如何将矢量重塑为 TensorFlow 的过滤器？
How to reshape a vector to TensorFlow's filters?
我想将一些由另一个网络训练的权重传输到 TensorFlow，权重存储在这样的单个向量中：
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]
通过使用 numpy，我可以将它重塑为两个 3 x 3 过滤器，如下所示：
1 2 3 9 10 11
3 4 5 12 13 14
6 7 8 15 16 17
因此，我的过滤器的形状是 (1,2,3,3)。然而，在 TensorFlow 中，过滤器的形状是 (3,3,2,1):
tf_weights = tf.Variable(tf.random_normal([3,3,2,1]))
将tf_weights重塑为预期的形状后，权值变得一团糟，得不到预期的卷积结果
具体来说，当图像或滤镜的形状为[number,channel,size,size]时，我写了一个卷积函数，它给出了正确答案，但它太慢了：
def convol(images,weights,biases,stride):
"""
Args:
images:input images or features, 4-D tensor
weights:weights, 4-D tensor
biases:biases, 1-D tensor
stride:stride, a float number
Returns:
conv_feature: convolved feature map
"""
image_num = images.shape[0] #the number of input images or feature maps
channel = images.shape[1] #channels of an image,images's shape should be like [n,c,h,w]
weight_num = weights.shape[0] #number of weights, weights' shape should be like [n,c,size,size]
ksize = weights.shape[2]
h = images.shape[2]
w = images.shape[3]
out_h = (h+np.floor(ksize/2)*2-ksize)/2+1
out_w = out_h
conv_features = np.zeros([image_num,weight_num,out_h,out_w])
for i in range(image_num):
image = images[i,...,...,...]
for j in range(weight_num):
sum_convol_feature = np.zeros([out_h,out_w])
for c in range(channel):
#extract a single channel image
channel_image = image[c,...,...]
#pad the image
padded_image = im_pad(channel_image,ksize/2)
#transform this image to a vector
im_col = im2col(padded_image,ksize,stride)
weight = weights[j,c,...,...]
weight_col = np.reshape(weight,[-1])
mul = np.dot(im_col,weight_col)
convol_feature = np.reshape(mul,[out_h,out_w])
sum_convol_feature = sum_convol_feature + convol_feature
conv_features[i,j,...,...] = sum_convol_feature + biases[j]
return conv_features
相反，通过像这样使用 tensorflow 的 conv2d：
img = np.zeros([1,3,224,224])
img = img - 1
img = np.rollaxis(img, 1, 4)
weight_array = googleNet.layers[1].weights
weight_array = np.reshape(weight_array,[64,3,7,7])
biases_array = googleNet.layers[1].biases
tf_weight = tf.Variable(weight_array)
tf_img = tf.Variable(img)
tf_img = tf.cast(tf_img,tf.float32)
tf_biases = tf.Variable(biases_array)
conv_feature = tf.nn.bias_add(tf.nn.conv2d(tf_img,tf_weight,strides=[1,2,2,1],padding='SAME'),tf_biases)
sess = tf.Session()
sess.run(tf.initialize_all_variables())
feautre = sess.run(conv_feature)
我得到的feature map是错误的
不要使用 np.reshape。它可能 mess up the order of your values.
改用np.rollaxis：
>>> a = np.array([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18])
>>> a = a.reshape((1,2,3,3))
>>> a
array([[[[ 1, 2, 3],
[ 4, 5, 6],
[ 7, 8, 9]],
[[10, 11, 12],
[13, 14, 15],
[16, 17, 18]]]])
>>> b = np.rollaxis(a, 1, 4)
>>> b.shape
(1, 3, 3, 2)
>>> b = np.rollaxis(b, 0, 4)
>>> b.shape
(3, 3, 2, 1)
请注意，尺寸为 3 的两个轴的顺序没有改变。如果我要标记它们，两个 rollaxis 操作导致形状更改为 (1, 2, 31, 32) -> (1, 31, 32, 2) -> (31 , 32, 2, 1).您的最终数组如下所示：
>>> b
array([[[[ 1],
[10]],
[[ 2],
[11]],
[[ 3],
[12]]],
[[[ 4],
[13]],
[[ 5],
[14]],
[[ 6],
[15]]],
[[[ 7],
[16]],
[[ 8],
[17]],
[[ 9],
[18]]]])
样本张量操作
我不知道这是否有帮助。考虑 Reshape、Gather、Dynamic_partition 和 Split 操作，并根据您的需要进行调整。
下面是这些操作的说明，可以根据您的情况进行调整。我从我的 git 仓库中复制了这个。我相信，如果您 运行 ipython 中的这个示例，您可以弄清楚您真正想要的是什么，并获得更好的洞察力。
重塑、聚集、Dynamic_partition和分裂
收集操作 ( tf.gather( ) )
生成数组并测试聚集操作。请注意这种快速原型制作方法：
- 我们在 Numpy 中生成一个数组，并在其上测试张量流的操作。
用途：根据索引从params中收集切片。
indices 必须是任意维度（通常为 0-D 或 1-D）的整数张量。这最好用一个例子来说明：
array = np.array([[1,2,3],[4,9,6],[2,3,4],[7,8,0]])
array.shape
(4, 3)
In [27]:
gather_output0 = tf.gather(array,1)
gather_output01 = tf.gather(array,2)
gather_output02 = tf.gather(array,3)
gather_output11 = tf.gather(array,[1,2])
gather_output12 = tf.gather(array,[1,3])
gather_output13 = tf.gather(array,[3,2])
gather_output = tf.gather(array,[1,0,2])
gather_output1 = tf.gather(array,[1,1,2])
gather_output2 = tf.gather(array,[1,2,1])
In [28]:
with tf.Session() as sess:
print (gather_output0.eval());print("\n")
print (gather_output01.eval());print("\n")
print (gather_output02.eval());print("\n")
print (gather_output11.eval());print("\n")
print (gather_output12.eval());print("\n")
print (gather_output13.eval());print("\n")
print (gather_output.eval());print("\n")
print (gather_output1.eval());print("\n")
print (gather_output2.eval());print("\n")
#print (gather_output2.eval());print("\n")
[4 9 6]
[2 3 4]
[7 8 0]
[[4 9 6]
[2 3 4]]
[[4 9 6]
[7 8 0]]
[[7 8 0]
[2 3 4]]
[[4 9 6]
[1 2 3]
[2 3 4]]
[[4 9 6]
[4 9 6]
[2 3 4]]
[[4 9 6]
[2 3 4]
[4 9 6]]
看看这个简单的例子：
- 初始化简单数组
测试收集操作
在[11]中：
array_simple = np.array([1,2,3])
In [15]:
print "shape of simple array is: ", array_simple.shape
shape of simple array is: (3,)
In [57]:
gather1 = tf.gather(array1,[0])
gather01 = tf.gather(array1,[1])
gather02 = tf.gather(array1,[2])
gather2 = tf.gather(array1,[1,2])
gather3 = tf.gather(array1,[0,1])
with tf.Session() as sess:
print (gather1.eval());print("\n")
print (gather01.eval());print("\n")
print (gather02.eval());print("\n")
print (gather2.eval());print("\n")
print (gather3.eval());print("\n")
[1]
[2]
[3]
[2 3]
[1 2]
tf.reshape( )
Note:
* Use the same array that was initiated
* Do reshape using tf.reshape( )
In [64]:
array.shape # Confirm array shape
Out[64]:
(4, 3)
In [74]:
print ("This is the array\n" ,array) # see the output and compare with the initial array,
This is the array
[[1 2 3]
[4 9 6]
[2 3 4]
[7 8 0]]
In [84]:
reshape_ops= tf.reshape(array,[-1,4]) # Note the parameters in reshpe
reshape_ops1= tf.reshape(array,[-1,3]) # Note the parameters in reshpe
reshape_ops2= tf.reshape(array,[-1,6]) # Note the parameters in reshpe
reshape_ops_back1= tf.reshape(array,[6,-1]) # Note the parameters in reshpe
reshape_ops_back2= tf.reshape(array,[3,-1]) # Note the parameters in reshpe
reshape_ops_back3= tf.reshape(array,[4,-1]) # Note the parameters in reshpe
In [86]:
with tf.Session() as sess:
print(reshape_ops.eval());print("\n")
print(reshape_ops1.eval());print("\n")
print(reshape_ops2.eval());print("\n")
print ("Output when we reverse the parameters:");print("\n")
print(reshape_ops_back1.eval());print("\n")
print(reshape_ops_back2.eval());print("\n")
print(reshape_ops_back3.eval());print("\n")
[[1 2 3 4]
[9 6 2 3]
[4 7 8 0]]
[[1 2 3]
[4 9 6]
[2 3 4]
[7 8 0]]
[[1 2 3 4 9 6]
[2 3 4 7 8 0]]
Output when we reverse the parameters:
[[1 2]
[3 4]
[9 6]
[2 3]
[4 7]
[8 0]]
[[1 2 3 4]
[9 6 2 3]
[4 7 8 0]]
[[1 2 3]
[4 9 6]
[2 3 4]
[7 8 0]]
注意：输入大小和输出大小必须相同。 ---否则会出错。检查这一点的简单方法是确保可以通过简单的乘法将输入分区到重塑参数中。
Dynamic_cell_partitions
This is declared as :
tf.dynamic_partition (array, partitions, num_partitions, name=None)
Note:
* we decalare number_partitions --- number of partitions
* Use our array initialised earlier
* We declare the partition as [0 1 0 1] . This signifies the partitions we want 0's fall to one partition and 1 the other partitions given that we have two num_partitions=2.
* The output is a list
In [96]:
print ("This is the array\n" ,array) # This is output array
This is the array
[[1 2 3]
[4 9 6]
[2 3 4]
[7 8 0]]
We show how to make two and three partitions below
In [123]:
num_partitions = 2
num_partitions1 = 3
partitions = [0, 0, 1, 1]
partitions1 = [0 ,1 ,1, 2 ]
In [119]:
dynamic_ops =tf.dynamic_partition(array, partitions, num_partitions, name=None) # 2 partitions
dynamic_ops1 =tf.dynamic_partition(array, partitions1, num_partitions1, name=None) # 3 partitions
In [125]:
with tf.Session() as sess:
run = sess.run(dynamic_ops)
run1 = sess.run(dynamic_ops1)
print("Output for 2 partitions: ")
print (run[0]);print("\n")
print(run[1]) ;print("\n")# Compare result with initial array. Out is list
print("Output for three partitions: ")
print (run1[0]);print("\n")
print (run1[1]);print("\n")
print (run1[2]);print("\n")
Output for 2 partitions:
[[1 2 3]
[4 9 6]]
[[2 3 4]
[7 8 0]]
Output for three partitions:
[[1 2 3]]
[[4 9 6]
[2 3 4]]
[[7 8 0]]
tf.split( )
确保您使用的是最新的 tensorflow 版本。否则在旧版本中，此实现将给出错误
这在文档中指定如下：
tf.split(值, num_or_size_splits, 轴=0, num=None, 名称='split').
它将张量拆分为子张量。这最好用一个例子来说明：
* we define (5,30) aray in numpy
* we split the array along axis 1
* We specify the number of splits as 1-Dimen Tensor along axis 1. So we have 3 splits.
Specify an array
Create a (5 by 30) numpy array. The syntax using numpy is shown below
In [2]:
ArrayBeforeSplitting = np.arange(150).reshape(5,30)
print ("Array shape without split operation is : " ,ArrayBeforeSplitting.shape)
('Array shape without split operation is : ', (5, 30))
specify number of splits
In [3]:
split_1D = tf.Variable([8,13,9])
print("specify number of partions using 1-Dimen Variable:" , tf.shape(split_1D))
('specify number of partions using 1-Dimen Variable:', <tf.Tensor 'Shape:0' shape=(1,) dtype=int32>)
Use tf.split
Make 3 splits aong y axis so that we have (5,8) ,(5,13),(5,9) splits. The axis 1 add up to give 30-- we can see axis 1 has 30 elements so the partition along that axis should add up to 30 otherwise it gives error.
In [6]:
split1,split2,split3 = tf.split(ArrayBeforeSplitting,split_1D,1)
# we have 3 splits along axis 1 specified spcifically
# by the split_1D . That is split axis 1D (with 30 elements) into partions with 8 ,13, and 9 elements while the x axis
#remains constant
In [7]:
#INitialise global variables. because split_ID is a variable and needs to be initialised before being
#used in a computational graph
init_op = tf.global_variables_initializer()
In [16]:
with tf.Session() as sess:
sess.run(init_op) # run variable initialisation.
result=split1.eval();print("\n")
print(result)
print("the shape of the first split operation is : ",result.shape)
result2=split2.eval();print("\n")
print(result2)
print("the shape of the second split operation is : ",result2.shape)
result3=split3.eval();print("\n")
print(result3)
print("the shape of the third split operation is : ",result3.shape)
[[ 0 1 2 3 4 5 6 7]
[ 30 31 32 33 34 35 36 37]
[ 60 61 62 63 64 65 66 67]
[ 90 91 92 93 94 95 96 97]
[120 121 122 123 124 125 126 127]]
('the shape of the first split operation is : ', (5, 8))
[[ 8 9 10 11 12 13 14 15 16 17 18 19 20]
[ 38 39 40 41 42 43 44 45 46 47 48 49 50]
[ 68 69 70 71 72 73 74 75 76 77 78 79 80]
[ 98 99 100 101 102 103 104 105 106 107 108 109 110]
[128 129 130 131 132 133 134 135 136 137 138 139 140]]
('the shape of the second split operation is : ', (5, 13))
希望对您有所帮助！
我想将一些由另一个网络训练的权重传输到 TensorFlow，权重存储在这样的单个向量中：
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]
通过使用 numpy，我可以将它重塑为两个 3 x 3 过滤器，如下所示：
1 2 3 9 10 11
3 4 5 12 13 14
6 7 8 15 16 17
因此，我的过滤器的形状是 (1,2,3,3)。然而，在 TensorFlow 中，过滤器的形状是 (3,3,2,1):
tf_weights = tf.Variable(tf.random_normal([3,3,2,1]))
将tf_weights重塑为预期的形状后，权值变得一团糟，得不到预期的卷积结果
具体来说，当图像或滤镜的形状为[number,channel,size,size]时，我写了一个卷积函数，它给出了正确答案，但它太慢了：
def convol(images,weights,biases,stride):
"""
Args:
images:input images or features, 4-D tensor
weights:weights, 4-D tensor
biases:biases, 1-D tensor
stride:stride, a float number
Returns:
conv_feature: convolved feature map
"""
image_num = images.shape[0] #the number of input images or feature maps
channel = images.shape[1] #channels of an image,images's shape should be like [n,c,h,w]
weight_num = weights.shape[0] #number of weights, weights' shape should be like [n,c,size,size]
ksize = weights.shape[2]
h = images.shape[2]
w = images.shape[3]
out_h = (h+np.floor(ksize/2)*2-ksize)/2+1
out_w = out_h
conv_features = np.zeros([image_num,weight_num,out_h,out_w])
for i in range(image_num):
image = images[i,...,...,...]
for j in range(weight_num):
sum_convol_feature = np.zeros([out_h,out_w])
for c in range(channel):
#extract a single channel image
channel_image = image[c,...,...]
#pad the image
padded_image = im_pad(channel_image,ksize/2)
#transform this image to a vector
im_col = im2col(padded_image,ksize,stride)
weight = weights[j,c,...,...]
weight_col = np.reshape(weight,[-1])
mul = np.dot(im_col,weight_col)
convol_feature = np.reshape(mul,[out_h,out_w])
sum_convol_feature = sum_convol_feature + convol_feature
conv_features[i,j,...,...] = sum_convol_feature + biases[j]
return conv_features
相反，通过像这样使用 tensorflow 的 conv2d：
img = np.zeros([1,3,224,224])
img = img - 1
img = np.rollaxis(img, 1, 4)
weight_array = googleNet.layers[1].weights
weight_array = np.reshape(weight_array,[64,3,7,7])
biases_array = googleNet.layers[1].biases
tf_weight = tf.Variable(weight_array)
tf_img = tf.Variable(img)
tf_img = tf.cast(tf_img,tf.float32)
tf_biases = tf.Variable(biases_array)
conv_feature = tf.nn.bias_add(tf.nn.conv2d(tf_img,tf_weight,strides=[1,2,2,1],padding='SAME'),tf_biases)
sess = tf.Session()
sess.run(tf.initialize_all_variables())
feautre = sess.run(conv_feature)
我得到的feature map是错误的
不要使用 np.reshape。它可能 mess up the order of your values.
改用np.rollaxis：
>>> a = np.array([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18])
>>> a = a.reshape((1,2,3,3))
>>> a
array([[[[ 1, 2, 3],
[ 4, 5, 6],
[ 7, 8, 9]],
[[10, 11, 12],
[13, 14, 15],
[16, 17, 18]]]])
>>> b = np.rollaxis(a, 1, 4)
>>> b.shape
(1, 3, 3, 2)
>>> b = np.rollaxis(b, 0, 4)
>>> b.shape
(3, 3, 2, 1)
请注意，尺寸为 3 的两个轴的顺序没有改变。如果我要标记它们，两个 rollaxis 操作导致形状更改为 (1, 2, 31, 32) -> (1, 31, 32, 2) -> (31 , 32, 2, 1).您的最终数组如下所示：
>>> b
array([[[[ 1],
[10]],
[[ 2],
[11]],
[[ 3],
[12]]],
[[[ 4],
[13]],
[[ 5],
[14]],
[[ 6],
[15]]],
[[[ 7],
[16]],
[[ 8],
[17]],
[[ 9],
[18]]]])
样本张量操作
我不知道这是否有帮助。考虑 Reshape、Gather、Dynamic_partition 和 Split 操作，并根据您的需要进行调整。 下面是这些操作的说明，可以根据您的情况进行调整。我从我的 git 仓库中复制了这个。我相信，如果您 运行 ipython 中的这个示例，您可以弄清楚您真正想要的是什么，并获得更好的洞察力。
重塑、聚集、Dynamic_partition和分裂
收集操作 ( tf.gather( ) )
生成数组并测试聚集操作。请注意这种快速原型制作方法：
- 我们在 Numpy 中生成一个数组，并在其上测试张量流的操作。
用途：根据索引从params中收集切片。
indices 必须是任意维度（通常为 0-D 或 1-D）的整数张量。这最好用一个例子来说明：
array = np.array([[1,2,3],[4,9,6],[2,3,4],[7,8,0]])
array.shape
(4, 3)
In [27]:
gather_output0 = tf.gather(array,1)
gather_output01 = tf.gather(array,2)
gather_output02 = tf.gather(array,3)
gather_output11 = tf.gather(array,[1,2])
gather_output12 = tf.gather(array,[1,3])
gather_output13 = tf.gather(array,[3,2])
gather_output = tf.gather(array,[1,0,2])
gather_output1 = tf.gather(array,[1,1,2])
gather_output2 = tf.gather(array,[1,2,1])
In [28]:
with tf.Session() as sess:
print (gather_output0.eval());print("\n")
print (gather_output01.eval());print("\n")
print (gather_output02.eval());print("\n")
print (gather_output11.eval());print("\n")
print (gather_output12.eval());print("\n")
print (gather_output13.eval());print("\n")
print (gather_output.eval());print("\n")
print (gather_output1.eval());print("\n")
print (gather_output2.eval());print("\n")
#print (gather_output2.eval());print("\n")
[4 9 6]
[2 3 4]
[7 8 0]
[[4 9 6]
[2 3 4]]
[[4 9 6]
[7 8 0]]
[[7 8 0]
[2 3 4]]
[[4 9 6]
[1 2 3]
[2 3 4]]
[[4 9 6]
[4 9 6]
[2 3 4]]
[[4 9 6]
[2 3 4]
[4 9 6]]
看看这个简单的例子：
- 初始化简单数组
测试收集操作
在[11]中：
array_simple = np.array([1,2,3]) In [15]: print "shape of simple array is: ", array_simple.shape shape of simple array is: (3,) In [57]: gather1 = tf.gather(array1,[0]) gather01 = tf.gather(array1,[1]) gather02 = tf.gather(array1,[2]) gather2 = tf.gather(array1,[1,2]) gather3 = tf.gather(array1,[0,1]) with tf.Session() as sess: print (gather1.eval());print("\n") print (gather01.eval());print("\n") print (gather02.eval());print("\n") print (gather2.eval());print("\n") print (gather3.eval());print("\n") [1] [2] [3] [2 3] [1 2] tf.reshape( ) Note: * Use the same array that was initiated * Do reshape using tf.reshape( ) In [64]: array.shape # Confirm array shape Out[64]: (4, 3) In [74]: print ("This is the array\n" ,array) # see the output and compare with the initial array, This is the array [[1 2 3] [4 9 6] [2 3 4] [7 8 0]] In [84]: reshape_ops= tf.reshape(array,[-1,4]) # Note the parameters in reshpe reshape_ops1= tf.reshape(array,[-1,3]) # Note the parameters in reshpe reshape_ops2= tf.reshape(array,[-1,6]) # Note the parameters in reshpe reshape_ops_back1= tf.reshape(array,[6,-1]) # Note the parameters in reshpe reshape_ops_back2= tf.reshape(array,[3,-1]) # Note the parameters in reshpe reshape_ops_back3= tf.reshape(array,[4,-1]) # Note the parameters in reshpe In [86]: with tf.Session() as sess: print(reshape_ops.eval());print("\n") print(reshape_ops1.eval());print("\n") print(reshape_ops2.eval());print("\n") print ("Output when we reverse the parameters:");print("\n") print(reshape_ops_back1.eval());print("\n") print(reshape_ops_back2.eval());print("\n") print(reshape_ops_back3.eval());print("\n") [[1 2 3 4] [9 6 2 3] [4 7 8 0]] [[1 2 3] [4 9 6] [2 3 4] [7 8 0]] [[1 2 3 4 9 6] [2 3 4 7 8 0]] Output when we reverse the parameters: [[1 2] [3 4] [9 6] [2 3] [4 7] [8 0]] [[1 2 3 4] [9 6 2 3] [4 7 8 0]] [[1 2 3] [4 9 6] [2 3 4] [7 8 0]]注意：输入大小和输出大小必须相同。 ---否则会出错。检查这一点的简单方法是确保可以通过简单的乘法将输入分区到重塑参数中。
Dynamic_cell_partitions
This is declared as :
tf.dynamic_partition (array, partitions, num_partitions, name=None)
Note:
* we decalare number_partitions --- number of partitions
* Use our array initialised earlier
* We declare the partition as [0 1 0 1] . This signifies the partitions we want 0's fall to one partition and 1 the other partitions given that we have two num_partitions=2.
* The output is a list
In [96]:
print ("This is the array\n" ,array) # This is output array
This is the array
[[1 2 3]
[4 9 6]
[2 3 4]
[7 8 0]]
We show how to make two and three partitions below
In [123]:
num_partitions = 2
num_partitions1 = 3
partitions = [0, 0, 1, 1]
partitions1 = [0 ,1 ,1, 2 ]
In [119]:
dynamic_ops =tf.dynamic_partition(array, partitions, num_partitions, name=None) # 2 partitions
dynamic_ops1 =tf.dynamic_partition(array, partitions1, num_partitions1, name=None) # 3 partitions
In [125]:
with tf.Session() as sess:
run = sess.run(dynamic_ops)
run1 = sess.run(dynamic_ops1)
print("Output for 2 partitions: ")
print (run[0]);print("\n")
print(run[1]) ;print("\n")# Compare result with initial array. Out is list
print("Output for three partitions: ")
print (run1[0]);print("\n")
print (run1[1]);print("\n")
print (run1[2]);print("\n")
Output for 2 partitions:
[[1 2 3]
[4 9 6]]
[[2 3 4]
[7 8 0]]
Output for three partitions:
[[1 2 3]]
[[4 9 6]
[2 3 4]]
[[7 8 0]]
tf.split( )
确保您使用的是最新的 tensorflow 版本。否则在旧版本中，此实现将给出错误
这在文档中指定如下：
tf.split(值, num_or_size_splits, 轴=0, num=None, 名称='split').
它将张量拆分为子张量。这最好用一个例子来说明：
* we define (5,30) aray in numpy
* we split the array along axis 1
* We specify the number of splits as 1-Dimen Tensor along axis 1. So we have 3 splits.
Specify an array
Create a (5 by 30) numpy array. The syntax using numpy is shown below
In [2]:
ArrayBeforeSplitting = np.arange(150).reshape(5,30)
print ("Array shape without split operation is : " ,ArrayBeforeSplitting.shape)
('Array shape without split operation is : ', (5, 30))
specify number of splits
In [3]:
split_1D = tf.Variable([8,13,9])
print("specify number of partions using 1-Dimen Variable:" , tf.shape(split_1D))
('specify number of partions using 1-Dimen Variable:', <tf.Tensor 'Shape:0' shape=(1,) dtype=int32>)
Use tf.split
Make 3 splits aong y axis so that we have (5,8) ,(5,13),(5,9) splits. The axis 1 add up to give 30-- we can see axis 1 has 30 elements so the partition along that axis should add up to 30 otherwise it gives error.
In [6]:
split1,split2,split3 = tf.split(ArrayBeforeSplitting,split_1D,1)
# we have 3 splits along axis 1 specified spcifically
# by the split_1D . That is split axis 1D (with 30 elements) into partions with 8 ,13, and 9 elements while the x axis
#remains constant
In [7]:
#INitialise global variables. because split_ID is a variable and needs to be initialised before being
#used in a computational graph
init_op = tf.global_variables_initializer()
In [16]:
with tf.Session() as sess:
sess.run(init_op) # run variable initialisation.
result=split1.eval();print("\n")
print(result)
print("the shape of the first split operation is : ",result.shape)
result2=split2.eval();print("\n")
print(result2)
print("the shape of the second split operation is : ",result2.shape)
result3=split3.eval();print("\n")
print(result3)
print("the shape of the third split operation is : ",result3.shape)
[[ 0 1 2 3 4 5 6 7]
[ 30 31 32 33 34 35 36 37]
[ 60 61 62 63 64 65 66 67]
[ 90 91 92 93 94 95 96 97]
[120 121 122 123 124 125 126 127]]
('the shape of the first split operation is : ', (5, 8))
[[ 8 9 10 11 12 13 14 15 16 17 18 19 20]
[ 38 39 40 41 42 43 44 45 46 47 48 49 50]
[ 68 69 70 71 72 73 74 75 76 77 78 79 80]
[ 98 99 100 101 102 103 104 105 106 107 108 109 110]
[128 129 130 131 132 133 134 135 136 137 138 139 140]]
('the shape of the second split operation is : ', (5, 13))
希望对您有所帮助！