Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is a good way to test the first row of a Matrix to check if one of the values equal a value. For instance, In this example, I want to get vector in the second row that has the bigger value in the first row, 10.

I can quickly do it with a lop, but don't think it's the best way...

enter image description here

share|improve this question
Perhaps MemberQ[m[[1]],n] ? – DavidC Oct 6 '12 at 20:56
What does bigger value mean in I want to get vector in the second row that has the bigger value ? – DavidC Oct 6 '12 at 20:59
I meant that in the example, I want to know that the bigger value is 10, and therefor the vector I'm interested in is {1,1,1}. In another example of a 3x3, the values at the first row may be 1, 4, 3 and I need to detect that the bigger value is now in position 2 (number 4), and will use that to get the vector in the second row, below the number 4. – whynot Oct 6 '12 at 21:28
up vote 2 down vote accepted


m = {{10, -2, 1}, {{1, 2, 3}, {4, 5, 6, 7}, {8, 9, 10}}}


This gives the vector in row 2 that is opposite (i.e. under) the largest value (or first of multiple largest values) in row 1:

m[[2, Position[m[[1]], Max[m[[1]]]][[1, 1]]]]

{1, 2, 3}


Max[m[[1]]] gives the largest value in row 1: 10

Position[m[[1]], Max[m[[1]]]][[1, 1]] gives the column of interest: 1

share|improve this answer

So for your general question you will want to use Position/Extract, let say for the example you give to get the vector that corresponds to 1 (which is neither the largest, nor the smallest ...) you could do:

Extract[mat[[2, All]], Position[mat[[1, All]], 1]]

Now if you want to generally get results that deal with numeric ordering, you can use b.gatessucks answer, or the one I prefer is:

mat[[2, Ordering[mat[[1, All]], -1]]]

Where -1 gives the largest value, 1 gives the smallest, and any other integer gives the relative largest/smallest (if negative)

share|improve this answer

If you define your input as

matrix = {{10, -2, 1}, {a, b, c}};

then you can do

Last@Last@SortBy[Transpose[matrix], #[[1]] &]
share|improve this answer
A variant is to use Ordering: Last@(Transpose@mat)[[Last@Ordering@First@mat]] – murray Oct 6 '12 at 21:34

The question of taking (or applying a transformation to) some part of a rectangular array based on the ordering of another part has come up in many variations. It is related to the general concept of Concomitants of order statistics (see, for example, Yang).

Using the built-in function Ordering (as in Gabriel's answer), one can define a function to extract the concomitants in a dataset as follows:

concomitantRows[list_, cols_, ordcol_, ranks_, orderingF_: Less] := 
  Part[list[[All, cols]], Ordering[list[[All, ordcol]], ranks, orderingF]]

which takes a list, and extracts cols associated with the ranks in the ordcol using the ordering function orderingF.

testdata = {RandomSample[Range[6]], RandomChoice[Range[5], 3] & /@ Range[6], 
  RandomChoice[CharacterRange["A", "F"], 2] & /@ Range[6], RandomSample[Range[6]]};
Transpose@testdata // Grid

enter image description here

Take the rows in the sub-array formed by columns {2,3,4} associated with the top 2 elements in col 1 based on default ordering Less:

concomitantRows[Transpose@testdata, {2, 3, 4}, 1, 2]
(* Out[] = {{{3, 3, 4}, {"B", "D"}, 5}, {{4, 4, 1}, {"C", "C"}, 4}} *)

Take the ones associated with the third element:

concomitantRows[Transpose@testdata, {2, 3, 4}, 1, {3}]
(* Out[] = {{{1, 1, 5}, {"A", "B"}, 3}}*)
share|improve this answer

One can use Pick[] for the purpose:

m = {{10, -2, 1}, {{1, 1, 1}, {-3, 1, 3}, {1, -2, 1}}};

First@Pick[Last[#], First[#], Max[First[#]]] &@m
   {1, 1, 1}
share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.