张量流中的基本一维卷积
python
tensorflow
5
0

好的,我想在Tensorflow中对时间序列数据进行一维卷积。根据这些 票证手册 ,显然可以使用tf.nn.conv2d支持此操作 。唯一的要求是设置strides=[1,1,1,1] 。听起来很简单!

但是,即使在非常小的测试用例中,我也无法弄清楚该如何做。我究竟做错了什么?

让我们进行设置。

import tensorflow as tf
import numpy as np
print(tf.__version__)
>>> 0.9.0

好的,现在在两个小的数组上生成基本的卷积测试。我将通过使用1的批量大小来简化该操作,并且由于时间序列是一维的,因此我的“图像高度”将为1。并且由于它是单变量时间序列,因此显然“通道”的数量也是1,所以这很简单,对吧?

g = tf.Graph()
with g.as_default():
    # data shape is "[batch, in_height, in_width, in_channels]",
    x = tf.Variable(np.array([0.0, 0.0, 0.0, 0.0, 1.0]).reshape(1,1,-1,1), name="x")
    # filter shape is "[filter_height, filter_width, in_channels, out_channels]"
    phi = tf.Variable(np.array([0.0, 0.5, 1.0]).reshape(1,-1,1,1), name="phi")
    conv = tf.nn.conv2d(
        phi,
        x,
        strides=[1, 1, 1, 1],
        padding="SAME",
        name="conv")

繁荣。错误。

ValueError: Dimensions 1 and 5 are not compatible

好的,首先,我不知道在任何维度上都应该发生这种情况,因为我已经指定要在卷积OP中填充参数。

但是很好,也许对此有限制。我一定对文档感到困惑,并且在张量的错误轴上设置了这种卷积。我将尝试所有可能的排列:

for i in range(4):
    for j in range(4):
        shape1 = [1,1,1,1]
        shape1[i] = -1
        shape2 = [1,1,1,1]
        shape2[j] = -1
        x_array = np.array([0.0, 0.0, 0.0, 0.0, 1.0]).reshape(*shape1)
        phi_array = np.array([0.0, 0.5, 1.0]).reshape(*shape2)
        try:
            g = tf.Graph()
            with g.as_default():
                x = tf.Variable(x_array, name="x")
                phi = tf.Variable(phi_array, name="phi")
                conv = tf.nn.conv2d(
                    x,
                    phi,
                    strides=[1, 1, 1, 1],
                    padding="SAME",
                    name="conv")
                init_op = tf.initialize_all_variables()
            sess = tf.Session(graph=g)
            sess.run(init_op)
            print("SUCCEEDED!", x_array.shape, phi_array.shape, conv.eval(session=sess))
            sess.close()
        except Exception as e:
            print("FAILED!", x_array.shape, phi_array.shape, type(e), e.args or e._message)

结果:

FAILED! (5, 1, 1, 1) (3, 1, 1, 1) <class 'ValueError'> ('Filter must not be larger than the input: Filter: (3, 1) Input: (1, 1)',)
FAILED! (5, 1, 1, 1) (1, 3, 1, 1) <class 'ValueError'> ('Filter must not be larger than the input: Filter: (1, 3) Input: (1, 1)',)
FAILED! (5, 1, 1, 1) (1, 1, 3, 1) <class 'ValueError'> ('Dimensions 1 and 3 are not compatible',)
FAILED! (5, 1, 1, 1) (1, 1, 1, 3) <class 'tensorflow.python.framework.errors.InvalidArgumentError'> No OpKernel was registered to support Op 'Conv2D' with these attrs
     [[Node: conv = Conv2D[T=DT_DOUBLE, data_format="NHWC", padding="SAME", strides=[1, 1, 1, 1], use_cudnn_on_gpu=true](x/read, phi/read)]]
FAILED! (1, 5, 1, 1) (3, 1, 1, 1) <class 'tensorflow.python.framework.errors.InvalidArgumentError'> No OpKernel was registered to support Op 'Conv2D' with these attrs
     [[Node: conv = Conv2D[T=DT_DOUBLE, data_format="NHWC", padding="SAME", strides=[1, 1, 1, 1], use_cudnn_on_gpu=true](x/read, phi/read)]]
FAILED! (1, 5, 1, 1) (1, 3, 1, 1) <class 'ValueError'> ('Filter must not be larger than the input: Filter: (1, 3) Input: (5, 1)',)
FAILED! (1, 5, 1, 1) (1, 1, 3, 1) <class 'ValueError'> ('Dimensions 1 and 3 are not compatible',)
FAILED! (1, 5, 1, 1) (1, 1, 1, 3) <class 'tensorflow.python.framework.errors.InvalidArgumentError'> No OpKernel was registered to support Op 'Conv2D' with these attrs
     [[Node: conv = Conv2D[T=DT_DOUBLE, data_format="NHWC", padding="SAME", strides=[1, 1, 1, 1], use_cudnn_on_gpu=true](x/read, phi/read)]]
FAILED! (1, 1, 5, 1) (3, 1, 1, 1) <class 'ValueError'> ('Filter must not be larger than the input: Filter: (3, 1) Input: (1, 5)',)
FAILED! (1, 1, 5, 1) (1, 3, 1, 1) <class 'tensorflow.python.framework.errors.InvalidArgumentError'> No OpKernel was registered to support Op 'Conv2D' with these attrs
     [[Node: conv = Conv2D[T=DT_DOUBLE, data_format="NHWC", padding="SAME", strides=[1, 1, 1, 1], use_cudnn_on_gpu=true](x/read, phi/read)]]
FAILED! (1, 1, 5, 1) (1, 1, 3, 1) <class 'ValueError'> ('Dimensions 1 and 3 are not compatible',)
FAILED! (1, 1, 5, 1) (1, 1, 1, 3) <class 'tensorflow.python.framework.errors.InvalidArgumentError'> No OpKernel was registered to support Op 'Conv2D' with these attrs
     [[Node: conv = Conv2D[T=DT_DOUBLE, data_format="NHWC", padding="SAME", strides=[1, 1, 1, 1], use_cudnn_on_gpu=true](x/read, phi/read)]]
FAILED! (1, 1, 1, 5) (3, 1, 1, 1) <class 'ValueError'> ('Dimensions 5 and 1 are not compatible',)
FAILED! (1, 1, 1, 5) (1, 3, 1, 1) <class 'ValueError'> ('Dimensions 5 and 1 are not compatible',)
FAILED! (1, 1, 1, 5) (1, 1, 3, 1) <class 'ValueError'> ('Dimensions 5 and 3 are not compatible',)
FAILED! (1, 1, 1, 5) (1, 1, 1, 3) <class 'ValueError'> ('Dimensions 5 and 1 are not compatible',)

嗯好的,看来现在有两个问题。首先,我猜ValueError是关于沿错误的轴应用过滤器的,尽管有两种形式。

但是,随后我可以沿其应用过滤器的轴也令人困惑-请注意,它实际上构造了具有输入形状(5,1,1,1,1)和过滤器形状(1,1,1,3)的图。根据文档中的AFAICT,这应该是一个过滤器,以批次为例,一个“像素”和一个“通道”,输出3个“通道”。那么,当其他人不起作用时,为什么一个人起作用呢?

无论如何,有时在构建图形时它不会失败。有时它会构造图;然后我们得到tensorflow.python.framework.errors.InvalidArgumentError 。从一些令人困惑的github票证中,我收集到这可能是由于我在CPU而不是GPU上运行的事实,反之亦然卷积运算仅针对32位浮点数而不是64位浮点数进行定义的事实。如果任何人都可以扔在哪个轴我应该对准什么上,为了与卷积内核时间序列一些轻,我会非常感激。

参考资料:
Stack Overflow
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抱歉,您的第一个代码几乎是正确的。您只是在tf.nn.conv2d反转了xphi

g = tf.Graph()
with g.as_default():
    # data shape is "[batch, in_height, in_width, in_channels]",
    x = tf.Variable(np.array([0.0, 0.0, 0.0, 0.0, 1.0]).reshape(1, 1, 5, 1), name="x")
    # filter shape is "[filter_height, filter_width, in_channels, out_channels]"
    phi = tf.Variable(np.array([0.0, 0.5, 1.0]).reshape(1, 3, 1, 1), name="phi")
    conv = tf.nn.conv2d(
        x,
        phi,
        strides=[1, 1, 1, 1],
        padding="SAME",
        name="conv")

更新:从版本r0.11开始,TensorFlow现在使用tf.nn.conv1d支持一维卷积。我以前在粘贴到此处的stackoverflow文档(现已不存在)中做了一个使用它们的指南:


一维卷积指南

考虑一个输入长度为10且尺寸为16的基本示例。批处理大小为32 。因此,我们有一个占位符,其输入形状为[batch_size, 10, 16]

batch_size = 32
x = tf.placeholder(tf.float32, [batch_size, 10, 16])

然后,我们创建一个宽度为3的过滤器,并以16通道作为输入,并输出16通道。

filter = tf.zeros([3, 16, 16])  # these should be real values, not 0

最后,我们在tf.nn.conv1d使用一个步幅和一个填充:- 步幅 :整数s 填充 :就像在2D中一样,您可以在SAMEVALID之间进行选择。 SAME将输出相同的输入长度,而VALID将不添加零填充。

在我们的示例中,跨度为2,有效填充为空白。

output = tf.nn.conv1d(x, filter, stride=2, padding="VALID")

输出形状应为[batch_size, 4, 16]
使用padding="SAME" ,我们的输出形状为[batch_size, 5, 16]

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在新版本的TF(从0.11开始)中,您具有conv1d ,因此无需使用2d卷积来进行1d卷积。这是有关如何使用conv1d的简单示例:

import tensorflow as tf
i = tf.constant([1, 0, 2, 3, 0, 1, 1], dtype=tf.float32, name='i')
k = tf.constant([2, 1, 3], dtype=tf.float32, name='k')

data   = tf.reshape(i, [1, int(i.shape[0]), 1], name='data')
kernel = tf.reshape(k, [int(k.shape[0]), 1, 1], name='kernel')

res = tf.squeeze(tf.nn.conv1d(data, kernel, stride=1, padding='VALID'))
with tf.Session() as sess:
    print sess.run(res)

要了解conv1d是如何计算的,请看各种示例

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我认为我可以满足我的需求。代码的注释/详细信息在代码上:

import numpy as np

import tensorflow as tf
from tensorflow.examples.tutorials.mnist import input_data

task_name = 'task_MNIST_flat_auto_encoder'
mnist = input_data.read_data_sets("MNIST_data/", one_hot=True)
X_train, Y_train = mnist.train.images, mnist.train.labels # N x D
X_cv, Y_cv = mnist.validation.images, mnist.validation.labels
X_test, Y_test = mnist.test.images, mnist.test.labels

# data shape is "[batch, in_height, in_width, in_channels]",
# X_train = N x D
N, D = X_train.shape
# think of it as N images with height 1 and width D.
X_train = X_train.reshape(N,1,D,1)
x = tf.placeholder(tf.float32, shape=[None,1,D,1], name='x-input')
#x = tf.Variable( X_train , name='x-input')
# filter shape is "[filter_height, filter_width, in_channels, out_channels]"
filter_size, nb_filters = 10, 12 # filter_size , number of hidden units/units
# think of it as having nb_filters number of filters, each of size filter_size
W = tf.Variable( tf.truncated_normal(shape=[1, filter_size, 1,nb_filters], stddev=0.1) )
stride_convd1 = 2 # controls the stride for 1D convolution
conv = tf.nn.conv2d(input=x, filter=W, strides=[1, 1, stride_convd1, 1], padding="SAME", name="conv")

with tf.Session() as sess:
    sess.run( tf.initialize_all_variables() )
    sess.run(fetches=conv, feed_dict={x:X_train})

感谢Olivier的帮助(有关进一步的说明,请参见他的评论中的讨论)。


手动检查:

X_train_org = np.array([[0,1,2,3]])
N, D = X_train_org.shape
X_train_1d = X_train_org.reshape(N,1,D,1)
#X_train = tf.constant( X_train_org )
# think of it as N images with height 1 and width D.
xx = tf.placeholder(tf.float32, shape=[None,1,D,1], name='xx-input')
#x = tf.Variable( X_train , name='x-input')
# filter shape is "[filter_height, filter_width, in_channels, out_channels]"
filter_size, nb_filters = 2, 2 # filter_size , number of hidden units/units
# think of it as having nb_filters number of filters, each of size filter_size
filter_w = np.array([[1,3],[2,4]]).reshape(1,filter_size,1,nb_filters)
#W = tf.Variable( tf.truncated_normal(shape=[1,filter_size,1,nb_filters], stddev=0.1) )
W = tf.Variable( tf.constant(filter_w, dtype=tf.float32) )
stride_convd1 = 2 # controls the stride for 1D convolution
conv = tf.nn.conv2d(input=xx, filter=W, strides=[1, 1, stride_convd1, 1], padding="SAME", name="conv")

#C = tf.constant( (np.array([[4,3,2,1]]).T).reshape(1,1,1,4) , dtype=tf.float32 ) #
#tf.reshape( conv , [])
#y_tf = tf.matmul(conv, C)


##
x = tf.placeholder(tf.float32, shape=[None,D], name='x-input') # N x 4
W1 = tf.Variable( tf.constant( np.array([[1,2,0,0],[3,4,0,0]]).T, dtype=tf.float32 ) ) # 2 x 4
y1 = tf.matmul(x,W1) # N x 2 = N x 4 x 4 x 2
W2 = tf.Variable( tf.constant( np.array([[0,0,1,2],[0,0,3,4]]).T, dtype=tf.float32 ))
y2 = tf.matmul(x,W2) # N x 2 = N x 4 x 4 x 2
C1 = tf.constant( np.array([[4,3]]).T, dtype=tf.float32 ) # 1 x 2
C2 = tf.constant( np.array([[2,1]]).T, dtype=tf.float32 )

p1 = tf.matmul(y1,C1)
p2 = tf.matmul(y2,C2)
y = p1 + p2
with tf.Session() as sess:
    sess.run( tf.initialize_all_variables() )
    print 'manual conv'
    print sess.run(fetches=y1, feed_dict={x:X_train_org})
    print sess.run(fetches=y2, feed_dict={x:X_train_org})
    #print sess.run(fetches=y, feed_dict={x:X_train_org})
    print 'tf conv'
    print sess.run(fetches=conv, feed_dict={xx:X_train_1d})
    #print sess.run(fetches=y_tf, feed_dict={xx:X_train_1d})

输出:

manual conv
[[ 2.  4.]]
[[  8.  18.]]
tf conv
[[[[  2.   4.]
   [  8.  18.]]]]
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