理解 Tensorflow 中的 while 循环 [英] Understanding the while loop in Tensorflow

问题描述

我正在使用

我目前的实现如下:

def Rosenbrock(data_tensor):列 = tf.unstack(data_tensor)总和 = 0对于范围内的 i (1, len(columns) - 1):first_term = tf.square(tf.subtract(columns[i + 1], tf.square(columns[i])))second_term = tf.square(tf.subtract(columns[i], 1.0))求和 += tf.add(tf.multiply(100.0, first_term), second_term)返回总和

我尝试在 tf.while_loop() 中实现求和代码>;但是，我发现 API 在使用索引整数时有点不直观，该整数旨在与数据保持分离.文档 中给出的示例使用数据作为索引(反之亦然)):

i = tf.constant(0)c = lambda i: tf.less(i, 10)b = lambda i: tf.add(i, 1)r = tf.while_loop(c, b, [i])

解决方案

这可以使用 tf.while_loop() 和标准 元组 根据 文档.

def Rosenbrock(data_tensor):列 = tf.unstack(data_tensor)# 以 (index, summation) 形式跟踪元组中的循环索引和求和index_summation = (tf.constant(1), tf.constant(0.0))# 循环条件，注意循环条件是 'i <n-1'定义条件(索引，求和):返回 tf.less(index, tf.subtract(tf.shape(columns)[0], 1))# 循环体，这将返回一个相同形式的结果元组(索引，求和)定义体(索引，求和):x_i = tf.gather(列，索引)x_ip1 = tf.gather(columns, tf.add(index, 1))first_term = tf.square(tf.subtract(x_ip1, tf.square(x_i)))second_term = tf.square(tf.subtract(x_i, 1.0))summand = tf.add(tf.multiply(100.0, first_term), second_term)返回 tf.add(index, 1), tf.add(summation, summand)# 这里我们不关心索引值，只返回求和return tf.while_loop(condition, body, index_summation)[1]

需要注意的是，索引增量应该出现在循环体中，类似于标准的 while 循环.在给出的解决方案中，它是 body() 函数返回的元组中的第一项.

此外，循环条件函数必须为求和分配一个参数，尽管在此特定示例中未使用该参数.

I am using the Python API for Tensorflow. I am trying to implement the Rosenbrock function given below without the use of a Python loop:

My current implementation is as follows:

def rosenbrock(data_tensor):
    columns = tf.unstack(data_tensor)

    summation = 0
    for i in range(1, len(columns) - 1):
        first_term = tf.square(tf.subtract(columns[i + 1], tf.square(columns[i])))
        second_term = tf.square(tf.subtract(columns[i], 1.0))
        summation += tf.add(tf.multiply(100.0, first_term), second_term)

    return summation

I have tried implementing the summation in a tf.while_loop(); however, I found the API somewhat unintuitive when it comes to using an index integer that is meant to remain separate from the data. The example given in the documentation uses the data as the index (or vice-versa):

i = tf.constant(0)
c = lambda i: tf.less(i, 10)
b = lambda i: tf.add(i, 1)
r = tf.while_loop(c, b, [i])

解决方案

This can be achieved using the tf.while_loop() and standard tuples as per the second example in the documentation.

def rosenbrock(data_tensor):
    columns = tf.unstack(data_tensor)

    # Track both the loop index and summation in a tuple in the form (index, summation)
    index_summation = (tf.constant(1), tf.constant(0.0))

    # The loop condition, note the loop condition is 'i < n-1'
    def condition(index, summation):
        return tf.less(index, tf.subtract(tf.shape(columns)[0], 1))

    # The loop body, this will return a result tuple in the same form (index, summation)
    def body(index, summation):
        x_i = tf.gather(columns, index)
        x_ip1 = tf.gather(columns, tf.add(index, 1))

        first_term = tf.square(tf.subtract(x_ip1, tf.square(x_i)))
        second_term = tf.square(tf.subtract(x_i, 1.0))
        summand = tf.add(tf.multiply(100.0, first_term), second_term)

        return tf.add(index, 1), tf.add(summation, summand)

    # We do not care about the index value here, return only the summation
    return tf.while_loop(condition, body, index_summation)[1]

It is important to note that the index increment should occur in the body of the loop similar to a standard while loop. In the solution given, it is the first item in the tuple returned by the body() function.

Additionally, the loop condition function must allot a parameter for the summation although it is not used in this particular example.

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