How to interpolate a string between two other lines in python

Note. I asked this question before, but it was closed as a duplicate, however I, like some others, believe that it was overly closed, I explain why editing in my original post . So I would like to ask this question again here.

Does anyone know a python library that can interpolate between two lines. For example, given the two solid lines below, I would like to create a dashed line in the middle. In other words, I would like to get the center line. The input is only two arrays of numpycoordinates with size N x 2and, M x 2respectively.

enter image description here

, , - python. .

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array([[ 1233.87375018,  1230.07095987],
       [ 1237.63559365,  1253.90749041],
       [ 1240.87500801,  1264.43925132],
       [ 1245.30875975,  1274.63795396],
       [ 1256.1449357 ,  1294.48254424],
       [ 1264.33600095,  1304.47893299],
       [ 1273.38192911,  1313.71468591],
       [ 1283.12411536,  1322.35942538],
       [ 1293.2559388 ,  1330.55873344],
       [ 1309.4817002 ,  1342.53074698],
       [ 1325.7074616 ,  1354.50276051],
       [ 1341.93322301,  1366.47477405],
       [ 1358.15898441,  1378.44678759],
       [ 1394.38474581,  1390.41880113]])

array([[ 1152.27115094,  1281.52899302],
       [ 1155.53345506,  1295.30515742],
       [ 1163.56506781,  1318.41642169],
       [ 1168.03497425,  1330.03181319],
       [ 1173.26135672,  1341.30559949],
       [ 1184.07110925,  1356.54121651],
       [ 1194.88086178,  1371.77683353],
       [ 1202.58908737,  1381.41765447],
       [ 1210.72465255,  1390.65097106],
       [ 1227.81309742,  1403.2904646 ],
       [ 1244.90154229,  1415.92995815],
       [ 1261.98998716,  1428.56945169],
       [ 1275.89219696,  1438.21626352],
       [ 1289.79440676,  1447.86307535],
       [ 1303.69661656,  1457.50988719],
       [ 1323.80994319,  1470.41028655],
       [ 1343.92326983,  1488.31068591],
       [ 1354.31738934,  1499.33260989],
       [ 1374.48879779,  1516.93734053],
       [ 1394.66020624,  1534.54207116]])

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+4
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import numpy as np

# Find the range of x values in a1
min_a1_x, max_a1_x = min(a1[:,0]), max(a1[:,0])
# Create an evenly spaced array that ranges from the minimum to the maximum
# I used 100 elements, but you can use more or fewer. 
# This will be used as your new x coordinates
new_a1_x = np.linspace(min_a1_x, max_a1_x, 100)
# Fit a 3rd degree polynomial to your data
a1_coefs = np.polyfit(a1[:,0],a1[:,1], 3)
# Get your new y coordinates from the coefficients of the above polynomial
new_a1_y = np.polyval(a1_coefs, new_a1_x)

# Repeat for array 2:
min_a2_x, max_a2_x = min(a2[:,0]), max(a2[:,0])
new_a2_x = np.linspace(min_a2_x, max_a2_x, 100)
a2_coefs = np.polyfit(a2[:,0],a2[:,1], 3)
new_a2_y = np.polyval(a2_coefs, new_a2_x)

, :

Fitted Arrays

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( 100, , ). x y . new_a1_x, new_a1_y, new_a2_x new_a2_y.

: x y

x y . np.mean:

midx = [np.mean([new_a1_x[i], new_a2_x[i]]) for i in range(100)]
midy = [np.mean([new_a1_y[i], new_a2_y[i]]) for i in range(100)]

midx midy 2 . ( ) Midpoint:

plt.plot(a1[:,0], a1[:,1],c='black')
plt.plot(a2[:,0], a2[:,1],c='black')
plt.plot(midx, midy, '--', c='black')
plt.show()

voilà:

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import matplotlib.pyplot as plt
import numpy as np

def interpolate(a1, a2, poly_deg=3, n_points=100, plot=True):

    min_a1_x, max_a1_x = min(a1[:,0]), max(a1[:,0])
    new_a1_x = np.linspace(min_a1_x, max_a1_x, n_points)
    a1_coefs = np.polyfit(a1[:,0],a1[:,1], poly_deg)
    new_a1_y = np.polyval(a1_coefs, new_a1_x)

    min_a2_x, max_a2_x = min(a2[:,0]), max(a2[:,0])
    new_a2_x = np.linspace(min_a2_x, max_a2_x, n_points)
    a2_coefs = np.polyfit(a2[:,0],a2[:,1], poly_deg)
    new_a2_y = np.polyval(a2_coefs, new_a2_x)

    midx = [np.mean([new_a1_x[i], new_a2_x[i]]) for i in range(n_points)]
    midy = [np.mean([new_a1_y[i], new_a2_y[i]]) for i in range(n_points)]

    if plot:
        plt.plot(a1[:,0], a1[:,1],c='black')
        plt.plot(a2[:,0], a2[:,1],c='black')
        plt.plot(midx, midy, '--', c='black')
        plt.show()

    return np.array([[x, y] for x, y in zip(midx, midy)])
+5

Source: https://habr.com/ru/post/1694259/


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