双三次下采样，批量处理数据集

学习…
今天在想对自己的数据集做一个双三次插值下采样的图像数据预处理，可能因为我是个小白 菜鸡一枚，所以自己写了一段代码，调用的pytorch torch.nn.functional.interpolate这个库函数，奈何最邻近下采样等都能用，但是就这个双三次下采样，就是不能用（因为有点菜，所以一直没有搞出来，有点打击自信心了哈哈哈）。
看了很多博客，关于这个库，俺发现，关于双三次下采样都是一语带过，多的就是说一句（如果采用双三次可能会出现问题，但是，就是不说解决方法）。博客中提出的示例都巧妙的避开了双三次这个方法。
咱就是说，为什么呢？
害，于是down了一种原始做法。不说了，直接上可用代码！！！
链接: link借鉴了这位大佬的！！！

import cv2
import numpy as np
import math
import sys, time

# Interpolation kernel
def u(s,a):
    if (abs(s) >=0) & (abs(s) <=1):
        return (a+2)*(abs(s)**3)-(a+3)*(abs(s)**2)+1
    elif (abs(s) > 1) & (abs(s) <= 2):
        return a*(abs(s)**3)-(5*a)*(abs(s)**2)+(8*a)*abs(s)-4*a
    return 0

#Paddnig
def padding(img,H,W,C):
    zimg = np.zeros((H+4,W+4,C))
    zimg[2:H+2,2:W+2,:C] = img
    #Pad the first/last two col and row
    zimg[2:H+2,0:2,:C]=img[:,0:1,:C]
    zimg[H+2:H+4,2:W+2,:]=img[H-1:H,:,:]
    zimg[2:H+2,W+2:W+4,:]=img[:,W-1:W,:]
    zimg[0:2,2:W+2,:C]=img[0:1,:,:C]
    #Pad the missing eight points
    zimg[0:2,0:2,:C]=img[0,0,:C]
    zimg[H+2:H+4,0:2,:C]=img[H-1,0,:C]
    zimg[H+2:H+4,W+2:W+4,:C]=img[H-1,W-1,:C]
    zimg[0:2,W+2:W+4,:C]=img[0,W-1,:C]
    return zimg

def get_progressbar_str(progress):
    END = 170
    MAX_LEN = 30
    BAR_LEN = int(MAX_LEN * progress)
    return ('Progress:[' + '=' * BAR_LEN +
            ('>' if BAR_LEN < MAX_LEN else '') +
            ' ' * (MAX_LEN - BAR_LEN) +
            '] %.1f%%' % (progress * 100.))

# Bicubic operation
def bicubic(img, ratio, a):
    #Get image size
    H,W,C = img.shape

    img = padding(img,H,W,C)
    #Create new image
    dH = math.floor(H*ratio)
    dW = math.floor(W*ratio)
    dst = np.zeros((dH, dW, 3))

    h = 1/ratio

    # print('Start bicubic interpolation')
    # print('It will take a little while...')
    inc = 0
    for c in range(C):
        for j in range(dH):
            for i in range(dW):
                x, y = i * h + 2 , j * h + 2

                x1 = 1 + x - math.floor(x)
                x2 = x - math.floor(x)
                x3 = math.floor(x) + 1 - x
                x4 = math.floor(x) + 2 - x

                y1 = 1 + y - math.floor(y)
                y2 = y - math.floor(y)
                y3 = math.floor(y) + 1 - y
                y4 = math.floor(y) + 2 - y

                mat_l = np.matrix([[u(x1,a),u(x2,a),u(x3,a),u(x4,a)]])
                mat_m = np.matrix([[img[int(y-y1),int(x-x1),c],img[int(y-y2),int(x-x1),c],img[int(y+y3),int(x-x1),c],img[int(y+y4),int(x-x1),c]],
                                   [img[int(y-y1),int(x-x2),c],img[int(y-y2),int(x-x2),c],img[int(y+y3),int(x-x2),c],img[int(y+y4),int(x-x2),c]],
                                   [img[int(y-y1),int(x+x3),c],img[int(y-y2),int(x+x3),c],img[int(y+y3),int(x+x3),c],img[int(y+y4),int(x+x3),c]],
                                   [img[int(y-y1),int(x+x4),c],img[int(y-y2),int(x+x4),c],img[int(y+y3),int(x+x4),c],img[int(y+y4),int(x+x4),c]]])
                mat_r = np.matrix([[u(y1,a)],[u(y2,a)],[u(y3,a)],[u(y4,a)]])
                dst[j, i, c] = np.dot(np.dot(mat_l, mat_m),mat_r)

                # Print progress
                inc = inc + 1
                sys.stderr.write('\r\033[K' + get_progressbar_str(inc/(C*dH*dW)))
                sys.stderr.flush()
    sys.stderr.write('\n')
    sys.stderr.flush()
    return dst

# Scale factor
ratio = 0.5
# Coefficient
a = -1/2

print('************开始***********')
for i in range(0,1020):
    print(i)
    img = cv2.imread('/************************************/{}.png'.format(i))
    dst = bicubic(img, ratio, a)
    cv2.imwrite('/****************************************lr/{}.png'.format(i), dst)

仅仅是个人记录，如果有大佬知道pytorch torch.nn.functional.interpolate这个库函数关于双三次下采样怎么调用，还请指教！！！