why does it take forever?

[greenflame]

I have the following function. It is supposed to multiply all the
elements of a row of a matrix my a certain factor. I make a matrix with
a list of lists. with each of the inner lists represents a row of a
matrix. So the matrix:

[1 2 3]
[4 5 6]
[7 8 9]

is coded as:

A = [
[1,2,3],
[4,5,6],
[7,8,9]
];

OK now heres the function:

function create2darr(rows,cols) {
output = new Array(rows);
for (i=0;i<rows;i++) {
output[i] = new Array(cols);
}
return output;
}
function copy2darr(input) {
output = create2darr(input.length,input[0].length);
for (i=0;i<input.length;i++) {
for (j=0;j<input[0].length;j++) {
output[i][j] = input[i][j];
}
}
return output;
}
function xrow(input,row,factor) {
output = copy2darr(input);
for (i=0;output[row-1].length;i++) {
output[row-1][i] *= factor;
}
return output;
}

[RobG]

greenflame wrote:

I have the following function. It is supposed to multiply all the
elements of a row of a matrix my a certain factor. I make a matrix with
a list of lists. with each of the inner lists represents a row of a
matrix. So the matrix:

[1 2 3]
[4 5 6]
[7 8 9]

is coded as:

A = [
[1,2,3],
[4,5,6],
[7,8,9]
];

OK now heres the function:

function create2darr(rows,cols) {
output = new Array(rows);
for (i=0;i<rows;i++) {
output[i] = new Array(cols);
}
return output;
}
function copy2darr(input) {
output = create2darr(input.length,input[0].length);
for (i=0;i<input.length;i++) {
for (j=0;j<input[0].length;j++) {
output[i][j] = input[i][j];
}
}
return output;
}
function xrow(input,row,factor) {
output = copy2darr(input);
for (i=0;output[row-1].length;i++) {
output[row-1][i] *= factor;
}
return output;
}

You create matrix A. You create a reference to it called 'input'
when calling the function xrow().

You then create a global variable 'output' that references a row of
A. You then do things with 'output' in other functions that are all
manipulating the same row of A - creating more references so that my
browser runs out of memory and never finishes.

There is no need to be so complex. If all you want to do is multiply
a row of A by some factor, then the following will do the job (note
that I have adjusted the index for rows to start from 1 rather than
zero, that seems to be what you were doing):

function xrow(input,row,factor) {
var j, r = row-1; // Adjust row index
// Make sure matrix and row exist and have length > 0
if ( !input || !input.length
|| !input[r] || !(j = input[r].length) ) {
return;
}
for (var i=0; i<j; i++) {
input[r][i] *= factor;
}
}

A = [
[1,2,3],
[4,5,6],
[7,8,9]
];

alert(A.join('\n'));
xrow(A,1,5);
alert(A.join('\n'));
Note that when you create A as a global object like this, if you
create a reference to some part of A and change it, then is it A
that is actually being changed.

It's the same as when you use:

x = document.getElementById('blah');

x is a reference to the element with id 'blah'. When you modify x,
you are actually modifying 'blah'.
--
Rob

[Jc]

greenflame wrote:

function xrow(input,row,factor) {
output = copy2darr(input);
for (i=0;output[row-1].length;i++) {
output[row-1][i] *= factor;
}
return output;
}

So I assume you are seeing the code perform slower than you expect?

Well, JavaScript isn't exactly a speed demon for mathematical
operations, but if you show how you are actually using xrow in an
actual working example, maybe someone can point out some optimizations.

I should also mention that you are likely not intending to create
global var's i and j (among others) in your for loops, which is what
happens when you use a var in a fuction without declaring it using the
var keyword (this does not apply to function parameter names of course).

[greenflame]

RobG wrote:

You create matrix A. You create a reference to it called 'input'
when calling the function xrow().

You then create a global variable 'output' that references a row of
A. You then do things with 'output' in other functions that are all
manipulating the same row of A - creating more references so that my
browser runs out of memory and never finishes.

There is no need to be so complex. If all you want to do is multiply
a row of A by some factor, then the following will do the job (note
that I have adjusted the index for rows to start from 1 rather than
zero, that seems to be what you were doing):

function xrow(input,row,factor) {
var j, r = row-1; // Adjust row index
// Make sure matrix and row exist and have length > 0
if ( !input || !input.length
|| !input[r] || !(j = input[r].length) ) {
return;
}
for (var i=0; i<j; i++) {
input[r][i] *= factor;
}
}

A = [
[1,2,3],
[4,5,6],
[7,8,9]
];

alert(A.join('\n'));
xrow(A,1,5);
alert(A.join('\n'));
Note that when you create A as a global object like this, if you
create a reference to some part of A and change it, then is it A
that is actually being changed.

It's the same as when you use:

x = document.getElementById('blah');

x is a reference to the element with id 'blah'. When you modify x,
you are actually modifying 'blah'.
--
Rob

ok but I want the function to return a new matrix and leave the
original untouched.

[Danny]

Keep it concise and precise, all I can tell is that you're doing a
2d-array set for some element iteration for multiplication, and it really
depends on how's coded really, no matter what language, if you don't use
resources sparingly and return them back as soon as not needed, it can bog
down the system, so, it isn't the language per se. Not sure what you're
specifically doing but I'd think it wouldn't take me 3 functions just to
take 2 arguments and process something for stdout.

Danny

On Sat, 11 Jun 2005 17:19:28 -0700, greenflame <al*********@yahoo.com>
wrote:

I have the following function. It is supposed to multiply all the
elements of a row of a matrix my a certain factor. I make a matrix with
a list of lists. with each of the inner lists represents a row of a
matrix. So the matrix:

[1 2 3]
[4 5 6]
[7 8 9]

is coded as:

A = [
[1,2,3],
[4,5,6],
[7,8,9]
];

OK now heres the function:

function create2darr(rows,cols) {
output = new Array(rows);
for (i=0;i<rows;i++) {
output[i] = new Array(cols);
}
return output;
}
function copy2darr(input) {
output = create2darr(input.length,input[0].length);
for (i=0;i<input.length;i++) {
for (j=0;j<input[0].length;j++) {
output[i][j] = input[i][j];
}
}
return output;
}
function xrow(input,row,factor) {
output = copy2darr(input);
for (i=0;output[row-1].length;i++) {
output[row-1][i] *= factor;
}
return output;
}

--
Using Opera's revolutionary e-mail client: http://www.opera.com/mail/

[RobG]

greenflame wrote:

[...]
ok but I want the function to return a new matrix and leave the
original untouched.

Then copy the matrix to a new object first:

function copyMatrix(A){
var j, i=A.length, B = [];
while (i--) {
B[i]= [];
j = A[i].length;
while ( j-- ){
B[i][j] = A[i][j];
}
}
return B;
}

function xrow(input,row,factor) {
var j, r = row-1; // Adjust row index
// Make sure matrix and row exist and have length > 0
if ( !input || !input.length
|| !input[r] || !(j = input[r].length) ) {
return;
}
for (var i=0; i<j; i++) {
input[r][i] *= factor;
}
}

A = [
[1,2,3],
[4,5,6],
[7,8,9]
];

B = copyMatrix(A);
xrow(B, 1, 5);
alert(A.join('\n'));
alert(B.join('\n'));

[greenflame]

ok thank you for all your help!