可以提供下仓库地址吗？

是的，w和b这些变量由tf.Variable来表示，你可以通过如下方式打印查看这些变量 vars = tf.trainable_variables() print(vars ) #打印tf格式变量 vars_vals = sess.run(vars) for var, val in zip(vars, vars_vals): print("var: {}, value: {}".format(var.name, val)) #打印变量值

哪篇文章呢？建议给作者私信沟通，回复更快捷喔。

楼主问题解决了吗？

用全黑(全蒙面)做一个mask。 用你的ROI形状的白色填充多边形 把mask和你的图结合起来，在其他地方获得黑色的ROI。 import cv2 import numpy as np # original image # -1 loads as-is so if it will be 3 or 4 channel as the original image = cv2.imread('image.png', -1) # mask defaulting to black for 3-channel and transparent for 4-channel # (of course replace corners with yours) mask = np.zeros(image.shape, dtype=np.uint8) roi_corners = np.array([[(10,10), (300,300), (10,300)]], dtype=np.int32) # fill the ROI so it doesn't get wiped out when the mask is applied channel_count = image.shape[2] # i.e. 3 or 4 depending on your image ignore_mask_color = (255,)*channel_count cv2.fillPoly(mask, roi_corners, ignore_mask_color) # from Masterfool: use cv2.fillConvexPoly if you know it's convex # apply the mask masked_image = cv2.bitwise_and(image, mask) # save the result cv2.imwrite('image_masked.png', masked_image)

有 In [47]: np.diag(np.fliplr(array)) Out[47]: array([ 4, 8, 12, 16, 20]) 或 In [48]: np.diag(np.rot90(array)) Out[48]: array([ 4, 8, 12, 16, 20]) np.diag(np.fliplr(array))更快： In [50]: %timeit np.diag(np.fliplr(array)) 100000 loops, best of 3: 4.29 us per loop In [51]: %timeit np.diag(np.rot90(array)) 100000 loops, best of 3: 6.09 us per loop

有 In [47]: np.diag(np.fliplr(array)) Out[47]: array([ 4, 8, 12, 16, 20]) 或 In [48]: np.diag(np.rot90(array)) Out[48]: array([ 4, 8, 12, 16, 20]) np.diag(np.fliplr(array))更快： In [50]: %timeit np.diag(np.fliplr(array)) 100000 loops, best of 3: 4.29 us per loop In [51]: %timeit np.diag(np.rot90(array)) 100000 loops, best of 3: 6.09 us per loop