fill in missing data for group by unique ID [duplicate]

问题

My clinical data structure looks like this:

patientid <- c(100,100,100,101,101,101,102,102,102,104,104,104)
group <- c(1,1,NA,2,NA,NA,1,1,1,2,2,NA)

Data<- data.frame(patientid=patientid,group=group)

If there is missing data then the NA should become the same value as the other group value for the same patient id. In other words a patient is always in the same group and the missing data needs to be filled in to reflect that. So it should look like this:

patientid <- c(100,100,100,101,101,101,102,102,102,104,104,104)
group <- c(1,1,1,2,2,2,1,1,1,2,2,2)

Data<- data.frame(patientid=patientid,group=group)

回答1:

You can write a little helper function like:

fun <- function(x) replace(x, is.na(x), x[!is.na(x)][1])

Then, you can use it in transform or within in base R:

transform(Data, group = ave(group, patientid, FUN = fun))
#    patientid group
# 1        100     1
# 2        100     1
# 3        100     1
# 4        101     2
# 5        101     2
# 6        101     2
# 7        102     1
# 8        102     1
# 9        102     1
# 10       104     2
# 11       104     2
# 12       104     2

Or even with other packages:

library(data.table)
as.data.table(Data)[, group := fun(group), patientid][]

This will work even if the group value is not the first value for each patient "patientid". Try, for example:

# First row of "group" is `NA`
Data <- Data[c(3, 1, 2, 4:nrow(Data)), ]

回答2:

We can use fill from tidyr after grouping by 'patientid'

library(dplyr)
library(tidyr)
Data %>%
     group_by(patientid) %>% 
     fill(group) %>%
     ungroup

-output

# A tibble: 12 x 2
#   patientid group
#       <dbl> <dbl>
# 1       100     1
# 2       100     1
# 3       100     1
# 4       101     2
# 5       101     2
# 6       101     2
# 7       102     1
# 8       102     1
# 9       102     1
#10       104     2
#11       104     2
#12       104     2

回答3:

A base R option with ave can make it

transform(
  Data,
  group = ave(group, patientid, FUN = function(x) unique(na.omit(x)))
)

which gives

   patientid group
1        100     1
2        100     1
3        100     1
4        101     2
5        101     2
6        101     2
7        102     1
8        102     1
9        102     1
10       104     2
11       104     2
12       104     2

A data.table option with nafill

setDT(Data)[, group := nafill(group, fill = unique(na.omit(group))), patientid]

which gives

> Data
    patientid group
 1:       100     1
 2:       100     1
 3:       100     1
 4:       101     2
 5:       101     2
 6:       101     2
 7:       102     1
 8:       102     1
 9:       102     1
10:       104     2
11:       104     2
12:       104     2

回答4:

dplyr:

Data%>%group_by(patientid)%>%mutate(group=(ifelse(is.na(group),max(group,na.rm=TRUE),group)))

回答5:

You can create a mapping between group and patientid, then use this mapping to fill in missing values.

# Create mapping btw group and patientid
df = data.frame(patientid, group)
patientid.unique = unique(patientid)
mapping = data.frame(pid=patientid.unique, group=NA)
for (pid in patientid.unique){
  mapping$group[mapping$pid == pid] = unique(df[df$patientid == pid & !is.na(df$group), "group"])
}

> mapping
  pid group
1 100     1
2 101     2
3 102     1
4 104     2

# Fill in missing values
group.filled = apply(df, 1, function(x) {mapping$group[mapping$pid == x[1]] })
df$group = group.filled

来源：https://stackoverflow.com/questions/65297927/fill-in-missing-data-for-group-by-unique-id

标签

missing-data