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Get cell labels of distance cell

Data:

I have a distance matrix

called mat with the following tables and values:

>mat
           179        175        176        181        249
175 147.451648                                            
176 174.527392  28.849183                                 
181 187.166092  40.007553  16.310747                      
249 194.644103  49.268129  20.637011  17.118671           
336 190.182183  44.613076  16.023090  13.697248   5.613544

Problem:

I would like to extract the row and column labels for the minimum value (5.613544) of the matrix.

The expected output looks something like this:

[1] 336 249

I tried to use colnames(mat)

and rownames(mat)

, but both returned NULL

.

I know that a function labels(mat)

can return labels:

[1] "179" "175" "176" "181" "249" "336"

But here I am stuck and I cannot lower my head. Maybe something easy, sorry in advance to ask this simple question. Any ideas on how to do this?

Here's my data with dput()

:

structure(c(147.451648169349, 174.527392333086, 187.166091923018, 
                194.644102830409, 190.182183344061, 28.849182654733, 40.007553207889, 
                49.2681293306533, 44.6130761331497, 16.3107466873124, 20.6370107209517, 
                16.0230895738958, 17.1186706020006, 13.6972482228087, 5.6135436361904
    ), Size = 6L, Labels = c("179", "175", "176", "181", "249", "336"
    ), Diag = FALSE, Upper = FALSE, method = "euclidean", class = "dist")
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1 answer


The field names are stored in labels(mat)

, as you can see in dput

:

labels(mat)[which(lower.tri(mat),arr.ind=TRUE)[which.min(mat),]]
# [1] 336 249
  • In the distance matrix, both fields are the same, so they dimnames

    are used instead labels

    .
  • which.min

    gives the vector position of the minimum value. However, it mat

    is a matrix, so ...
  • which(lower.tri(mat),arr.ind=TRUE)

    gives a link between matrix and vector positions.


@ Alternative to David Arenburg

names(which(as.matrix(mat) == min(mat), arr.ind = TRUE)[, 1])
  • as.matrix(mat)

    takes labels

    how dimnames

    . These names carry over to as.matrix(mat) == min(mat)

    and are saved as a result which(...,arr.ind=TRUE)

    .
  • Since the distance matrix is ​​symmetric, c d(a,b)==d(b,a)

    , we have two minima (assuming a single minimizing pair of points). We can select a couple of points by taking the column positions [,1]

    , which, again, are stored on mat

    original labels

    , but now in its attribute names

    .
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