Question

我使用numpy创建了一个滑动窗口算法，该算法在wav音频文件上滑动并将其切片输入到我的NN中，在tensorflow中检测音频切片中的功能。一旦张量流完成它，它就会将其输出返回到numpy land，在那里我将切片重新组合成一个与原始文件的每个样本位置匹配的预测数组：

import tensorflow as tf
import numpy as np
import nn

def slide_predict(layers, X, modelPath):
    output = None

    graph = tf.Graph()
    with graph.as_default():
        input_layer_size, hidden_layer_size, num_labels = layers

        X_placeholder = tf.placeholder(tf.float32, shape=(None, input_layer_size), name='X')
        Theta1 = tf.Variable(nn.randInitializeWeights(input_layer_size, hidden_layer_size), name='Theta1')
        bias1 = tf.Variable(nn.randInitializeWeights(hidden_layer_size, 1), name='bias1')
        Theta2 = tf.Variable(nn.randInitializeWeights(hidden_layer_size, num_labels), name='Theta2')
        bias2 = tf.Variable(nn.randInitializeWeights(num_labels, 1), name='bias2')
        hypothesis = nn.forward_prop(X_placeholder, Theta1, bias1, Theta2, bias2)

        sess = tf.Session(graph=graph)
        saver = tf.train.Saver()
        init = tf.global_variables_initializer()
        sess.run(init)

        saver.restore(sess, modelPath)

        window_size = layers[0]

        pad_amount = (window_size * 2) - (X.shape[0] % window_size)
        X = np.pad(X, (pad_amount, 0), 'constant')

        for w in range(window_size):
            start = w
            end = -window_size + w
            X_shifted = X[start:end]
            X_matrix = X_shifted.reshape((-1, window_size))

            prediction = sess.run(hypothesis, feed_dict={X_placeholder: X_matrix})

            output = prediction if (output is None) else np.hstack((output, prediction))

        sess.close()

    output.shape = (X.size, -1)

    return output

不幸的是，这个算法很慢。我沿途放置了一些日志，到目前为止，最慢的部分是我实际运行张量流图的部分。这可能是由于实际的张量流量计算很慢（如果是这样，我可能只是SOL），但我想知道是否有很大一部分因为我正在反复传输大型音频文件张量流入和流出张量流。所以我的问题是：

1）像这样反复喂食占位符会明显慢于喂食一次并计算张量流内X的值吗？

2）如果是，那么在tensorflow中实现滑动窗口算法的最佳方法是进行此计算吗？

Answer 1

第一个问题是你的算法在window_size中具有二次时间复杂度，因为在每次迭代中调用np.hstack()来构建output数组，该数组同时复制当前值output和prediction成新阵列：

for w in range(window_size):
    # ...
    output = prediction if (output is None) else np.hstack((output, prediction))

不是在每次迭代中调用np.hstack()，而是在循环终止后构建prediction数组的Python列表并在其上调用np.hstack()更高效：

output_list = []
for w in range(window_size):
    # ...
    prediction = sess.run(...)
    output_list.append(prediction)
output = np.hstack(output_list)

第二个问题是，如果sess.run()调用中的计算量很小，那么向TensorFlow提供大值可能效率低，因为这些值（当前）被复制到C ++中（结果被复制出来）一个有用的策略是尝试使用tf.map_fn()构造将滑动窗口循环移动到TensorFlow图形中。例如，您可以按如下方式重构程序：

# NOTE: If you call this function often, you may want to (i) move the `np.pad()`
# into the graph as `tf.pad()`, and (ii) replace `X_t` with a placeholder.
X = np.pad(X, (pad_amount, 0), 'constant')
X_t  = tf.convert_to_tensor(X)

def window_func(w):
    start = w
    end = w - window_size
    X_matrix = tf.reshape(X_t[start:end], (-1, window_size))
    return nn.forward_prop(X_matrix, Theta1, bias1, Theta2, bias2)

output_t = tf.map_fn(window_func, tf.range(window_size))
# ...
output = sess.run(output_t)

如何在张量流中实现滑动窗口？

1 个答案: