Choosing the smallest number in a matrix except for those on the main diagonal line

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I'm trying to identify the similarity of words contained in column A and produced a matrix using the adist function. I want to choose the words with the shortest distance except on the diagonal line. Therefore, I applied a function that would identifies the second-to-smallest number, but it sometimes include the 0 on the diagonal line, because all numbers on the main diagonal of the matrix is 0 by definition. I want to get the smallest number except for the one on the main diagonal.

a1 = df['A']
a2 = df['A'] <- adist(a1$A, a2$A, partial = TRUE, = TRUE) <- apply(, 1, function(x)(sort(x)[2]))

Below is an example of the matrix.

    a1  a2  a3  a4  a5
a1  0   3   0   3   1
a2  1   0   3   5   0
a3  2   3   0   0   5
a4  3   0   1   0   5
a5  0   3   2   1   0

I want to be able to choose [a1, a3] rather than [a1, a1].

Consider mat is your data.frame:

> mat2 <- as.matrix(mat)
> diag(mat2) <- NA

Alternative 1

  > apply(mat2, 1, min, na.rm=TRUE) 
    a1 a2 a3 a4 a5 
     0  0  0  0  0

Alternative 2

> which(mat2==min(mat2, na.rm = TRUE), arr.ind = TRUE)
   row col
a5   5   1
a4   4   2
a1   1   3
a3   3   4
a2   2   5

r, If you want the smallest number of a matrix (or df) not on the diagonal, you can always do this by insuring that the min value isn't on the  In a matrix, all the entries are complex numbers. If we set the main diagonal entries to be zero, then the matrix will be Hermitian. Does this matrix has a name or some nice properties, especially

If you want the smallest number of a matrix (or df) not on the diagonal, you can always do this by insuring that the min value isn't on the diagonal. One way to do this is- df2 = df + diag(max(df)+1,nrow(df),ncol(df) ). Then min(df2) will not be on the diagonal.

minimum value of a matrix with the main diagonal elements equals , I have a 10x10 matrix where the main diagonal is zeros. How the minimum value of the elements is calculated, excluding the zeros? Thank you. Best,. Pavlos  D = diag (v) returns a square diagonal matrix with the elements of vector v on the main diagonal. D = diag (v,k) places the elements of vector v on the k th diagonal. k=0 represents the main diagonal, k>0 is above the main diagonal, and k<0 is below the main diagonal. x = diag (A) returns a column vector of the main diagonal elements of A.

Lacking any copy of df or even a clear description of it this is a guess. The code:

 !diag(dim([1] )

... creates a logical square matrix of same dimensions as and when used as an index selects the non-diagonal elements, thusly: <- min([!diag(dim([1] )]

After rebuilding a copy of the matrix we see that it still finds all the 0's that were in the off-diagonal elements. So I perhaps odidn't I understand the goal, but negating diag(5) does remove all the diagonal zeroes: <- matrix( scan(text="
   0   3   0   3   1
   1   0   3   5   0
   2   3   0   0   5
   3   0   1   0   5
   0   3   2   1   0"), 5,5,byrow=TRUE, dimnames=list(c(    'a1',  'a2',  'a3',  'a4',  'a5'), c(    'a1',  'a2',  'a3',  'a4',  'a5')) )

Find smallest and largest element from square matrix diagonals , Given a square matrix of order n*n, find the smallest and largest elements from both cout << ( "Secondary Diagonal Smallest Element: " ) static void Main(). It is based on fact that smallest number is in left upper corner of the array and largest is in right bottom one. Let's take largest number equal to X. Algorithm takes first smallest number, replaces it with X+1 and then sorts given NxM array. This can be done with O(M+N) operations. As a result, smallest number will appear again in left top corner.

Applied Algebra, Algebraic Algorithms, and Error-correcting Codes: , Let the diagonal of Q be the entries q°, q\ q^\ . Let w, be the maximum column weight of the matrix Q and w2the maximum column Hence, we should choose a polynomial P(x) of lowest weight. However, the number of irreducible polynomials grows quickly with m. In these cases the selection criterion is simply: C3. The number of zero-valued elements divided by the total number of elements (e.g., m × n for an m × n matrix) is called the sparsity of the matrix (which is equal to 1 minus the density of the matrix). Using those definitions, a matrix will be sparse when its sparsity is greater than 0.5.

Computational Science and Its Applications, This work leverages on a detailed knowledge of the characteristics of the linear row of the matrix of the system) with the minimum number of nonzero entries. suited for these multiphysics systems traces back to the structure of the matrix of blocks, and, moreover, choosing the pivot as the diagonal entry of an FS row  Create a matrix and return the smallest value between each of its elements compared to a scalar. A = [1 7 3; 6 2 9] A = 2×3 1 7 3 6 2 9

The Phylogenetic Handbook: A Practical Approach to DNA and Protein , Then a diagonal line is drawn connecting the points with the highest score. Most programs allow the user to select an appropriate weight matrix to score either The mutation-cost matrix scores the minimum number of base changes  In the case of a square matrix, the main or principal diagonal is the diagonal line of entries running from the top-left corner to the bottom-right corner. [9] [10] [11] For a matrix A {\displaystyle A} with row index specified by i {\displaystyle i} and column index specified by j {\displaystyle j} , these would be entries A i j {\displaystyle

  • I think I can do it but I don't have any df
  • which.min(`diag<-`(m, Inf))? (Not tested since the example is not easy to copy-paste and desired output is not clear)
  • what is the expected output?
  • Thanks for the answer. A description of df is not important in this context. What's important is the matrix.
  • OK. I think this is correct to that goal then. And it is a more "viRtuous" answer in the sense of using all vectorized function (no apply).