[tensorflow] softmax函数解析

def softmax(logits, axis=None, name=None, dim=None):
  """Computes softmax activations.

  This function performs the equivalent of

      softmax = tf.exp(logits) / tf.reduce_sum(tf.exp(logits), axis)

  Args:
    logits: A non-empty `Tensor`. Must be one of the following types: `half`,
      `float32`, `float64`.
    axis: The dimension softmax would be performed on. The default is -1 which
      indicates the last dimension.
    name: A name for the operation (optional).
    dim: Deprecated alias for `axis`.

  Returns:
    A `Tensor`. Has the same type and shape as `logits`.

  Raises:
    InvalidArgumentError: if `logits` is empty or `axis` is beyond the last
      dimension of `logits`.
  """
  axis = deprecation.deprecated_argument_lookup("axis", axis, "dim", dim)
  if axis is None:
    axis = -1
  return _softmax(logits, gen_nn_ops.softmax, axis, name)

softmax函数的返回结果和输入的tensor有相同的shape，既然没有改变tensor的形状，那么softmax究竟对tensor做了什么？

答案就是softmax会以某一个轴的下标为索引，对这一轴上其他维度的值进行激活 + 归一化处理。

一般来说，这个索引轴都是表示类别的那个维度（tf.nn.softmax中默认为axis=-1,也就是最后一个维度）

举例：

def softmax(X, theta = 1.0, axis = None):
    """
    Compute the softmax of each element along an axis of X.

    Parameters
    ----------
    X: ND-Array. Probably should be floats.
    theta (optional): float parameter, used as a multiplier
        prior to exponentiation. Default = 1.0
    axis (optional): axis to compute values along. Default is the
        first non-singleton axis.

    Returns an array the same size as X. The result will sum to 1
    along the specified axis.
    """

    # make X at least 2d
    y = np.atleast_2d(X)

    # find axis
    if axis is None:
        axis = next(j[0] for j in enumerate(y.shape) if j[1] > 1)

    # multiply y against the theta parameter,
    y = y * float(theta)

    # subtract the max for numerical stability
    y = y - np.expand_dims(np.max(y, axis = axis), axis)

    # exponentiate y
    y = np.exp(y)

    # take the sum along the specified axis
    ax_sum = np.expand_dims(np.sum(y, axis = axis), axis)

    # finally: divide elementwise
    p = y / ax_sum

    # flatten if X was 1D
    if len(X.shape) == 1: p = p.flatten()

    return p
c = np.random.randn(2,3)
print(c)
# 假设第0维是类别，一共有里两种类别
cc = softmax(c,axis=0)
# 假设最后一维是类别，一共有3种类别
ccc = softmax(c,axis=-1)
print(cc)
print(ccc)

结果：

c:
[[-1.30022268  0.59127472  1.21384177]
 [ 0.1981082  -0.83686108 -1.54785864]]
cc:
[[0.1826746  0.80661068 0.94057075]
 [0.8173254  0.19338932 0.05942925]]
ccc:
[[0.0500392  0.33172426 0.61823654]
 [0.65371718 0.23222472 0.1140581 ]]

可以看到，对axis=0的轴做softmax时，输出结果在axis=0轴上和为1(eg: 0.1826746+0.8173254)，同理在axis=1轴上做的话结果的axis=1轴和也为1(eg: 0.0500392+0.33172426+0.61823654)。

这些值是怎么得到的呢？

以cc为例（沿着axis=0做softmax）：

$cc[0,0] = 0.1826746 = \frac{e^{cc[0,0]}}{\sum_{k=0}^{1}e^{cc[k,0]}}$

其中 $e^{cc[0,0]}$ 属于激活操作， $\frac{1}{\sum_{k=0}^{1}e^{cc[k,0]}}$ 属于归一化操作。

以ccc为例（沿着axis=1做softmax）：

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$ccc[0,0] = 0.0500392 = \frac{e^{ccc[0,0]}}{\sum_{k=0}^{2}e^{cc[0,k]}}$

知道了计算方法，现在我们再来讨论一下这些值的实际意义：

cc[0,0]实际上表示这样一种概率： P( label = 0 | value = [-1.30022268 0.1981082] = c[*,0] ) = 0.1826746

cc[1,0]实际上表示这样一种概率： P( label = 1 | value = [-1.30022268 0.1981082] = c[*,0] ) = 0.8173254

ccc[0,0]实际上表示这样一种概率： P( label = 0 | value = [-1.30022268 0.59127472 1.21384177] = c[0]) = 0.0500392

ccc[0,1]实际上表示这样一种概率： P( label = 1 | value = [-1.30022268 0.59127472 1.21384177] = c[0]) = 0.33172426

ccc[0,2]实际上表示这样一种概率： P( label = 2 | value = [-1.30022268 0.59127472 1.21384177] = c[0]) = 0.61823654

将他们扩展到更多维的情况：假设c是一个[batch_size , timesteps, categories]的三维tensor

output = tf.nn.softmax(c,axis=-1)

那么 output[1, 2, 3] 则表示 P(label =3 | value = c[1,2] )

参考：

[tensorflow] softmax函数解析

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