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Mathematical working procedure of duplicated() function in r

3 messages · K Purna Prakash, Rui Barradas, Greg Snow

K Purna Prakash
# 
Dear Sir(s),
I request you to provide the detailed* internal mathematical working
mechanism of the following function *for better understanding.
*x[duplicated(x) | duplicated(x, fromLast=TRUE), ]*
I am having some confusion in understanding how duplicates are being
identified when thousands of records are there.
I will look for a positive response.
Thank you,
K.Purna Prakash.
Rui Barradas
# 
Hello,

R is open source, you can see exactly what is the internal working of 
any function. You can have access to the code by typing the function's 
name without parenthesis at an R command line.

 > duplicated
function (x, incomparables = FALSE, ...)
UseMethod("duplicated")
<bytecode: 0x55e5ef683040>
<environment: namespace:base>

Now, this tells users that duplicated is a generic function, and that 
there are methods written to handle the different S3 classes of objects x.
When this happens, there is always a default method, duplicated.default

 > duplicated.default
function (x, incomparables = FALSE, fromLast = FALSE, nmax = NA,
     ...)
.Internal(duplicated(x, incomparables, fromLast, if (is.factor(x)) 
min(length(x),
     nlevels(x) + 1L) else nmax))
<bytecode: 0x55e5ef6826a0>
<environment: namespace:base>


The default method calls .Internal(duplicated, etc). So you'll have to 
download the R sources, if you haven't done it yet, and search for a 
file where that function might be. The file is

src/main/duplicate.c


Good reading.
Also, like the posting guide asks R-Help users to do, please post in 
plain text, not in HTML.

Hope this helps,

Rui Barradas

?s 12:54 de 04/08/20, K Purna Prakash escreveu:
Greg Snow
# 
Rui pointed out that you can examine the source yourself.  FAQ 7.40
has a link to an article with detail on finding and examining the
source code.

A general algorithm for checking for duplicates follows (I have not
examined to R source code to see if they use something more clever).

Create an empty object (I will call it seen).  This could be a simple
vector, but for efficiency it is better to use an object type that has
fast lookup, e.g. binary tree, associative array/hash/dictionary, etc.

Create an empty vector of logicals the same length as x (I will call it result).

loop from 1 to the length of x (or from the length to 1 if
fromLast=TRUE), on each iteration
 check to see if the value of x[i] is in seen
   If it is: set result[i] to TRUE
   If it is not: add the current value to seen and set result[i] to false

After the loop finishes, throw away seen and reclaim the memory, then
return result.

Since it looks like you are using this on a matrix or data frame,
there is probably a preprocessing step that combines all the values on
each row into a single character string.
On Tue, Aug 4, 2020 at 6:45 AM K Purna Prakash <prakash.nani at gmail.com> wrote: