如何在Python中插值2D曲线

Question

如何在Python中插值2D曲线

pythonscipyinterpolation

19

我有一组x和y坐标，它是一个曲线/形状，我想要平滑曲线/形状并绘制图形。

我尝试了不同的插值方法来平滑曲线/形状，但仍然无法达到我的期望。使用点来绘制平滑的曲线/形状。

像下面这样，使用x、y点来获取平滑的圆/曲线

然而，我得到了如下结果：

circle.jpg

curve.jpg

square.jpg

我在样条插值和RBF插值方面遇到了麻烦。

对于cubic_spline_interpolation，我得到了以下错误信息：

ValueError: Error on input data

对于univariate_spline_interpolated，我得到了以下错误信息：

ValueError: x must be strictly increasing

对于rbf，我得到了以下错误信息：

numpy.linalg.linalg.LinAlgError: Matrix is singular.

我不知道如何修复它们并获得正确的尖锐和曲线。非常感谢您的帮助。

编辑：对于那些无法下载源代码和x、y坐标文件的人，我在问题中发布了代码和x、y坐标。

#!/usr/bin/env python3
from std_lib import *

import os
import numpy as np
import cv2

from scipy import interpolate
import matplotlib.pyplot as plt

CUR_DIR = os.getcwd()
CIRCLE_FILE = "circle.txt"
CURVE_FILE  = "curve.txt"
SQUARE_FILE = "square.txt"

#test
CIRCLE_NAME = "circle"
CURVE_NAME  = "curve"
SQUARE_NAME = "square"

SYS_TOKEN_CNT = 2   # x, y

total_pt_cnt = 0        # total no. of points      
x_arr = np.array([])    # x position set
y_arr = np.array([])    # y position set

def convert_coord_to_array(file_path):
    global total_pt_cnt
    global x_arr
    global y_arr

    if file_path == "":
        return FALSE

    with open(file_path) as f:
        content = f.readlines()

    content = [x.strip() for x in content] 

    total_pt_cnt = len(content)

    if (total_pt_cnt <= 0):
        return FALSE

    ##
    x_arr = np.empty((0, total_pt_cnt))
    y_arr = np.empty((0, total_pt_cnt))

    #compare the first and last x 
    # if ((content[0][0]) > (content[-1])):
        # is_reverse = TRUE

    for x in content:
        token_cnt = get_token_cnt(x, ',') 

        if (token_cnt != SYS_TOKEN_CNT):
            return FALSE

        for idx in range(token_cnt):
            token_string = get_token_string(x, ',', idx)
            token_string = token_string.strip()
            if (not token_string.isdigit()): 
                return FALSE

            # save x, y set
            if (idx == 0):
                x_arr = np.append(x_arr, int(token_string))
            else:
                y_arr = np.append(y_arr, int(token_string))

    return TRUE

def linear_interpolation(fig, axs):
    xnew = np.linspace(x_arr.min(), x_arr.max(), len(x_arr))
    f = interpolate.interp1d(xnew , y_arr)

    axs.plot(xnew, f(xnew))
    axs.set_title('linear')

def cubic_interpolation(fig, axs):
    xnew = np.linspace(x_arr.min(), x_arr.max(), len(x_arr))
    f = interpolate.interp1d(xnew , y_arr, kind='cubic')

    axs.plot(xnew, f(xnew))
    axs.set_title('cubic')

def cubic_spline_interpolation(fig, axs):
    xnew = np.linspace(x_arr.min(), x_arr.max(), len(x_arr))
    tck = interpolate.splrep(x_arr, y_arr, s=0) #always fail (ValueError: Error on input data)
    ynew = interpolate.splev(xnew, tck, der=0)

    axs.plot(xnew, ynew)
    axs.set_title('cubic spline')

def parametric_spline_interpolation(fig, axs):
    xnew = np.linspace(x_arr.min(), x_arr.max(), len(x_arr))
    tck, u = interpolate.splprep([x_arr, y_arr], s=0)
    out = interpolate.splev(xnew, tck)

    axs.plot(out[0], out[1])
    axs.set_title('parametric spline')

def univariate_spline_interpolated(fig, axs):   
    s = interpolate.InterpolatedUnivariateSpline(x_arr, y_arr)# ValueError: x must be strictly increasing
    xnew = np.linspace(x_arr.min(), x_arr.max(), len(x_arr))
    ynew = s(xnew)

    axs.plot(xnew, ynew)
    axs.set_title('univariate spline')

def rbf(fig, axs):
    xnew = np.linspace(x_arr.min(), x_arr.max(), len(x_arr))
    rbf = interpolate.Rbf(x_arr, y_arr) # numpy.linalg.linalg.LinAlgError: Matrix is singular.
    fi = rbf(xnew)

    axs.plot(xnew, fi)
    axs.set_title('rbf')

def interpolation():
    fig, axs = plt.subplots(nrows=4)
    axs[0].plot(x_arr, y_arr, 'r-')
    axs[0].set_title('org')

    cubic_interpolation(fig, axs[1])
    # cubic_spline_interpolation(fig, axs[2]) 
    parametric_spline_interpolation(fig, axs[2])
    # univariate_spline_interpolated(fig, axs[3])
    # rbf(fig, axs[3])        
    linear_interpolation(fig, axs[3])

    plt.show()

#------- main -------
if __name__ == "__main__":
    # np.seterr(divide='ignore', invalid='ignore')

    file_name = CUR_DIR + "/" + CIRCLE_FILE 
    convert_coord_to_array(file_name)  
    #file_name = CUR_DIR + "/" + CURVE_FILE 
    #convert_coord_to_array(file_name) 
    #file_name = CUR_DIR + "/" + SQUARE_FILE 
    #convert_coord_to_array(file_name) 
    #
    interpolation()

圆形的x，y坐标

已解决

def linear_interpolateion(self, x, y):

        points = np.array([x, y]).T  # a (nbre_points x nbre_dim) array

        # Linear length along the line:
        distance = np.cumsum( np.sqrt(np.sum( np.diff(points, axis=0)**2, axis=1 )) )
        distance = np.insert(distance, 0, 0)

        alpha = np.linspace(distance.min(), int(distance.max()), len(x))
        interpolator =  interpolate.interp1d(distance, points, kind='slinear', axis=0)
        interpolated_points = interpolator(alpha)

        out_x = interpolated_points.T[0]
        out_y = interpolated_points.T[1]

        return out_x, out_y

- shadow dk

请在问题中发布您算法的代码行。 - Roi Danton

x.min()和x.max()在原始数据上表示什么？我认为最小值和最大值在原始数据和插值点上必须完全相同。只有“内部”曲线上的点被添加... - xdze2

1个回答

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- xdze2 · Accepted Answer

由于插值是针对通用的2d曲线，即(x, y)=f(s)其中s是沿着曲线的坐标，而不是y=f(x)，因此必须首先计算沿线s的距离。然后，每个坐标的插值相对于s执行。（例如，在圆形情况下，y=f(x)有两个解）

在此代码中，s（或distance）被计算为给定点之间每个段的长度的累积和。

import numpy as np
from scipy.interpolate import interp1d
import matplotlib.pyplot as plt

# Define some points:
points = np.array([[0, 1, 8, 2, 2],
                   [1, 0, 6, 7, 2]]).T  # a (nbre_points x nbre_dim) array

# Linear length along the line:
distance = np.cumsum( np.sqrt(np.sum( np.diff(points, axis=0)**2, axis=1 )) )
distance = np.insert(distance, 0, 0)/distance[-1]

# Interpolation for different methods:
interpolations_methods = ['slinear', 'quadratic', 'cubic']
alpha = np.linspace(0, 1, 75)

interpolated_points = {}
for method in interpolations_methods:
    interpolator =  interp1d(distance, points, kind=method, axis=0)
    interpolated_points[method] = interpolator(alpha)

# Graph:
plt.figure(figsize=(7,7))
for method_name, curve in interpolated_points.items():
    plt.plot(*curve.T, '-', label=method_name);

plt.plot(*points.T, 'ok', label='original points');
plt.axis('equal'); plt.legend(); plt.xlabel('x'); plt.ylabel('y');

这将得到：

关于图表，似乎您正在寻找平滑方法而不是点的插值。在此，类似的方法被用来分别对给定曲线的每个坐标进行样条拟合（参见Scipy UnivariateSpline）：

import numpy as np
import matplotlib.pyplot as plt

from scipy.interpolate import UnivariateSpline

# Define some points:
theta = np.linspace(-3, 2, 40)
points = np.vstack( (np.cos(theta), np.sin(theta)) ).T

# add some noise:
points = points + 0.05*np.random.randn(*points.shape)

# Linear length along the line:
distance = np.cumsum( np.sqrt(np.sum( np.diff(points, axis=0)**2, axis=1 )) )
distance = np.insert(distance, 0, 0)/distance[-1]

# Build a list of the spline function, one for each dimension:
splines = [UnivariateSpline(distance, coords, k=3, s=.2) for coords in points.T]

# Computed the spline for the asked distances:
alpha = np.linspace(0, 1, 75)
points_fitted = np.vstack( spl(alpha) for spl in splines ).T

# Graph:
plt.plot(*points.T, 'ok', label='original points');
plt.plot(*points_fitted.T, '-r', label='fitted spline k=3, s=.2');
plt.axis('equal'); plt.legend(); plt.xlabel('x'); plt.ylabel('y');

这将产生: