row-by-row operations and updates in data.table

Question

I ended up with a big data.table and I have to do operations per row. (yes... I know that this is clearly not what data.table are for)

R) set.seed(1)
R) DT=data.table(matrix(rnorm(100),nrow=10))
R) DT[,c('a','b'):=list(1:10,2:11)]
R) DT
               V1             V2             V3             V4            V5            V6             V7              V8            V9           V10  a  b
 1: -0.6264538107  1.51178116845  0.91897737161  1.35867955153 -0.1645235963  0.3981058804  2.40161776050  0.475509528900 -0.5686687328 -0.5425200310  1  2
 2:  0.1836433242  0.38984323641  0.78213630073 -0.10278772734 -0.2533616801 -0.6120263933 -0.03924000273 -0.709946430922 -0.1351786151  1.2078678060  2  3
 3: -0.8356286124 -0.62124058054  0.07456498337  0.38767161156  0.6969633754  0.3411196914  0.68973936245  0.610726353489  1.1780869966  1.1604026157  3  4
 4:  1.5952808021 -2.21469988718 -1.98935169586 -0.05380504058  0.5566631987 -1.1293630961  0.02800215878 -0.934097631644 -1.5235668004  0.7002136495  4  5
 5:  0.3295077718  1.12493091814  0.61982574789 -1.37705955683 -0.6887556945  1.4330237017 -0.74327320888 -1.253633400239  0.5939461876  1.5868334545  5  6
 6: -0.8204683841 -0.04493360902 -0.05612873953 -0.41499456330 -0.7074951570  1.9803998985  0.18879229951  0.291446235517  0.3329503712  0.5584864256  6  7
 7:  0.4874290524 -0.01619026310 -0.15579550671 -0.39428995371  0.3645819621 -0.3672214765 -1.80495862889 -0.443291873218  1.0630998373 -1.2765922085  7  8
 8:  0.7383247051  0.94383621069 -1.47075238390 -0.05931339671  0.7685329245 -1.0441346263  1.46555486156  0.001105351632 -0.3041839236 -0.5732654142  8  9
 9:  0.5757813517  0.82122119510 -0.47815005511  1.10002537198 -0.1123462122  0.5697196274  0.15325333821  0.074341324152  0.3700188099 -1.2246126149  9 10
10: -0.3053883872  0.59390132122  0.41794156020  0.76317574846  0.8811077265 -0.1350546039  2.17261167036 -0.589520946188  0.2670987908 -0.4734006364 10 11

Say I want the min across of all the Vi columns row by row, I used to use apply when I was using data.frame.

apply(DT[,paste0('V',1:10),with=FALSE],FUN=min,MAR=1)
 [1] -0.6264538107 -0.7099464309 -0.8356286124 -2.2146998872 -1.3770595568 -0.8204683841 -1.8049586289 -1.4707523839 -1.2246126149 -0.5895209462

So I can update easily.

Ok, now say that I want to update the min and max at once (off course this is an example so I took just 2 things but in real life that would be 10...)

 f = function(x){return(c(max=max(x),min=min(x)))}
 new=apply(DT[,paste0('V',1:10),with=FALSE],FUN=f,MAR=1)
             [,1]          [,2]          [,3]         [,4]         [,5]          [,6]         [,7]         [,8]         [,9]         [,10]
max  2.4016177605  1.2078678060  1.1780869966  1.595280802  1.586833455  1.9803998985  1.063099837  1.465554862  1.100025372  2.1726116704
min -0.6264538107 -0.7099464309 -0.8356286124 -2.214699887 -1.377059557 -0.8204683841 -1.804958629 -1.470752384 -1.224612615 -0.5895209462

I would like to write

DT[,rownames(new):=new]

but this does not work, so here are my questions

1. Using my method, how can I transform new such that I can update DT at once ?
2. Are there some better approach (that would allow me to update multiple columns at once, with the result of a by-row calculation)

EDIT: I found a solution for 1 but that's UGLY, actually It is strange that := do not handle matrix, I am pretty sure it used to be the case

DT[,c('a1','a2'):=data.table(matrix(apply(DT[,paste0('V',1:10),with=FALSE],FUN=f,MAR=1),byrow=T,nrow=10))]
R) DT
               V1             V2             V3             V4            V5            V6             V7              V8            V9           V10  a  b
 1: -0.6264538107  1.51178116845  0.91897737161  1.35867955153 -0.1645235963  0.3981058804  2.40161776050  0.475509528900 -0.5686687328 -0.5425200310  1  2
 2:  0.1836433242  0.38984323641  0.78213630073 -0.10278772734 -0.2533616801 -0.6120263933 -0.03924000273 -0.709946430922 -0.1351786151  1.2078678060  2  3
 3: -0.8356286124 -0.62124058054  0.07456498337  0.38767161156  0.6969633754  0.3411196914  0.68973936245  0.610726353489  1.1780869966  1.1604026157  3  4
 4:  1.5952808021 -2.21469988718 -1.98935169586 -0.05380504058  0.5566631987 -1.1293630961  0.02800215878 -0.934097631644 -1.5235668004  0.7002136495  4  5
 5:  0.3295077718  1.12493091814  0.61982574789 -1.37705955683 -0.6887556945  1.4330237017 -0.74327320888 -1.253633400239  0.5939461876  1.5868334545  5  6
 6: -0.8204683841 -0.04493360902 -0.05612873953 -0.41499456330 -0.7074951570  1.9803998985  0.18879229951  0.291446235517  0.3329503712  0.5584864256  6  7
 7:  0.4874290524 -0.01619026310 -0.15579550671 -0.39428995371  0.3645819621 -0.3672214765 -1.80495862889 -0.443291873218  1.0630998373 -1.2765922085  7  8
 8:  0.7383247051  0.94383621069 -1.47075238390 -0.05931339671  0.7685329245 -1.0441346263  1.46555486156  0.001105351632 -0.3041839236 -0.5732654142  8  9
 9:  0.5757813517  0.82122119510 -0.47815005511  1.10002537198 -0.1123462122  0.5697196274  0.15325333821  0.074341324152  0.3700188099 -1.2246126149  9 10
10: -0.3053883872  0.59390132122  0.41794156020  0.76317574846  0.8811077265 -0.1350546039  2.17261167036 -0.589520946188  0.2670987908 -0.4734006364 10 11
             a1            a2
 1: 2.401617761 -0.6264538107
 2: 1.207867806 -0.7099464309
 3: 1.178086997 -0.8356286124
 4: 1.595280802 -2.2146998872
 5: 1.586833455 -1.3770595568
 6: 1.980399899 -0.8204683841
 7: 1.063099837 -1.8049586289
 8: 1.465554862 -1.4707523839
 9: 1.100025372 -1.2246126149
10: 2.172611670 -0.5895209462

EDIT2: It looks on my data that using DT[, (newColnames):=f(.DT), by=IDX, .SDcols=someIdx] is much slower than the apply way, is that expected ?

Solution 1

Creating .SD on each row could be a very costly operation, especially if your data.table consists of rows >> columns. I'd advice using pmin and pmax across columns with a loop. I'll illustrate this with a bigger data (along the rows).

Data:

set.seed(1)
require(data.table)
DT1 <- data.table(matrix(rnorm(1e6),ncol=10))
DT1[, a := 1:1e5]
DT2 <- copy(DT1)
DT3 <- copy(DT1)

Functions:

arun <- function(DT) {
    # assign first column (dummy)
    DT[, `:=`(min = DT[, V1], max = DT[, V1])]
    # get all other column names and use pmin and pmax 
    # and replace min and max columns
    cols <- names(DT)[2:10]
    for (i in cols) {
        DT[, `:=`(min = pmin(min, DT[[i]]), max = pmax(max, DT[[i]]))]
    }
    DT
}

eddi <- function(DT) {
    DT[, `:=`(min = min(.SD), max = max(.SD)), by = a, .SDcols = paste0("V", 1:10)]
}

frank <- function(DT) {
    cols    <- names(DT)[grepl('^V[[:digit:]]+$',names(DT))]
    newcols <- c("min","max")
    myfun   <- range
    DT[,(newcols):=as.list(myfun(.SD)),.SDcols=cols,by=1:nrow(DT)]
}

Benchmarking:

require(microbenchmark)
microbenchmark(o1 <- arun(DT1), o2 <- eddi(DT2), o3 <- frank(DT3), times=2)

Unit: milliseconds
             expr        min         lq     median          uq         max neval
  o1 <- arun(DT1)   204.4417   204.4417   250.5205    296.5992    296.5992     2
  o2 <- eddi(DT2) 92343.5321 92343.5321 96706.1622 101068.7923 101068.7923     2
 o3 <- frank(DT3) 49083.7000 49083.7000 49521.9296  49960.1592  49960.1592     2

identical(o1, o2) # TRUE
identical(o1, o3) # TRUE

--

As @Frank points out under comments, you could replace the for-loop with do.call as:

DT[, c("min", "max") := { z <- dt[, 1:10]; 
             list(do.call(pmin, z), do.call(pmax, z))}]

Solution 2

Since you already have the row numbers as a column in your data.table*, you could just do:

DT[, `:=`(a1 = max(.SD), a2 = min(.SD)), by = a, .SDcols = paste0("V", 1:10)]

or

setkey(DT, a)
DT[J(a), `:=`(a1 = max(.SD), a2 = min(.SD)), .SDcols = paste0("V", 1:10)]

The second option uses the silent by-without-by.

*of course you could also just use row.names or 1:nrow(DT)

Solution 3

This spells out the steps in case you want to use a different function:

cols    <- names(DT)[grepl('^V[[:digit:]]+$',names(DT))]
newcols <- c("min","max")
myfun   <- range
DT[,(newcols):=as.list(myfun(.SD)),.SDcols=cols,by=1:nrow(DT)]

Solution 4

Am I missing something, doesn't this give the min across row

set.seed(1)
DT=data.table(matrix(rnorm(100),nrow=10))
DT[,c('a','b'):=list(1:10,2:11)]
DT
cols<-c("V1", "V2", "V3", "V4", "V5", "V6", "V7", "V8", "V9", "V10")

Method 1

DT[,Min_Vi:=do.call(pmin, c(.SD, na.rm=TRUE)), .SDcols=cols]

Method 2

transform(DT,Min_Vi=pmin(get(cols)))

View more solutions

Comments

[Frank]

I tried by=row.names(DT) and got an error. To clarify, it says "by=eval(row.names(DT)) should work."

[eddi]

@Frank use by=list(row.names(DT))

[eddi]

@statquant - sorry I don't get what you want? just add whatever other calculation you need

[Frank]

I was going to post an answer just like yours, but spelling it out: cols <- names(DT)[grepl('^V[[:digit:]]$',names(DT))];newcols <- c("min","max");myfun <- range;DT[,(newcols):=as.list(myfun(.SD)),.SDcols=cols,by=1:n‌​row(DT)].

[eddi]

@Frank - I like the 1:nrow(DT) better, I'm going to steal that and add it above :)

[Frank]

Hrm, okay. I worry that it's somewhat redundant, but I guess it's rearranged a little.

[Arun]

you'd need to use [[:digit:]]+$ I suppose otherwise, It'll skip all columns with numbers > 9 (V10 for example).

[Arun]

Sure. Unfortunately, this'll be terribly slow on a data.table with a LOT of rows.

[Frank]

That could be why the OP hasn't accepted an answer yet. I don't know how to speed it up without just storing the data in a matrix instead of a data.table. I'd be interested to see how it could be done.

[Frank]

Cool stuff. This performs about the same on my computer and lacks a loop: colrange = (1:ncol(DT))[grepl('^V[[:digit:]]+$',names(DT))]; DT[,c("min","max"):={z <- "[.listof"(DT,colrange); list(do.call(pmin,z),do.call(pmax,z))}]. I had to look up methods("[") to figure that out.

[Arun]

That's even better, but you don't need the z <- "[.listof".... do.call expects a list and data.table is internally a list. So, you could just do: DT[,c("min","max"):= list(do.call(pmin,DT[, 1:10, with=FALSE),do.call(pmax,DT[, 1:10, with=FALSE))]

[Frank]

Yes, but I need to only include the relevant columns (not "a").

[Frank]

(Replying in R Public chat.)