how to find index of matrix if value is known?

Kv= 212.640075234138  212.941159818301  213.241495908758  213.541083505509
211.539179873813  211.843738655644  212.147548943768  212.450610738187
210.439033007195  210.747065986692  211.054350472484  211.360886464570
209.339634634282  209.651141811447  209.961900494907        210.271910684660
208.240984755075  208.555966129908  208.870199011035  209.183683398456
207.143083369574  207.461538942074  207.779246020869  208.096204605957
206.045930477779  206.367860247947  206.689041524409  207.009474307164
NaN  NaN  NaN  NaN
NaN  NaN  NaN  NaN

If i know thw column is 1 and min value is 206 how would i find row index?

[Rik]

What do you mean? Do you want to find a minimum value in some specific column?

[jonas]

the second output of the min() function is the index

[Shubham Mohan Tatpalliwar]

HELLO RIK yes i want to know the min value of a specific column

as the input would be column

and as a output

the minimum value and row is expected

Answer 1

[KSSV]

[val,idx] = min(kv(:,1))

Or if you want (row,col) of the minimum element in a matrix use:

A = rand(3) ;
[val,idx] = min(A(:)) ;
[i,j] = ind2sub(size(A),idx) ;  % row, column position of min 
[A(i,j) val]

[Shubham Mohan Tatpalliwar]

Actually i a calculating the index of column and there i am getting as 6.3 or something in decimal i=d-1000 i=column d is input

how should i get a whole number

[KSSV]

YOu will never get index in decimal...check your code.

[Shubham Mohan Tatpalliwar]

no its a value which is coming from some calculation

i just want to approximate it

[KSSV]

Read about round, fix, floor and ceil

[Image Analyst]

You accepted this, so does it work or not?

I would not use min() or ind2sub() to find the row. I'd use find() like in my solution. Take a look at it.

[Shubham Mohan Tatpalliwar]

yes it worked for my case

Answer 2

[Luna]

[minValue, indexofMin] = min(Kv(:,1)) % for the first column

[minValue, indexofMin] = min(Kv) % for all columns

Answer 3

[Image Analyst]

I would not use min() to get the index because it finds ONLY the first occurrence of the min and if the min occurs more than once, it will miss it. I'd use find, like in this demo:

Kv = [212.640075234138  212.941159818301  213.241495908758  213.541083505509
211.539179873813  211.843738655644  212.147548943768  212.450610738187
210.439033007195  210.747065986692  211.054350472484  211.360886464570
209.339634634282  209.651141811447  209.961900494907        210.271910684660
208.240984755075  208.555966129908  208.870199011035  209.183683398456
207.143083369574  207.461538942074  207.779246020869  208.096204605957
206.045930477779  206.367860247947  206.689041524409  207.009474307164
NaN  NaN  NaN  NaN
NaN  NaN  NaN  NaN
212.640075234138  212.941159818301  213.241495908758  213.541083505509
211.539179873813  211.843738655644  212.147548943768  212.450610738187
210.439033007195  210.747065986692  211.054350472484  211.360886464570
209.339634634282  209.651141811447  209.961900494907        210.271910684660
208.240984755075  208.555966129908  208.870199011035  209.183683398456
207.143083369574  207.461538942074  207.779246020869  208.096204605957
206.045930477779  206.367860247947  206.689041524409  207.009474307164
NaN  NaN  NaN  NaN
NaN  NaN  NaN  NaN]
% Find min value in column 1
minValue = min(Kv(:, 1))
% Use find() instead of min() to get the row
% because the min may occur more than once.
rows = find(Kv(:, 1) == minValue)

You'll see

minValue =
          206.045930477779
rows =
     7
    16

find() finds both occurrences of the min in column 1, not just the first one like min() would.

[Shubham Mohan Tatpalliwar]

thats nice.. thank you for the info.

Answer 4

[Vikas Singh]

how i can find the index of matrix if i know the value of matrix ??

[Image Analyst]

[rows, columns] = find(m == value)

[Vikas Singh]

Thnk you ....

[Rik]

Note that this may be sensitive to rounding errors. You may consider comparing the absolute difference to a tolerance or using ismembertol.

[Vikas Singh]

Thank you it works for me

[Image Analyst]

@Rik, I don't think it's subject to rounding errors if the value is actually in the matrix and not one you arrived at some other way. For example

value = max(m(:)) % Gauranteed to be in m and exactly match.
[rows, columns] = find(m == value)

Now it should always find all occurrences as far as I know. Of course if the max was 1.00000000001 and you put m == 1, then of course the value 1 is not in there exactly and rows and columns will be empty. But @Vikas Singh said that he does "know the value of matrix" exactly so it should work if the value actually came from the matrix.