Python 避免循环在三维阵列上运行线性回归

我需要在3d数组中运行线性回归，例如：-

数组=np.arange（3*4*5）。重塑（3,4,5）

我需要在[0,20,40]、[1,21,41]…[5,25,45]…[19,39,59]上运行LR 我正在使用以下代码：-

LrCalcRate = np.array([])
col_nums = 3
for j in range(Array.shape[1]):
    for i in range(Array.shape[2]):
        y = Array[:, j, i].T
        x = np.arange(col_nums)
        A = np.vstack([x, np.ones(len(x))]).T
        m, c = np.linalg.lstsq(A, y)[0]
        LrCalcRate = np.append(LrCalcRate, -m)

---

两个for循环需要花费大量时间，不使用for循环是否可以对其进行优化。

您编写的内容与

x = np.arange(col_nums)
A = np.vstack([x, np.ones(len(x))]).T
-np.linalg.lstsq(A, a.reshape(a.shape[0], -1))[0][0]

例如，使用为每个目标生成不同坡度的数据：

In [9]: a = (np.linspace(0, 2, num=3*4*5)**2).reshape(3,4,5)                                

In [11]: lr_calc_rate = np.array([]) 
    ...: col_nums = 3 
    ...: for j in range(a.shape[1]): 
    ...:     for i in range(a.shape[2]): 
    ...:         y = a[:, j, i].T 
    ...:         x = np.arange(col_nums) 
    ...:         A = np.vstack([x, np.ones(len(x))]).T 
    ...:         m, c = np.linalg.lstsq(A, y)[0] 
    ...:         lr_calc_rate = np.append(lr_calc_rate, -m) 
    ...: lr_calc_rate

Out[11]: 
array([-0.91927607, -0.96523987, -1.01120368, -1.05716748, -1.10313128,
       -1.14909509, -1.19505889, -1.24102269, -1.2869865 , -1.3329503 ,
       -1.37891411, -1.42487791, -1.47084171, -1.51680552, -1.56276932,
       -1.60873312, -1.65469693, -1.70066073, -1.74662453, -1.79258834])

In [12]: x = np.arange(col_nums)
    ...: A = np.vstack([x, np.ones(len(x))]).T
    ...: -np.linalg.lstsq(A, a.reshape(3, -1))[0][0]

Out[12]: 
array([-0.91927607, -0.96523987, -1.01120368, -1.05716748, -1.10313128,
       -1.14909509, -1.19505889, -1.24102269, -1.2869865 , -1.3329503 ,
       -1.37891411, -1.42487791, -1.47084171, -1.51680552, -1.56276932,
       -1.60873312, -1.65469693, -1.70066073, -1.74662453, -1.79258834])

---

这个例子是不可复制的。请定义

col\u nums

，

noisemeasuresnp

等。这看起来需要使用并从那里开始线性回归。但是这个例子太混乱了，无法理解，为什么你不能适当地重塑呢？嗨@yatu，我已经更正了代码谢谢fuglede，你能解释一下它是如何工作的，或者你能告诉我怎么做的吗？嗯，它正是你用手所做的。From:“如果b是二维的，则计算b的K列的最小二乘解”。

In [9]: a = (np.linspace(0, 2, num=3*4*5)**2).reshape(3,4,5)                                

In [11]: lr_calc_rate = np.array([]) 
    ...: col_nums = 3 
    ...: for j in range(a.shape[1]): 
    ...:     for i in range(a.shape[2]): 
    ...:         y = a[:, j, i].T 
    ...:         x = np.arange(col_nums) 
    ...:         A = np.vstack([x, np.ones(len(x))]).T 
    ...:         m, c = np.linalg.lstsq(A, y)[0] 
    ...:         lr_calc_rate = np.append(lr_calc_rate, -m) 
    ...: lr_calc_rate

Out[11]: 
array([-0.91927607, -0.96523987, -1.01120368, -1.05716748, -1.10313128,
       -1.14909509, -1.19505889, -1.24102269, -1.2869865 , -1.3329503 ,
       -1.37891411, -1.42487791, -1.47084171, -1.51680552, -1.56276932,
       -1.60873312, -1.65469693, -1.70066073, -1.74662453, -1.79258834])

In [12]: x = np.arange(col_nums)
    ...: A = np.vstack([x, np.ones(len(x))]).T
    ...: -np.linalg.lstsq(A, a.reshape(3, -1))[0][0]

Out[12]: 
array([-0.91927607, -0.96523987, -1.01120368, -1.05716748, -1.10313128,
       -1.14909509, -1.19505889, -1.24102269, -1.2869865 , -1.3329503 ,
       -1.37891411, -1.42487791, -1.47084171, -1.51680552, -1.56276932,
       -1.60873312, -1.65469693, -1.70066073, -1.74662453, -1.79258834])