tf.nn.l2_loss和 tf.nn.l2_normalize

一，L2正则化：

tf.nn.l2_loss形如 1/2Σw2 1 / 2 Σ w 2 ,一般用于优化的目标函数中的正则项，防止参数太多复杂容易过拟合。

tf.nn.l2_loss(var)
Computes half the L2 norm of a tensor without the `sqrt`:

      output = sum(t ** 2) / 2

不过在这里又有疑问？什么是L1，L2范数和L1，L2正则化项，都听过但是名字感觉非常相似，下面来讨论下：
L1范数 – (Lasso Regression)

L1范数表示向量中每个元素绝对值的和：

‖x⃗ ‖1=(|x1|+⋯+|xm|))(87) (87) ‖ x → ‖ 1 = ( | x 1 | + ⋯ + | x m | ) )

L2范数 – (Ridge Regression)

L2范数即欧氏距离：

‖x⃗ ‖2=(|x1|2+⋯+|xm|2)12=x⃗ Tx⃗ ‾‾‾‾√(88) (88) ‖ x → ‖ 2 = ( | x 1 | 2 + ⋯ + | x m | 2 ) 1 2 = x → T x →

• L1正则化是指权值向量w中各个元素的绝对值之和，通常表示为||w||1
• L2正则化是指权值向量w中各个元素的平方和然后再求平方根（可以看到Ridge回归的L2正则化项有平方符号），通常表示为||w||2
  也就是说Lx范数应用于优化的目标函数就叫做Lx正则化。

二，tf.nn.l2_normalize

这个是用L2范数的结果来对输入进行scale，输出的维度不变。
源码如下：

def l2_normalize(x, dim, epsilon=1e-12, name=None):
  """Normalizes along dimension `dim` using an L2 norm.

  For a 1-D tensor with `dim = 0`, computes

      output = x / sqrt(max(sum(x**2), epsilon))

  For `x` with more dimensions, independently normalizes each 1-D slice along
  dimension `dim`.

  Args:
    x: A `Tensor`.
    dim: Dimension along which to normalize.  A scalar or a vector of
      integers.
    epsilon: A lower bound value for the norm. Will use `sqrt(epsilon)` as the
      divisor if `norm < sqrt(epsilon)`.
    name: A name for this operation (optional).

  Returns:
    A `Tensor` with the same shape as `x`.
  """
  with ops.name_scope(name, "l2_normalize", [x]) as name:
    x = ops.convert_to_tensor(x, name="x")
    square_sum = math_ops.reduce_sum(math_ops.square(x), dim, keep_dims=True)
    x_inv_norm = math_ops.rsqrt(math_ops.maximum(square_sum, epsilon))
    return math_ops.multiply(x, x_inv_norm, name=name)

l2_normalize(x, dim, epsilon=1e-12, name=None)
主要参数就是dim，从上面代码看到，dim的计算是按照reduce_sum的方式来的，reduce可以理解为减少维度的意思，通过实验来验证dim的方式。

x = tf.nn.l2_normalize(x_init, dim=[0, 1, 2])
"Normalizes along dimension `dim` using an L2 norm