using dplyr that are equally slow.

GRP = DT %>% group_by(id)
data.table(GRP %>% mutate_each(funs(blah=na.locf(., na.rm=FALSE))))

I was hopeful that I could come up with a rolling 'self' join using the data.table functionality, but I just can't seem to get it right (I suspect I would need to use .N but I just haven't figured it out).

At this point I'm thinking I'll have to write something in Rcpp to efficiently apply the grouped locf.

I'm new to R, but I'm not new to C++ - so I'm confident I can do it. I just feel like there should be an efficient way to do this in R using data.table.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

A very simple na.locf can be built by forwarding (cummax) the non-NA indices ((!is.na(x)) * seq_along(x)) and subsetting accordingly:

x = c(1, NA, NA, 6, 4, 5, 4, NA, NA, 2)
x[cummax((!is.na(x)) * seq_along(x))]
# [1] 1 1 1 6 4 5 4 4 4 2

This replicates na.locf with an na.rm = TRUE argument, to get na.rm = FALSE behavior we simply need to make sure the first element in the cummax is TRUE:

x = c(NA, NA, 1, NA, 2)
x[cummax(c(TRUE, tail((!is.na(x)) * seq_along(x), -1)))]
#[1] NA NA  1  1  2

In this case, we need to take into account not only the non-NA indices but, also, of the indices where the (ordered, or to be ordered) "id" column changes value:

id = c(10, 10, 11, 11, 11, 12, 12, 12, 13, 13)
c(TRUE, id[-1] != id[-length(id)])
# [1]  TRUE FALSE  TRUE FALSE FALSE  TRUE FALSE FALSE  TRUE FALSE

Combining the above:

id = c(10, 10, 11, 11, 11, 12, 12, 12, 13, 13)
x =  c(1,  NA, NA, 6,  4,  5,  4,  NA, NA, 2)

x[cummax(((!is.na(x)) | c(TRUE, id[-1] != id[-length(id)])) * seq_along(x))]
# [1]  1  1 NA  6  4  5  4  4 NA  2

Note, that here we OR the first element with TRUE, i.e. make it equal to TRUE, thus getting the na.rm = FALSE behavior.

And for this example:

id_change = DT[, c(TRUE, id[-1] != id[-.N])]
DT[, lapply(.SD, function(x) x[cummax(((!is.na(x)) | id_change) * .I)])]
#    id aa bb cc
# 1:  1  A NA  1
# 2:  1  A NA  1
# 3:  1  B NA  1
# 4:  1  C NA  1
# 5:  2 NA NA NA
# 6:  2 NA NA  4
# 7:  2  D NA  4
# 8:  2  E NA  5
# 9:  3  F NA  6
#10:  3  F NA  6
#11:  3  F NA  7
#12:  3  F NA  7

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...