Question

我正在尝试创建一个q为3的矩阵。在这种情况下，q = 10.或每行我希望这三个值是我下面的代码中描述的三角函数的结果。

问题是我一直收到错误，说列表索引超出范围。我不明白为什么它说它超出了范围。在我看来，我的循环似乎是正确的。谁能告诉我我在忽视/做错了什么？

# Input az matrix
    az = [142.243258152,116.039625836,80.1585056414,139.614063776,87.2093336287,94.1433825229,35.5599100744,11.0328982848,177.717968103,19.0072693362]

# Construct frame of X matrix
X = [[0 for x in range(10)] for y in range(3)]

# Use az matrix to complete X matrix
f=0
for bear in az:
    X[f][0] = (M.cos(bear))**2
    X[f][1] = 2*M.cos(bear)*M.sin(bear)
    X[f][2] = (M.sin(bear))**2
    f=f+1

print X

Answer 1

OP的输入列表az有10个元素，而不是8个元素，应该交换矩阵的范围。

此外，sin和cos函数通常将弧度视为输入，而az似乎包含以度为单位的误差角度。

此片段：

from math import radians, cos, sin

# Input az matrix
az = [142.243258152, 116.039625836, 80.1585056414, 139.614063776, 87.2093336287, 94.1433825229, 35.5599100744, 11.0328982848, 177.717968103, 19.0072693362]

# Construct frame of X matrix
X = [[0 for x in range(3)] for y in range(10)]

# Use az matrix to complete X matrix
f=0
for bear in az:
    r = radians(bear)
    c = cos(r)
    s = sin(r)
    X[f][0] = c**2
    X[f][1] = 2*c*s
    X[f][2] = s**2
    f=f+1

print(X)

给出这个输出：

[[0.6250760791021176, -0.9682065367191874, 0.37492392089788235], [0.19271454590900655, -0.7888615840667916, 0.8072854540909934], [0.029214706063653385, 0.3368157182393228, 0.9707852939363467], [0.5801828858777331, -0.9870576575100736, 0.41981711412226685], [0.0023704299165554724, 0.09725864441922212, 0.9976295700834447], [0.0052204459914281754, -0.14412762309951216, 0.9947795540085718], [0.6617950612456389, 0.9461973539521655, 0.33820493875436103], [0.9633765287676627, 0.3756710933102597, 0.0366234712323373], [0.9984144917844932, -0.07957372378380607, 0.001585508215506806], [0.893927252777247, 0.615861411421014, 0.10607274722275291]]

试图在python中构建一个带循环的矩阵

1 个答案: