为R中的每个ID每月添加一行

Question

So I have a list of IDs.

I would like to create a table consisting of rows for each month of a year for each id in the list.

I tried to use a rbind in a for loop but this takes forever... as such:

for (k in seq_along(members))
{
  for (i in seq(1,12))
  {
    df1<-rbind(df1, data.frame(MemYearMo=paste(members[k],"_",year,formatC(i,width=2,flag=0), sep="")))
  }
}

where members is obviously my list of id's.

My desired output is: XXX_201701 XXX_201702 XXX_201703 . . . XXX 201712

where XXX is one of my id's.

What would be the fastest way to do something like this?

Answer 1

I'm not sure I understand your desired output but you can use expand.grid to generate the different combinations of member ids with the months of the year.

Example

df <- as.data.frame(expand.grid(members = 901:903, ym = 201701:201712))
df$MemYearMo <- paste(df$members, df$ym, sep = "_")
df
#>    members     ym  MemYearMo
#> 1      901 201701 901_201701
#> 2      902 201701 902_201701
#> 3      903 201701 903_201701
#> 4      901 201702 901_201702
#> 5      902 201702 902_201702
#> 6      903 201702 903_201702
#> ...(and so on)...
#> 34     901 201712 901_201712
#> 35     902 201712 902_201712
#> 36     903 201712 903_201712

Answer 2

Using data.table and paste0 you can do this, assuming that your id's are unique.

id <- as.data.table(letters)
id <- id[, .(output = paste0(letters, paste0("_2017", c(paste0(0, 1:9), 10:12)))), by = .(letters)]

Answer 3

Using rep :

id <- letters[1:10]
month <- 201701:201712
ids <- rep(ids, each=length(month))
months <- rep(months, length.out=length(id))
data.frame(id=ids, ym=months, id_ym = paste(ids, months, sep="_"))

gives

    id     ym    id_ym
1    a 201701 a_201701
2    a 201702 a_201702
3    a 201703 a_201703
...