[R] Mathematical working procedure of duplicated() function in r

[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

Às 12:54 de 04/08/20, K Purna Prakash escreveu:
> 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.
> 
> 	[[alternative HTML version deleted]]
> 
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