Given the matrix:

matrix = ConstantArray[0, {3, 2}]

and date list:

list = {a, b, c, d, e, f}

a way to enter the list elements in the matrix is as follow:

matrix[[1,1]] = list[[1]];
matrix[[1,2]] = list[[2]];
matrix[[2,1]] = list[[3]];
matrix[[2,2]] = list[[4]];
matrix[[3,1]] = list[[5]];
matrix[[3,2]] = list[[6]];

getting what you want:

matrix == {{a, b}, {c, d}, {e, f}}

You could show a more elegant and effective way to do this?

Thanks so much.

Similarly, I can write:

matrix = ConstantArray[0, {3, 2}];

list1 = {a, b, c};
list2 = {d, e, f};
list3 = {g, h, i};
list4 = {j, k, l};
list5 = {m, n, o};
list6 = {p, q, r};

list = {list1, list2, list3, list4, list5, list6};

matrix[[1, 1]] = list[[1, 1]];
matrix[[1, 2]] = list[[2, 1]];
matrix[[2, 1]] = list[[3, 1]];
matrix[[2, 2]] = list[[4, 1]];
matrix[[3, 1]] = list[[5, 1]];
matrix[[3, 2]] = list[[6, 1]];

and obtain:

matrix == {{a, d}, {g, j}, {m, p}}

but I can not take advantage of the command that you kindly showed me.

In this other case, how can you do?

One last thing, instead of writing:

matrix[[1, 1]] = Grid[list[[1, 1]], Frame -> All, ItemSize -> All];
matrix[[1, 2]] = Grid[list[[2, 1]], Frame -> All, ItemSize -> All];
matrix[[3, 2]] = Grid[list[[6, 1]], Frame -> All, ItemSize -> All];

how can I do? I thought I was able to generalize the previous simple cases but I can not.

  • 2
    $\begingroup$ matrix = ArrayReshape[list, {3, 2}]? $\endgroup$
    – Kuba
    Feb 12, 2017 at 13:29
  • $\begingroup$ Notice that you don't have to define matrix with ConstantArray before. Unless you want to fill part of a bigger matrix, but that is a different question. $\endgroup$
    – Kuba
    Feb 12, 2017 at 13:31
  • 2
    $\begingroup$ Partition[list, 2] $\endgroup$
    – Bob Hanlon
    Feb 12, 2017 at 13:44
  • $\begingroup$ I have marked this question as a duplicate i.e. "already has an answer here." If you feel that your problem is not adequately covered there please edit your question to describe specifically how your needs differ and add a link to that question for reference. $\endgroup$
    – Mr.Wizard
    Feb 13, 2017 at 10:40

2 Answers 2


The way you are thinking about the problem is hindering you -- you ask: how do I fill the entries of a matrix with certain values? This means you first create a matrix and then proceed to set its values. A more idiomatic way to envision this problem is: I have a list in a particular form, how do I change its form (select elements, partition them, reshape the array) to achieve the desired form? Proceeding this way, for your first case you want to partition the original list:

Partition[list, 2]

For the second, you want to select the elements of the list you want to keep, and then partition:

Partition[list[[All, 1]], 2]
{{a, d}, {g, j}, {m, p}}

And now for the third: here you apply (Map) the Grid function to the selected elements and then partition.

Partition[Map[Grid, list, {2}][[All, 1]], 2]
{{Grid[a], Grid[d]}, {Grid[g], Grid[j]}, {Grid[m], Grid[p]}}

Or, keeping all the options:

Partition[Map[Grid[#, Frame -> All, ItemSize -> All] &, list, {2}][[All, 1]], 2]

Frankly, these final versions do not make a lot of sense. Grid is a display command... filling a matrix with Grid elements is not going to give you a matrix that can be used for anything.

  • 1
    $\begingroup$ Manu -- you are doing it again. In this case, you do not want to make a matrix of elements, each of which is a Grid object. Grid is a display command... so you make the thing you want to display and then apply the display command(s) to the complete item. Nonetheless, I have shown above how to do this. $\endgroup$
    – bill s
    Feb 12, 2017 at 15:45
matrix = ArrayReshape[list, Dimensions[matrix]]


matrix[[All, All]] = ArrayReshape[list, Dimensions[matrix]]

assuming list is of the appropriate length.


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