矩陣運算, 每次遇到這個問題就頭痛想半天, 要搞這個就要先理解術語
台灣跟大陸定義剛好是反過來, 看得很煩躁
m = row = y = height = 高 = 列 = 行(簡)
n = column = x = width = 寬 = (欄/行) = 列(簡)
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| height = len(matrix) # y 的最大範圍 (Rows)
width = len(matrix[0]) # x 的最大範圍 (Cols)
for y in range(height): # 外層跑垂直高度 (y 方向)
for x in range(width): # 內層跑水平寬度 (x 方向)
# 注意：矩陣存取通常是 [y][x]
print(f"座標 ({x}, {y}) 的值為: {matrix[y][x]}")
rows = len(matrix)
cols = len(matrix[0])
for i in range(rows):
for j in range(cols):
print(matrix[i][j])
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搞定術語後就可以進入正題
可以參考這裡 mathsisfun
或是用這個線上工具來算 matrixmultiplication
(1, 2, 3) • (7, 9, 11)= 1×7 + 2×9 + 3×11 = 58
(1, 2, 3) • (8, 10, 12) = 1×8 + 2×10 + 3×12 = 64
(4, 5, 6) • (7, 9, 11) = 4×7 + 5×9 + 6×11 = 139
(4, 5, 6) • (8, 10, 12) = 4×8 + 5×10 + 6×12 = 154
關鍵在於先把解答矩陣大小算出來
A (2 * 3) = 2 h * 3 w
B (3 * 2) = 3 h * 2 w
只取 A 的 h 跟 B 的 w
C (2 * 2) = C(Ay * Bx)
所以這裡會得到一個 m = 2, n 也 = 2 的 matrix
計算時跟平時操作 2d array 一樣, aWidth 等於就是次數
第三個迴圈是這題另外一個關鍵, 他第一次會長這樣以此類推
C[0][0] += A[0][0] * B[0][0] //17
C[0][0] += A[0][1] * B[1][0] //29
C[0][0] += A[0][2] * B[2][0] //3*11
結果為 58
C[0][1] += A[0][0] * B[0][1] //18
C[0][1] += A[0][1] * B[1][1] //210
C[0][1] += A[0][2] * B[1][1] //3*12
結果為 64
最後要注意到矩陣維度問題
(3,2) * (3,2) not work
(2,3) * (3,2) work
(3,2) * (2,3) work
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| let A = [
[1,2,3],
[4,5,6]
]
let B = [
[7,8],
[9,10],
[11,12],
]
ay = aHeight = A.length
ax = aWidth = A[0].length
by = bHeight = B.length
bx = bWidth = B[0].length
console.log(`ay ${ay}`)
console.log(`ax ${ax}`)
console.log(`by ${by}`)
console.log(`bx ${bx}`)
let C = []
for (let y = 0; y < ay; y++) {
C[y] = []
for (let x = 0; x < bx; x++) {
C[y][x] = 0
}
}
//立即列印, 防止 chrome 之後印參考
console.log(JSON.stringify(C))
/*
let C = [
[0,0],
[0,0],
]
*/
//計算結果
for(let y = 0; y < aHeight; y++){
for(let x = 0; x < bWidth; x++){
for(let z = 0; z < aWidth; z++){
C[y][x] += A[y][z] * B[z][x]
}
}
}
/*
C[0][0] += A[0][0] * B[0][0] //1*7
C[0][0] += A[0][1] * B[1][0] //2*9
C[0][0] += A[0][2] * B[2][0] //3*11
結果為 58
C[0][1] += A[0][0] * B[0][1] //1*8
C[0][1] += A[0][1] * B[1][1] //2*10
C[0][1] += A[0][2] * B21][1] //3*12
結果為 64
*/
console.log(C)
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偷懶用 chatgpt 搞成 python XD
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| # 定義矩陣
A = [
[1, 2, 3],
[4, 5, 6]
]
B = [
[7, 8],
[9, 10],
[11, 12]
]
# 矩陣尺寸
aHeight = len(A)
aWidth = len(A[0])
bHeight = len(B)
bWidth = len(B[0])
print(f"ay {aHeight}")
print(f"ax {aWidth}")
print(f"by {bHeight}")
print(f"bx {bWidth}")
# 初始化結果矩陣 C
C = [[0 for _ in range(bWidth)] for _ in range(aHeight)]
# 立即列印，防止引用問題
import json
print(json.dumps(C))
# 計算結果
for y in range(aHeight):
for x in range(bWidth):
for z in range(aWidth):
C[y][x] += A[y][z] * B[z][x]
print(C)
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順帶一提要算 dot product 可以這樣寫
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| A = [[0, 3, 2], [-3, -3, 1], [1, 0, 2]]
B = [[1, 0, 6], [2, -1, 0], [5, 1, 4]]
total = 0
for y in range(len(A)):
for x in range(len(A[0])):
total += A[y][x] * B[y][x]
print(f"{A[y][x] * B[y][x]:>2} ", end="")
print()
print()
print(f"total:{total}")
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