Get the Row and Column Name of the Minimum Element of a Matrix

R: Get the row and column name of the minimum element of a matrix but with minimum != 0

Seems like there could be a shorter answer but u can replace the zeros with NA and use na.rm=T

test = matrix(c(0:9,0:9),nrow =4,ncol=5)
test[which(test==0)] = NA
minValue = min(test,na.rm=T) 
rows = which(apply(test,1,min,na.rm=T)==minValue)
cols = which(apply(test,2,min,na.rm=T)==minValue)

Allows for duplicates

Print Column Name of Smallest Element of Each Row in R

A base R option with apply

apply(b, 1, FUN = function(x) names(x)[which.min(x)])
#[1] "2"  "4"  "2"  "12" "7"  "12" "4"  "9"  "2"

Find 2 or more indices (row and column) of minimum element of a matrix

You can use find():

[rows, columns] = find(A == min(min(A)));

Furthermore, you can also specify how many matches to find:

Nmatches = 1;
[rows, columns] = find(A == min(min(A)), Nmatches);

For completeness, you can also specify which matches to find, as in the first or last N matches, by using the 'first' or 'last' flag.

How to find rows in a matrix with only the minimum element in the first column?

Create a logical vector with the min of the first column and subset the matrix

out1 <- out[out[,1] == min(out[,1]),]

Minimum element on each row and column

// Minimum in Column
 public static void colMin(double[][] n){
    double[] result = new double[n.length];
    double min = 0;
    System.out.println("Column min values: ");
    for (int i = 0; i < n.length; ++i) {
        min = Integer.MAX_VALUE;
        for (int j = 1; j < n[i].length; ++j)
            if (n[j][i] < min){
                min = n[j][i];
                result[i] = min;
            }

        System.out.printf("%.2f ", result[i]);
        System.out.println();
    }

}

Just Assign your computed min to result[i] as shown above in my code.
and display.

Cheers.

Pandas locate minimum of DataFrame matrix: index, col

Try this:

df.stack().idxmin()

Out[108]: ('index0', 'col0')

Getting the name of the column with the minimum value for a row in a dataframe

We can use max.col which will return column number with max value in each row and to handle NAs we can replace them with 0. We use ties.method = "first" so in case if we have a tie in a row for max value we select the first max always.

soo$max_col <- names(soo)[max.col(replace(soo, is.na(soo),0),ties.method = "first")]

soo
#   Mycol1 Mycol2 max_col
#1    11.9   10.2  Mycol1
#2     7.8    2.7  Mycol1
#3     8.2    6.4  Mycol1
#4    14.7   55.1  Mycol2
#5     1.0    1.0  Mycol1
#6    21.7    8.6  Mycol1
#7     0.0    0.0  Mycol1
#8    48.6    0.0  Mycol1
#9     0.0    0.0  Mycol1
#10    2.8    1.9  Mycol1

Index row and index column of the minimum value matrix A in 2nd until 3rd row and 2nd column

Since you are inputting the column index so there is no need to find that again. The better approach would be to just do:

col_ind = 2;
[minimum, row_ind]  = min(A(2:3,col_ind)); 
row_ind = row_ind+1;      %1 is added since first row is skipped in above line

In your code,

A(2:3, 2) has only 1 column. So columns will always be 1.
If A(1,2) is also the same as the minimum value, your code will return the row index of that first value.
The fact that the first row and column are skipped in calculations is not incorporated in outputs.

Fixing above problems:

[rows,columns]=find(A(2:3,2)==min(A(2:3,2)));
rows=rows+1;     columns=columns+1;