jf328
jf328

Reputation: 7351

R data.table dot product with matching column names (for each group)

I have a datatable of data and a datatable of fitted coefficients. I want to calculate the fitted value for each row.

dt = data.table(a = rep(c("x","y"), each = 5), b = rnorm(10), c = rnorm(10), d = rnorm(10))
coefs = data.table(a = c("x","y"), b = c(0, 1), d = c(2,3))
dt
#    a           b          c           d
# 1: x -0.25174915 -0.2130797 -0.67909764
# 2: x -0.35569766  0.6014930  0.35201386
# 3: x -0.31600957  0.4398968 -1.15475814
# 4: x -0.54113762 -2.3497952  0.64503654
# 5: x  0.11227873  0.0233775 -0.96891456
# 6: y  1.24077566 -1.2843439  1.98883516
# 7: y -0.23819626  0.9950835 -0.17279980
# 8: y  1.49353589  0.3067897 -0.02592004
# 9: y  0.01033722 -0.5967766 -0.28536224
#10: y  0.69882444  0.8702424  1.24131062

coefs # NB no "c" column
#   a b d
#1: x 0 2
#2: y 1 3

For each a=="x" row in dt, I want 0*b+2*d; and for each a=="y" row in dt, I want 1*b+3*d.

Is there a datatable way to do this without hardcode the column name? I'm happy to put the column names in a variable cols = colnames(coefs)[-1].

It's easy to loop over groups and rbind together, so if the grouping is causing trouble, please ignore that part.

Upvotes: 7

Views: 1192

Answers (3)

Roland
Roland

Reputation: 132706

Join the data.tables:

dt[coefs, res := b * i.b + d * i.d, on = "a"]
 #   a           b            c          d        res
 #1: x  0.09901786 -0.362080111 -0.5108862 -1.0217723
 #2: x -0.16128422  0.169655945  0.3199648  0.6399295
 #3: x -0.79648896 -0.502279345  1.3828633  2.7657266
 #4: x -0.26121421  0.480548972 -1.1559392 -2.3118783
 #5: x  0.54085591 -0.601323442  1.3833795  2.7667590
 #6: y  0.83662761  0.607666970  0.6320762  2.7328562
 #7: y -1.92510391 -0.050515610 -0.3176544 -2.8780671
 #8: y  1.65639926 -0.167090105  0.6830158  3.7054466
 #9: y  1.48772354 -0.349713539 -1.2736467 -2.3332166
#10: y  1.49065993  0.008198885 -0.1923361  0.9136516

Usually you would use the matrix product here, but that would mean that you had to coerce the respective subset to a matrix. That would result in a copy being made and since data.tables are mainly used for larger data, you want to avoid copies.

If you need dynamic column names, the most simple solution that comes to mind is actually an eval/parse construct:

cols = colnames(coefs)[-1]
expr <- parse(text = paste(paste(cols, paste0("i.", cols), sep = "*"), collapse = "+"))
#expression(b*i.b+d*i.d)

dt[coefs, res := eval(expr), on = "a"] 

Maybe someone else can suggest a better solution.

Here is a solution using matrix multiplication:

dt[, res := as.matrix(.SD) %*% unlist(coefs[a == .BY, .SD, .SDcols = cols]), 
  by = "a", .SDcols = cols]

Of course this makes copies, which is potentially less efficient then the eval solution.

Upvotes: 11

Mike H.
Mike H.

Reputation: 14360

Another alternative (but slower) approach is:

dt$res <- unsplit(Map(function(x,y){x$b*y$b + x$d*y$d}, split(dt, dt$a=="x"), 
                 split(coefs,coefs$a=="x")),dt$a=="x")

    dt
    a           b          c          d        res
 1: x  0.47859729  1.3479271  0.5691897  1.1383794
 2: x  0.28491505 -0.3291934  1.8621365  3.7242730
 3: x -1.43894695  1.5555413  0.3685772  0.7371544
 4: x  0.04360066  0.1358920  0.5240700  1.0481400
 5: x -1.39897890 -0.0175886 -0.6876451 -1.3752901
 6: y -0.60952146  1.2331907 -0.3582176 -1.6841742
 7: y  0.31777772  1.4090295 -0.4053615 -0.8983067
 8: y  0.42758431 -0.3746061  2.1208417  6.7901094
 9: y -0.60701063 -0.9232092  1.9386482  5.2089341
10: y -1.52042316 -0.8871454 -0.9314232 -4.3146927

This same code would work in base R as well if your data was already data.frames.

Upvotes: 0

jf328
jf328

Reputation: 7351

I found out that data.table of all numerical type columns can do arithmetic operations (+,-,*,/), but no name matching -- just order matching.

> coefs
   a b d
1: x 0 2
2: y 1 3
> coefs[, .(b,d)] * coefs[, .(b,d)]
   b d
1: 0 4
2: 1 9
> coefs[, .(b,d)] * coefs[, .(d,b)]
   b d
1: 0 0
2: 3 3

so a solution based on this

> cols = colnames(coefs)[-1]
> zz = rowSums(coefs[dt[,.(a)], .SD, on = 'a', .SDcols = cols] * dt[, .SD, .SDcols = cols])
> dt[, newcol := zz]

Upvotes: 0

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