opencv打卡27: 双三次插值（Bicubic Interpolation)--需要再次理解

1、问题

2、代码

import cv2
import numpy as np
import matplotlib.pyplot as plt


# Bi-cubic interpolation
def bc_interpolate(img, ax=1., ay=1.):
    H, W, C = img.shape
    aH = int(ay * H)
    aW = int(ax * W)

    # get position of resized image
    y = np.arange(aH).repeat(aW).reshape(aW, -1)
    x = np.tile(np.arange(aW), (aH, 1))

    # get position of original position
    y = y / ay
    x = x / ax

    ix = np.floor(x).astype(np.int)
    iy = np.floor(y).astype(np.int)

    ix = np.minimum(ix, W - 2)
    iy = np.minimum(iy, H - 2)

    # get distance
    dx2 = x - ix
    dy2 = y - iy
    dx1 = dx2 + 1
    dy1 = dy2 + 1
    dx3 = 1 - dx2
    dy3 = 1 - dy2
    dx4 = 1 + dx3
    dy4 = 1 + dy3

    dxs = [dx1, dx2, dx3, dx4]
    dys = [dy1, dy2, dy3, dy4]

    # bi-cubic weight
    def weight(t):
        a = -1.
        at = np.abs(t)
        w = np.zeros_like(t)
        ind = np.where(at <= 1)
        w[ind] = ((a + 2) * np.power(at, 3) - (a + 3) * np.power(at, 2) + 1)[ind]
        ind = np.where((at > 1) & (at <= 2))
        w[ind] = (a * np.power(at, 3) - 5 * a * np.power(at, 2) + 8 * a * at - 4 * a)[ind]
        return w

    w_sum = np.zeros((aH, aW, C), dtype=np.float32)
    out = np.zeros((aH, aW, C), dtype=np.float32)

    # interpolate
    for j in range(-1, 3):
        for i in range(-1, 3):
            ind_x = np.minimum(np.maximum(ix + i, 0), W - 1)
            ind_y = np.minimum(np.maximum(iy + j, 0), H - 1)

            wx = weight(dxs[i + 1])
            wy = weight(dys[j + 1])
            wx = np.repeat(np.expand_dims(wx, axis=-1), 3, axis=-1)
            wy = np.repeat(np.expand_dims(wy, axis=-1), 3, axis=-1)

            w_sum += wx * wy
            out += wx * wy * img[ind_y, ind_x]

    out /= w_sum
    out = np.clip(out, 0, 255)
    out = out.astype(np.uint8)

    return out


img = cv2.imread('lena.jpg').astype(np.float)

out = bc_interpolate(img, 1.5, 1.5)

cv2.imshow('result', out)
cv2.waitKey(0)
cv2.destroyAllWindows()

3、反思
后续补上。。。