R Loop To New Data Frame Summary Weighted

Question

I have a tall data frame as such:

data = data.frame("id"=c(1,2,3,4,5,6,7,8,9,10),
                  "group"=c(1,1,2,1,2,2,2,2,1,2),
                  "type"=c(1,1,2,3,2,2,3,3,3,1),
                  "score1"=c(sample(1:4,10,r=T)),
                  "score2"=c(sample(1:4,10,r=T)),
                  "score3"=c(sample(1:4,10,r=T)),
                  "score4"=c(sample(1:4,10,r=T)),
                  "score5"=c(sample(1:4,10,r=T)),
                  "weight1"=c(173,109,136,189,186,146,173,102,178,174),
                  "weight2"=c(147,187,125,126,120,165,142,129,144,197),
                  "weight3"=c(103,192,102,159,128,179,195,193,135,145),
                  "weight4"=c(114,182,199,101,111,116,198,123,119,181),
                  "weight5"=c(159,125,104,171,166,154,197,124,180,154))

library(reshape2)
library(plyr)

data1 <- reshape(data, direction = "long",
                 varying = list(c(paste0("score",1:5)),c(paste0("weight",1:5))),
                 v.names = c("score","weight"),
                 idvar = "id", timevar = "count", times = c(1:5))
data1 <- data1[order(data1$id), ]

And what I want to create is a new data frame like so:

want = data.frame("score"=rep(1:4,6),
                  "group"=rep(1:2,12),
                  "type"=rep(1:3,8),
                  "weightedCOUNT"=NA) # how to calculate this? count(data1, score, wt = weight)

I am just not sure how to calculate weightedCOUNT which should apply the weights to the score variable so then it gives in column 'weightedCOUNT' a weighted count that is aggregated by score and group and type.

Answer 1

An option would be to melt (from data.table - which can take multiple measure patterns, and then grouped by 'group', 'type' get the count

library(data.table)
library(dplyr)
melt(setDT(data), measure = patterns('^score', "^weight"), 
   value.name = c("score", "weight")) %>% 
   group_by(group, type) %>%
   count(score, wt = weight)

---

If we need to have a complete set of combinations

library(tidyr)
melt(setDT(data), measure = patterns('^score', "^weight"), 
       value.name = c("score", "weight")) %>%      
   group_by(group, type) %>%
   ungroup %>% 
   complete(group, type, score, fill = list(n = 0))

Answer 2

If I understand correctly, weightedCOUNT is the sum of weights grouped by score, group, and type.

For the sake of completeness, I would like to show how the accepted solution would look like when implemented in pure base R and pure data.table syntax, resp.

Base R

The OP was almost there. He has already reshaped data from wide to long format for multiple value variables. Only the final aggregation step was missing:

data1 <- reshape(data, direction = "long",
                 varying = list(c(paste0("score",1:5)),c(paste0("weight",1:5))),
                 v.names = c("score","weight"),
                 idvar = "id", timevar = "count", times = c(1:5))

result <- aggregate(weight ~ score + group + type, data1, FUN = sum)

result

   score group type weight
1      1     1    1    479
2      3     1    1    558
3      4     1    1    454
4      1     2    1    378
5      2     2    1    154
6      3     2    1    174
7      4     2    1    145
8      1     2    2    535
9      2     2    2    855
10     3     2    2    248
11     4     2    2    499
12     1     1    3    189
13     2     1    3    351
14     3     1    3    600
15     4     1    3    362
16     1     2    3    596
17     2     2    3    265
18     3     2    3    193
19     4     2    3    522

result can be reordered by

with(result, result[order(score, group, type), ])

   score group type weight
1      1     1    1    479
12     1     1    3    189
4      1     2    1    378
8      1     2    2    535
16     1     2    3    596
13     2     1    3    351
5      2     2    1    154
9      2     2    2    855
17     2     2    3    265
2      3     1    1    558
14     3     1    3    600
6      3     2    1    174
10     3     2    2    248
18     3     2    3    193
3      4     1    1    454
15     4     1    3    362
7      4     2    1    145
11     4     2    2    499
19     4     2    3    522

data.table

As shown by akrun, melt() from the data.table package can be combined with dplyr. Alternatively, we can stay with the data.table syntax for aggregation:

library(data.table)
cols <- c("score", "weight") # to save typing
melt(setDT(data), measure = patterns(cols), value.name = cols)[
  , .(weightedCOUNT = sum(weight)), keyby = .(score, group, type)]

    score group type weightedCOUNT
 1:     1     1    1           479
 2:     1     1    3           189
 3:     1     2    1           378
 4:     1     2    2           535
 5:     1     2    3           596
 6:     2     1    3           351
 7:     2     2    1           154
 8:     2     2    2           855
 9:     2     2    3           265
10:     3     1    1           558
11:     3     1    3           600
12:     3     2    1           174
13:     3     2    2           248
14:     3     2    3           193
15:     4     1    1           454
16:     4     1    3           362
17:     4     2    1           145
18:     4     2    2           499
19:     4     2    3           522

The keyby parameter is used for grouping and ordering the output in one step.

Completion of missing combinations of the grouping variables is also possible in data.table syntax using the cross join function CJ():

melt(setDT(data), measure = patterns(cols), value.name = cols)[
  , .(weightedCOUNT = sum(weight)), keyby = .(score, group, type)][
    CJ(score, group, type, unique = TRUE), on = .(score, group, type)][
      is.na(weightedCOUNT), weightedCOUNT := 0][]

    score group type weightedCOUNT
 1:     1     1    1           479
 2:     1     1    2             0
 3:     1     1    3           189
 4:     1     2    1           378
 5:     1     2    2           535
 6:     1     2    3           596
 7:     2     1    1             0
 8:     2     1    2             0
 9:     2     1    3           351
10:     2     2    1           154
11:     2     2    2           855
12:     2     2    3           265
13:     3     1    1           558
14:     3     1    2             0
15:     3     1    3           600
16:     3     2    1           174
17:     3     2    2           248
18:     3     2    3           193
19:     4     1    1           454
20:     4     1    2             0
21:     4     1    3           362
22:     4     2    1           145
23:     4     2    2           499
24:     4     2    3           522
    score group type weightedCOUNT