我正在做一个将 2 个矩阵相乘的函数。
矩阵将始终具有相同的行数和列数。 (2x2, 5x5, 23x23, ...)
当我打印它时,它不起作用。为什么?
例如,如果我创建两个 2x2 矩阵:
矩阵A:
[1][2]
[3][4]
矩阵B:
[5][6]
[7][8]
结果应该是:
[19][22]
[43][50]
(http://ncalculators.com/matrix/2x2-matrix-multiplication-calculator.htm)
但是,我得到:
[19][undefined]
[22][indefined]
function multiplyMatrix(matrixA, matrixB) {
var result = new Array(); //declare an array
//var numColsRows=$("#matrixRC").val();
numColsRows = 2;
//iterating through first matrix rows
for (var i = 0; i < numColsRows; i++) {
//iterating through second matrix columns
for (var j = 0; j < numColsRows; j++) {
var matrixRow = new Array(); //declare an array
var rrr = new Array();
var resu = new Array();
//calculating sum of pairwise products
for (var k = 0; k < numColsRows; k++) {
rrr.push(parseInt(matrixA[i][k]) * parseInt(matrixB[k][j]));
} //for 3
resu.push(parseInt(rrr[i]) + parseInt(rrr[i + 1]));
result.push(resu);
//result.push(matrixRow);
} //for 2
} //for 1
return result;
} // function multiplyMatrix
您对各种临时数组感到困惑。
undefined
值是由最内层循环下方的行上的越界访问引起的。
我建议您坚持为乘法结果制作一个数组。您可能已经知道,问题在于 JavaScript 不允许您初始化多维数组。要制作二维数组,您必须初始化一维数组,然后遍历其元素并将每个元素初始化为一维数组。
function multiply(a, b) {
var aNumRows = a.length, aNumCols = a[0].length,
bNumRows = b.length, bNumCols = b[0].length,
m = new Array(aNumRows); // initialize array of rows
for (var r = 0; r < aNumRows; ++r) {
m[r] = new Array(bNumCols); // initialize the current row
for (var c = 0; c < bNumCols; ++c) {
m[r][c] = 0; // initialize the current cell
for (var i = 0; i < aNumCols; ++i) {
m[r][c] += a[r][i] * b[i][c];
}
}
}
return m;
}
function display(m) {
for (var r = 0; r < m.length; ++r) {
document.write(' '+m[r].join(' ')+'<br />');
}
}
var a = [[8, 3], [2, 4], [3, 6]],
b = [[1, 2, 3], [4, 6, 8]];
document.write('matrix a:<br />');
display(a);
document.write('matrix b:<br />');
display(b);
document.write('a * b =<br />');
display(multiply(a, b));
您可以使用来自以下网址的 multiplyMatrices() 函数:http://tech.pro/tutorial/1527/matrix-multiplication-in-functional-javascript 它就像魅力一样。示例(您可以使用 console.table() 在 Chrome 和 Firefox 控制台中打印带有样式的矩阵):
function multiplyMatrices(m1, m2) {
var result = [];
for (var i = 0; i < m1.length; i++) {
result[i] = [];
for (var j = 0; j < m2[0].length; j++) {
var sum = 0;
for (var k = 0; k < m1[0].length; k++) {
sum += m1[i][k] * m2[k][j];
}
result[i][j] = sum;
}
}
return result;
}
var m1 = [[1,2],[3,4]]
var m2 = [[5,6],[7,8]]
var mResult = multiplyMatrices(m1, m2)
/*In Google Chrome and Firefox you can do:*/
console.table(mResult) /* it shows the matrix in a table */
我知道这是一个老问题,但我recommend切换到我的答案。
我的解决方案有很好的性能,因为它使用了
Map
功能Reduce
//The chosen one
function matrixDot (A, B) {
var result = new Array(A.length).fill(0).map(row => new Array(B[0].length).fill(0));
return result.map((row, i) => {
return row.map((val, j) => {
return A[i].reduce((sum, elm, k) => sum + (elm*B[k][j]) ,0)
})
})
}
var print = m => m.forEach(r => document.write(` ${r.join(' ')}<br/>`))
var a = [[8, 3], [2, 4], [3, 6]]
var b = [[1, 2, 3], [4, 6, 8]]
document.write('matrix a:<br />');
print(a);
document.write('matrix b:<br />');
print(b);
document.write('a * b =<br />');
print(matrixDot(a,b));
对于那些对纯功能解决方案感兴趣的人:
let MatrixProd = (A, B) =>
A.map((row, i) =>
B[0].map((_, j) =>
row.reduce((acc, _, n) =>
acc + A[i][n] * B[n][j], 0
)
)
)
浏览器测试代码:
let A = [[8, 3], [2, 4], [3, 6]];
let B = [[1, 2, 3], [4, 6, 8]];
console.table(MatrixProd(A,B));
此版本将行存储为临时行,从而减少索引查找的有效数量。通过这个 benchmark 与没有存储行的版本相比,实现的性能几乎快 2 倍。
function multiply(a, b) {
let aRows = a.length;
let aCols = a[0].length;
let bCols = b[0].length;
let result = new Array(aRows);
for (let r = 0; r < aRows; ++r) {
const row = new Array(bCols);
result[r] = row;
const ar = a[r];
for (let c = 0; c < bCols; ++c) {
let sum = 0.;
for (let i = 0; i < aCols; ++i) {
sum += ar[i] * b[i][c];
}
row[c] = sum;
}
}
return result;
}
const m = multiply(
[[8, 3], [2, 4], [3, 6]],
[[1, 2, 3], [4, 6, 8]]
);
console.log(m);
function display(m) {
for (var r = 0; r < m.length; ++r) {
document.write(' '+m[r].join(' ')+'<br />');
}
}
var a = [[8, 3], [2, 4], [3, 6]],
b = [[1, 2, 3], [4, 6, 8]];
document.write('matrix a:<br />');
display(a);
document.write('matrix b:<br />');
display(b);
document.write('a * b =<br />');
display(multiply(a, b));
这是我的带有数学错误处理的 ES6 解决方案:
const matrixDot = (A, B) => {
// Error handling
const mx = [A, B];
const cols = mx.map((matrix) => matrix[0].length);
if (!mx.every((matrix, i) => matrix.every((row) => row.length === cols[i]))) {
throw new Error(
'All rows in a matrix must have the same number of columns'
);
} else if (cols[0] !== B.length) {
throw new Error(
'The number of columns in the 1st matrix must be equal to the number of rows in the 2nd matrix'
);
}
// Calculations
return A.map((rowA) =>
B[0].map((_, xb) =>
rowA.reduce((acc, itemA, yb) => acc + itemA * B[yb][xb], 0)
)
);
};
// Example
const A = [
[3, 2, 5],
[6, 4, 1],
];
const B = [
[2, 6],
[5, 3],
[1, 4],
];
console.log(matrixDot(A, B));
希望对某人有所帮助;)
聚会有点晚了,但我想我找到了一个很好的解决方案。
恕我直言,我们在尝试解决此问题时发现的主要挑战是将
MatrixA.row1
与 MatrixB.col1
联系起来,这是原因之一,大多数解决方案都使用类似 MatrixB[0].length
的方法来获取结果的列数矩阵,而且,就我个人而言,我不喜欢这种“解决方法”(我没有充分的理由,我只是不喜欢它)。但我确实喜欢 map()
s 和 reduce()
的组合,正如 Jan Turoň 所提出的那样。
然后,受到这个矩阵乘法网站的启发,我想:如果生成的矩阵总是
MatrixA.number_of_rows
乘以MatrixB.number_of_columns
并且痛苦基本上是通过列“迭代”,为什么不转置第二个矩阵?
谢谢hobs,对于transpose function.
最终结果如下(我对结果很满意,因为执行时间非常接近Duloren的解决方案):
你会注意到我无法摆脱
的东西,因为我需要这种方法来转置矩阵。好吧...我仍然对结果感到满意:)Matrix[0]
const a = [
[1, 2, 0],
[6, 3, 8]
];
const b = [
[4, 6],
[1, 9],
[4, 8]
];
function multiplyMatrix(a, b) {
const tB = transpose(b);
// Return the matrix (array of rows)
return a.map((row_a) => {
// Return the rows with the values (array of values where the length
// will be the number of columns of 'b', which is the same as
// the length of `tB` (transposed `b`))
return tB.map((row_b) => {
// Return the sum of the products, which is the final value itself
// (therefore, defines the columns)
return row_a.reduce((carry, value_of_a, index_of_column_of_a) => {
// Because we transposed `b` the value in b that corresponds to a specific
// value in `a` will have the same `column_index`.
const corresponding_b = row_b[index_of_column_of_a];
return carry + (value_of_a * corresponding_b);
}, 0);
});
});
}
function transpose(m) {
return Object.keys(m[0]).map(columnIndex => {
return m.map(row => row[columnIndex])
});
}
function printMatrix(m, h = '') {
// console.table(m);
// For those that don't support the console.table()
let output = h + '\n--------\n';
output += m.reduce((carry, row) => {
return carry += row.join(' ') + '\n';
},'');
console.log(output);
}
printMatrix(a, 'A');
printMatrix(b, 'B');
printMatrix(multiplyMatrix(a, b), 'A times B');
console.log('==========');
printMatrix(transpose(b), 'Transposed B');
如果你想走疯狂的路线,你也可以在一些现代浏览器中可用的 WebGL 工具中做一些顶点转换。
不太确定这是否会以与 OpenCL 中的向量转换相同的方式工作(**实际上它们是类型等效/可互操作的),但总体思路是:
将您的值添加到缓冲区
“假装”它是一个顶点数组
使用 GPU 引擎转换 en-mass
从向量中检索修改后的值
(看这里的演示) http://www.html5rocks.com/en/tutorials/webgl/webgl_transforms/
只是通常的循环方法的替代方法。老实说,考虑到 OpenCL 是为这种事情设计的,有点麻烦
在 OpenCL 1.2 规范中,可以使用 OpenCL 加载和转换来自 OpenGL 的顶点缓冲区(参见。https://software.intel.com/en-us/articles/opencl-and-opengl-interoperability-tutorial)
您可以使用 Memoization 通过 动态规划 解决此问题。这是一个描述优化技术的术语,您可以在其中缓存先前计算的结果,并在再次需要相同的计算时返回缓存的结果。
let mat1 = [[1, 2, 3], [2, 1, 2]];
let mat2 = [[1, 2], [1, 2], [1, 2]];
function matrixMulti(x, y) {
let saveComputation = {};
let finalMat = [],
length=x.length,
length1 = y[0].length,
length2 = y.length;
for (let i = 0; i < length; i++) {
finalMat.push([]);
for (let j = 0; j < length1; j++) {
finalMat[i][j] = 0;
for (let k = 0; k < length2; k++) {
// check if we already computed this calculation or not
if (saveComputation[y[k][j] + '*' + x[i][k]] || saveComputation[x[i][k] + '*' + y[k][j]]) {
finalMat[i][j] = finalMat[i][j] + saveComputation[y[k][j] + '*' + x[i][k]];
} else {
// save if not computed
saveComputation[x[i][k] + '*' + y[k][j]] = x[i][k] * y[k][j]; // check format below how it is saved.
saveComputation[y[k][j] + '*' + x[i][k]] = x[i][k] * y[k][j];
finalMat[i][j] = finalMat[i][j] + saveComputation[y[k][j] + '*' + x[i][k]];
}
}
}
}
console.log(finalMat);
}
matrixMulti(mat1, mat2);
对于上述saveComputation的输入值将是
{ '1*1': 1,
'2*1': 2,
'1*2': 2,
'3*1': 3,
'1*3': 3,
'2*2': 4,
'3*2': 6,
'2*3': 6 }
const getDot = (arrA, arrB, row, col) => {
return arrA[row].map((val, i) => (val * arrB[i][col]))
.reduce((valA, valB) => valA + valB);
}
const multiplyMatricies = (a, b) => {
let matrixShape = new Array(a.length).fill(0)
.map(() => new Array(b[0].length).fill(0));
return matrixShape.map((row, i) =>
row.map((val, j) => getDot(a, b, i, j)));
}
const arrA = [
[1, 3, 0],
[2, 1, 1]
];
const arrB = [
[1, 2, 0, 1],
[2, 3, 1, 2],
[1, 2, 1, 1]
];
let product = multiplyMatricies(arrA, arrB);
console.log("product:", product);
您可以now 使用 TensorFlow.js 实现此目的:
<!DOCTYPE html>
<html>
<head>
<script src="https://cdn.jsdelivr.net/npm/@tensorflow/[email protected]/dist/tf.min.js"></script>
<script>
const matrix_1 = tf.tensor([[1,2],[3,4]]);
const matrix_2 = tf.tensor([[5,6],[7,8]]);
const matrix_result = tf.matMul(matrix_1, matrix_2);
matrix_result.print()
</script>
</head>
<body>
<h1>Check the console log!</h1>
</body>
</html>
向上=)
function multiplyMatrices(m1, m2) {
const result = new Array(m1.length)
.fill(0)
.map(() => new Array(m2[0].length)
.fill(0));
return result
.map((row, i) => row
.map((_, j) => m1[i]
.reduce((sum, e, k) => sum + (e * m2[k][j]), 0)));
}
这也可以与mathjs一起使用。
A = [[1, 2],
[3, 4]]
B = [[5, 6],
[7, 8]]
math.multiply(A, B)
npm 安装快递
节点服务器.js
var express = require('express');
var app = express();
var A=new Array(3);
var B=new Array(3);
var preA = [ 1, 2, 3, 4, 5, 6,7, 8, 9 ];
var preB = [ 1,1 ,1,2,2, 2,3, 3, 3 ];
//#########################preparing blank 3*3 matrix A and B###############
for(i=0;i<3;i++){
A[i]=new Array(3);
B[i]=new Array(3);
}
//#####################Assigning values to matrix places from predefine arrays preA and preB #####
var k=0;
for(i=0;i<3;i++){
for(j=0;j<3;j++){
A[i][j]=preA[k];
B[i][j]=preB[k];
k++;
}
};
console.log('################################');
console.log('First matrix:');
console.log(A[0]);
console.log(A[1]);
console.log(A[2]);
console.log('');
console.log('################################');
console.log('Second matrix:');
console.log(B[0]);
console.log(B[1]);
console.log(B[2]);
//###################### multiplication logic as disscussed ################
var result =[];
for (var i = 0; i < 3; i++) {
result[i] = new Array(3);
for (var j = 0; j < 3; j++) {
var sum = 0;
for (var k = 0; k < 3; k++) {
sum += A[i][k] * B[k][j];
}
result[i][j] = sum;
}
}
console.log('');
console.log('################################');
console.log('################################');
console.log('After Multiplication');
console.log(result[0]);
console.log(result[1]);
console.log(result[2]);
app.listen(9999);