TensorFlow如何用边值填充张量



如何使用边值填充张量(尺寸为 WxHxC(?

例如:

[1, 2, 3]
[4, 5, 6]
[7, 8, 9]

成为:

[1, 1, 2, 3, 3]
[1, 1, 2, 3, 3]
[4, 4, 5, 6, 6]
[7, 7, 8, 9, 9]
[7, 7, 8, 9, 9]

使用tf.pad()和模式"对称" - 它将反映边缘上的值,但如果您只执行 1 个深度填充,则等效于重复边缘值。如果你需要更多的填充,你必须重复这个操作,但你可以成倍地进行(先 1,然后是 2,然后是 4,依此类推(。此代码(经过测试(:

import tensorflow as tf
a = tf.reshape( tf.constant( range( 1, 10 ) ), ( 3, 3 ) )
b = tf.pad( a, [ [ 1, 1 ], [ 1, 1 ] ], "SYMMETRIC" )
with tf.Session() as sess:
print( sess.run( b ) )

输出:

[[1 1 2 3 3] [1 1 2 3 3] [4 4 5 6 6] [7 7 8 9 9] [7 7 8 9 9]]



如愿以偿。

作为补充,如果您想使用像opencv这样的复制模式填充图像,以下可以做到,dst_image是要填充的图像。pad_h_up,pad_h_down,pad_w_left,pad_w_right,是四个论点:

def pad_replica(image_pad, up,down, left, right):
paddings_up = tf.constant([[1, 0],[0,0],[0,0]])
paddings_down = tf.constant([[0, 1],[0,0],[0,0]])
paddings_left = tf.constant([[0, 0],[1,0],[0,0]])
paddings_right = tf.constant([[0, 0],[0, 1],[0 ,0]])
i = tf.constant(0)
c = lambda i,pad_len,pad_mode, image: tf.less(i, pad_len)
def body(i,pad_len,pad_mode,image):
i = i+1
image = tf.pad(image, pad_mode,"SYMMETRIC")
return [i, pad_len,pad_mode, image]
[_, _, _, image_pad_up] = tf.while_loop(c, body, 
[i, up, paddings_up, image_pad])
i = tf.constant(0)
[_, _, _, image_pad_down] = tf.while_loop(c, body, [i, down,paddings_down, image_pad_up])
i = tf.constant(0)
[_, _, _, image_pad_left] = tf.while_loop(c, body, [i, left, paddings_left, image_pad_down])
i = tf.constant(0)
[_, _, _, image_pad_right] = tf.while_loop(c, body, [i, right,paddings_right, image_pad_left])
i = tf.constant(0)
return image_pad_right
dst_image.set_shape([None, None, None])
dst_image = pad_replica(dst_image,
tf.cast(pad_h_up, tf.int32),
tf.cast(pad_h_down,tf.int32),
tf.cast(pad_w_left, tf.int32),
tf.cast(pad_w_right,tf.int32)
)

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