numpy - NumPy 数组中每两个连续点的线方程

Question

我想在 NumPy 数组中找到每两个连续点（二维）的线方程。我知道如何稳健地做到这一点（即使用循环），但我想知道是否有更复杂的方法。

谢谢。

下面给出了一个稳健的方法：

import numpy as np

a = np.array([[1, 2], [2, 4], [3, 8], [5, 1]])
N = int(max(a.shape))
b = []
for i in range(N - 1):
    x = [a[i,0], a[i + 1,0]]
    y = [a[i,1], a[i + 1,1]]
    b.append(tuple(np.polyfit(x, y, 1)))

print(b)

标签： numpylinenumpy-ndarray

Answer 1

对于直线，请直接进行计算。y = mx + c 其中 m 是梯度 = y 的变化 / x 的变化，c 是常数 c = y0 - m*x0

import numpy as np

a = np.array([[1, 2], [2, 4], [3, 8], [5, 1]])

x = a[:,0]
y = a[:,1]

dx = np.diff(x)  # Change in x
dy = np.diff(y)  # Change in y

# Amended for @Nan's comment below.
# If any dx is zero this will now return +-inf in m and c without raising a warning
# The code using m and c will need to handle this if it can occur.
with np.errstate( divide = 'ignore' ):
    m = dy/dx  # Gradient
c = y[1:] - m * x[1:]   # Constant

m, c
# (array([ 2. ,  4. , -3.5]), array([ 0. , -4. , 18.5]))
# y = mx + c

# Rerunning with              x=3     x=3
a = np.array([[1, 2], [2, 4], [3, 8], [3, 12]])
m, c
(array([ 2.,  4., inf]), array([  0.,  -4., -inf]))