views:

106

answers:

3

Hi all,

I have the following 2 data.frames:

a1 <- data.frame(a = 1:5, b=letters[1:5])
a2 <- data.frame(a = 1:3, b=letters[1:3])

I want to find the row a1 has that a2 doesn't.

Is there a built in function for this type of operation?

(p.s: I did write a solution for it, I am simply curious if someone already made a more crafted code)

Here is my solution:

a1 <- data.frame(a = 1:5, b=letters[1:5])
a2 <- data.frame(a = 1:3, b=letters[1:3])

rows.in.a1.that.are.not.in.a2  <- function(a1,a2)
{
    a1.vec <- apply(a1, 1, paste, collapse = "")
    a2.vec <- apply(a2, 1, paste, collapse = "")
    a1.without.a2.rows <- a1[!a1.vec %in% a2.vec,]
    return(a1.without.a2.rows)
}
rows.in.a1.that.are.not.in.a2(a1,a2)
+3  A: 

It is certainly not efficient for this particular purpose, but what I often do in these situations is to insert indicator variables in each data.frame and then merge:

a1$included_a1 <- TRUE
a2$included_a2 <- TRUE
res <- merge(a1, a2, all=TRUE)

missing values in included_a1 will note which rows are missing in a1. similarly for a2.

One problem with your solution is that the column orders must match. Another problem is that it is easy to imagine situations where the rows are coded as the same when in fact are different. The advantage of using merge is that you get for free all error checking that is necessary for a good solution.

Eduardo Leoni
Good points Eduardo - thank you :)
Tal Galili
+2  A: 

This doesn't answer your question directly, but it will give you the elements that are in common. This can be done with Paul Murrell's package compare:

> library(compare)
> a1 <- data.frame(a = 1:5, b = letters[1:5])
> a2 <- data.frame(a = 1:3, b = letters[1:3])
> comparison <- compare(a1,a2,allowAll=TRUE)
> comparison$tM
  a b
1 1 a
2 2 b
3 3 c

The function compare gives you a lot of flexibility in terms of what kind of comparisons are allowed (e.g. changing order of elements of each vector, changing order and names of variables, shortening variables, changing case of strings). From this, you should be able to figure out what was missing from one or the other. For example (this is not very elegant):

> difference <-
+   data.frame(lapply(1:ncol(a1),function(i)setdiff(a1[,i],comparison$tM[,i])))
> colnames(difference) <- colnames(a1)
> difference
  a b
1 4 d
2 5 e
nullglob
This is the type of answer I was looking for - thank you for teaching me something new today :) (I will give this a bit more time for people to offer more answers, and then except your answer)
Tal Galili
+2  A: 

I adapted the merge function to get this functionality. On larger dataframes it uses less memory than the full merge solution. And I can play with the names of the key columns.

Another solution is to use the library prob.

#  Derived from src/library/base/R/merge.R
#  Part of the R package, http://www.R-project.org
#
#  This program is free software; you can redistribute it and/or modify
#  it under the terms of the GNU General Public License as published by
#  the Free Software Foundation; either version 2 of the License, or
#  (at your option) any later version.
#
#  This program is distributed in the hope that it will be useful,
#  but WITHOUT ANY WARRANTY; without even the implied warranty of
#  MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
#  GNU General Public License for more details.
#
#  A copy of the GNU General Public License is available at
#  http://www.r-project.org/Licenses/

XinY <-
    function(x, y, by = intersect(names(x), names(y)), by.x = by, by.y = by,
             notin = FALSE, incomparables = NULL,
             ...)
{
    fix.by <- function(by, df)
    {
        ## fix up 'by' to be a valid set of cols by number: 0 is row.names
        if(is.null(by)) by <- numeric(0L)
        by <- as.vector(by)
        nc <- ncol(df)
        if(is.character(by))
            by <- match(by, c("row.names", names(df))) - 1L
        else if(is.numeric(by)) {
            if(any(by < 0L) || any(by > nc))
                stop("'by' must match numbers of columns")
        } else if(is.logical(by)) {
            if(length(by) != nc) stop("'by' must match number of columns")
            by <- seq_along(by)[by]
        } else stop("'by' must specify column(s) as numbers, names or logical")
        if(any(is.na(by))) stop("'by' must specify valid column(s)")
        unique(by)
    }

    nx <- nrow(x <- as.data.frame(x)); ny <- nrow(y <- as.data.frame(y))
    by.x <- fix.by(by.x, x)
    by.y <- fix.by(by.y, y)
    if((l.b <- length(by.x)) != length(by.y))
        stop("'by.x' and 'by.y' specify different numbers of columns")
    if(l.b == 0L) {
        ## was: stop("no columns to match on")
        ## returns x
        x
    }
    else {
        if(any(by.x == 0L)) {
            x <- cbind(Row.names = I(row.names(x)), x)
            by.x <- by.x + 1L
        }
        if(any(by.y == 0L)) {
            y <- cbind(Row.names = I(row.names(y)), y)
            by.y <- by.y + 1L
        }
        ## create keys from 'by' columns:
        if(l.b == 1L) {                  # (be faster)
            bx <- x[, by.x]; if(is.factor(bx)) bx <- as.character(bx)
            by <- y[, by.y]; if(is.factor(by)) by <- as.character(by)
        } else {
            ## Do these together for consistency in as.character.
            ## Use same set of names.
            bx <- x[, by.x, drop=FALSE]; by <- y[, by.y, drop=FALSE]
            names(bx) <- names(by) <- paste("V", seq_len(ncol(bx)), sep="")
            bz <- do.call("paste", c(rbind(bx, by), sep = "\r"))
            bx <- bz[seq_len(nx)]
            by <- bz[nx + seq_len(ny)]
        }
        comm <- match(bx, by, 0L)
        if (notin) {
            res <- x[comm == 0,]
        } else {
            res <- x[comm > 0,]
        }
    }
    ## avoid a copy
    ## row.names(res) <- NULL
    attr(res, "row.names") <- .set_row_names(nrow(res))
    res
}


XnotinY <-
    function(x, y, by = intersect(names(x), names(y)), by.x = by, by.y = by,
             notin = TRUE, incomparables = NULL,
             ...)
{
    XinY(x,y,by,by.x,by.y,notin,incomparables)
}
Henrico
Thank you Henrico :)
Tal Galili