【问题标题】:How to reshape a vector to TensorFlow's filters?如何将向量重塑为 TensorFlow 的过滤器?
【发布时间】:2016-04-15 23:25:09
【问题描述】:

我想将一些由另一个网络训练的权重转移到 TensorFlow,这些权重存储在单个向量中,如下所示:

[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]

通过使用 numpy,我可以将其重塑为两个 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)。但是在TensorFlow中,filters的shape是(3,3,2,1)

tf_weights = tf.Variable(tf.random_normal([3,3,2,1]))

将 tf_weights 重塑为预期形状后,权重变得一团糟,无法得到预期的卷积结果。

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,它给出了正确的答案,但是速度太慢了:

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)

我得到的特征图是错误的。

【问题讨论】:

  • 恐怕您的编辑使您的问题非常混乱。我不完全理解您要做什么,而且这里的变量太多。您可以尝试制作minimal reproducible example 吗?

标签: python numpy tensorflow


【解决方案1】:

张量操作示例

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,这些操作可以适应您的情况。我从我的 git repo 复制了这个。我相信,如果您在 ipython 中运行此示例,您可以弄清楚您真正想要什么并获得更好的洞察力。

重塑、聚集、动态分区和拆分

收集操作(tf.gather())

生成一个数组并测试收集操作。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:根据索引从参数中收集切片。

索引必须是任意维度的整数张量(通常为 0-D 或 1-D)。最好用一个例子来说明这一点:

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输入可以通过简单的乘法划分为重塑参数。

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,这个实现会报错

这在文档中指定如下:

tf.split(value, num_or_size_splits, axis=0, num=None, name='split')。

它将张量拆分为子张量。最好用一个例子来说明这一点:

* 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))

希望这会有所帮助!

【讨论】:

    【解决方案2】:

    不要使用np.reshape。可能是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,大小为 3 的两个轴的顺序没有改变。如果我要标记它们,两个 rollaxis 操作会导致形状更改为 (1, 2, 31, 32) -> (1, 3 1, 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]]]])
    

    【讨论】:

    • 感谢您的回答!我试过这段代码。但重塑后的阵列无法按预期工作。我认为 tensorflow 的维度顺序是如此的连贯。在 numpy 中,我们将图像重塑为像 [number, channel, height, width] 的 4-D 张量,但在 tensorflow 中就像 [number, height,width,channel]。所以在重塑之后,权重或图像变成了一个质量,我无法按预期得到正确的卷积特征
    • 好吧,根据你的评论,你可能只需要b = np.rollaxis(a, 1, 4)。这也不行吗?不幸的是,我不熟悉张量流。但这绝对是 在 numpy 中移动轴的一种方式。
    • 是的,它给了我正确的重量形状。但不知何故,它质量了权重的顺序,所以在卷积之后,我无法得到正确的答案。我已经更新了上面的问题。
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