How to pick an element of a row corresponding to the minimum in a particular column?

Question

How can I find what value in column 1 of my list corresponds to the minimum value in column 2?

Answer 1

Pick[] is one way to go about it:

test = {{15, 11}, {5, 14}, {2, 13}, {3, 5}, {13, 15}, {6, 10}, {8, 15}, {0, 2},
        {10, 13}, {2, 5}};

Pick[#1, #2, Min[#2]] & @@ Transpose[test]
   {0}

Answer 2

Edit

In my earlier submission, I mistakenly took "corresponds to" as meaning "has the same value as". Here I interpret "corresponds to" as "occupies the same row as". This variation relies on Position, as before.

#1[[Position[#2, Min@#2][[1, 1]]]] & @@ Transpose@t

---

Exemplification

t happens to be 15x7:

t={{0, 10, 4, 19, 1, 3, 2}, {3, 18, 1, 12, 7, 14, 16}, {11, 8, 17, 18, 7, 12, 17}, {12, 16, 0, 16, 8, 3, 11}, {8, 2, 14, 3, 18, 7, 6}, {5, 15, 14, 9, 9, 3, 2}, {9, 10, 17, 6, 19, 14, 0}, {3, 5, 18, 11, 10, 12, 6}, {7, 13, 7, 13, 16, 14, 16}, {14, 12, 4, 19, 18, 20, 7}, {18,3, 19, 15, 16, 18, 8}, {1, 18, 5, 11, 3, 5, 2}, {16, 11, 7, 11, 2, 2, 19}, {1, 8, 7, 7, 15, 1, 20}, {11, 9, 2, 7, 2, 18, 4}}

The lowest value in column 2 is 2. It sits next to the value 8 in column one.

#1[[Position[#2, Min@#2][[1, 1]]]] & @@ Transpose@t

8

Explanation

Let's explain what the following does:

#1[[Position[#2, Min@#2][[1, 1]]]] 

First note that Min@#2returns the minimum value in the second column: 2

Now plug 2 into

Position[#2, Min@#2][[1, 1]]

obtaining

Position[#2, 2][[1, 1]]

returning `5'.

Substituting again, #1[[5]] returns the value the cell at column 1, row 5:

8

Answer 3

Here's a straightforward way using Cases:

list = {{1, 2}, {2, 2}, {3, 1}, {4, 1}, {2, 4}}; (* example list *)
With[{min = Min@#[[All, 2]]}, Cases[#, {x_, min} :> x]] &@list
(* {3, 4} *)

You can also use Select or Pick, depending on the usage. You should read through the documentation to see how to use them.

Answer 4

Here's another way, stealing @J.M's test list

test = {{15, 11}, {5, 14}, {2, 13}, {3, 5}, {13, 15}, {6, 10}, {8, 15}, {0, 2},
        {10, 13}, {2, 5}};

First@Extract[test, test~Reverse~2~Ordering~1]

0

Remove First to see the whole row

Or probably faster is

Extract[#1, #2~Ordering~1] & @@ Transpose[test]

Answer 5

Sure.

Not sure about "asking" but here's a way to do it.

I'll build some data for an example:

a = RandomReal[{0, 10}, {10, 2}]

And to find the value in the position in column 1 of the minimum of column 2:

a[[Flatten[Position[a[[All, 2]], Min[a[[All, 2]]]]], 1]]

Answer 6

Beware of the possibility of multiple minima, especially in long lists of discrete elements. I think for a novice J.M.'s solution is the more instructive. I have changed its first element to demonstrate two identical minima:

test = {{15, 2}, {5, 14}, {2, 13}, {3, 5}, {13, 15}, {6, 10}, {8, 15}, {0, 2}, {10, 13}, {2, 5}};

Pick[#1, #2, Min[#2]] & @@ Transpose[test]

out: {15, 0}

Finding the position of the minimum/minima:

min = Min@test[[All, 2]]
pos = Position[test[[All, 2]], min]

out: 2
out: {{1}, {8}}

In order to grab the corresponding values in Col. 1, the "pos" list must be Flattened first:

test[[Flatten@pos, 1]]

out: {15, 0}

Answer 7

With

test = {{15, 11}, {5, 14}, {2, 13}, {3, 5}, {13, 15}, {6, 10}, {8, 15},
    {0,2}, {10, 13}, {2, 5}};

variations with Ordering in Rojo's answer using Part instead of Extract:

test[[Ordering[test[[All, 2]], 1]]][[1, 1]]

or

First[#[[1]][[Ordering[#[[2]], 1]]] &[Transpose[test]]]

or

test[[Ordering[test, 1, #1[[2]] < #2[[2]] &]]][[1, 1]]
(* 0 *)