如何在R中的data.table中插入连续行（给出的示例）？

Question

df is data.table and df_expected is desired data.table . I want to add hour column from 0 to 23 and visits value would be filled as 0 for hours newly added .

df<-data.table(customer=c("x","x","x","y","y"),location_id=c(1,1,1,2,3),hour=c(2,5,7,0,4),visits=c(40,50,60,70,80))






df_expected<-data.table(customer=c("x","x","x","x","x","x","x","x","x","x","x","x","x","x","x","x","x","x","x","x","x","x","x","x",
                               "y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y",
                               "y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y","y"),

                    location_id=c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
                                  2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,
                                  3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3),

                    hour=c(0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,
                           0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,
                           0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23),

                    visits=c(0,0,40,0,0,50,0,60,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
                             70,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
                             0,0,0,0,80,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0))

This is what I tried to obtain my result , but it did not work

df1<-df[,':='(hour=seq(0:23)),by=(customer)]
Error in `[.data.table`(df, , `:=`(hour = seq(0L:23L)), by = (customer)) : 
Type of RHS ('integer') must match LHS ('double'). To check and coerce would impact 
performance too much for the fastest cases. Either change the type of the target column, or 
coerce the RHS of := yourself (e.g. by using 1L instead of 1)

Answer 1

Here's an approach that creates the target and then uses a join to add in the visits information. The ifelse statement just helps up clean up the NA from the merge. You could also leave them in and replace them with := in the new data.table.

target <- data.table(
  customer = rep(unique(df$customer), each = 24),
  hour = 0:23)

df_join <- df[target, on = c("customer", "hour"), 
   .(customer, hour, visits = ifelse(is.na(visits), 0, visits))
   ]

all.equal(df_expected, df_join)

Edit:

This addresses the request to include the location_id column. One way to do this is with by=location in the creation of the target. I've also added in some of the code from chinsoon12's answer.

target <- df[ , .("customer" = rep(unique(customer), each = 24L),
                  "hour" = rep(0L:23L, times = uniqueN(customer))),
              by = location_id]

df_join <- df[target, on = .NATURAL, 
              .(customer, location_id, hour, visits = fcoalesce(visits, 0))]

all.equal(df_expected, df_join)

Answer 2

Another option using CJ to generate your universe, on=.NATURAL for joining on identically named columns, and fcoalesce to handle NAs:

df[CJ(customer, hour=0L:23L, unique=TRUE), on=.NATURAL, allow.cartesian=TRUE, 
    .(customer=i.customer, hour=i.hour, visits=fcoalesce(visits, 0))]

Answer 3

here's a for-loop answer.

df_final <- data.table()
for(i in seq(24)){
  if(i %in% df[,hour]){
    a <- df[hour==i]
  }else{
    a <- data.table(customer="x", hour=i, visits=0)}

  df_final <- rbind(df_final, a)
}
df_final

You can wrap this in another for-loop to have your multiple customers x, y, etc. (the following loop isnt very clean but gets the job done).

df_final <- data.table()

for(j in unique(df[,customer])){

  for(i in seq(24)){

    if(i %in% df[,hour]){
      if(df[hour==i,customer] %in% j){
        a <- df[hour==i]
      }else{
        a <- data.table(customer=j, hour=i, visits=0)
      }
    }else{
      a <- data.table(customer=j, hour=i, visits=0)
    }

    df_final <- rbind(df_final, a)
  }
}

df_final