按照我原来的想法，Triplet Loss 三元组应该是这样选择的：

(1) 前馈操作： cnn 先执行forward 操作，获取到embedding 的具体值后，再去用非 tf 函数去处理embedding并生成三元组

(2) 训练操作： 用(1) 生成的三元组进行训练

今天看了下别人的实现代码，直接用Tensorflow 在一次cnn操作中，直接选取并获取最终的loss值。(而我想象的那样，不是一次前馈+一次训练) 

很多很巧妙的思路(如果让我写，我想不出来的那种)：

(1) 获取所有三元组的loss，已知d[i,j] 是第i个和第j embedding 的距离，dx[i,j,1] = d[i,1,k] = d[i,j] 

     通过以下一句就可以实现获取所有三元组的loss值

#triplet_loss[i,j,k] = dx[i,j,1] - dy[i,1,k] +margin = d[i,j] - d[i,k] +margin
triplet_loss = anchor_positive_dist - anchor_negative_dist + margin

(2) 通过reduce_max / reduce_min 等操作，省略选取三元组中的各种if 操作

代码来源：https://github.com/omoindrot/tensorflow-triplet-loss/blob/master/model/triplet_loss.py

以下代码加了我的注释:

"""Define functions to create the triplet loss with online triplet mining."""

import tensorflow as tf

# (*) 返回embeddings 两两之间的距离
def _pairwise_distances(embeddings, squared=False):
    """Compute the 2D matrix of distances between all the embeddings.

    Args:
        embeddings: tensor of shape (batch_size, embed_dim)
        squared: Boolean. If true, output is the pairwise squared euclidean distance matrix.
                 If false, output is the pairwise euclidean distance matrix.

    Returns:
        pairwise_distances: tensor of shape (batch_size, batch_size)
    """
    # Get the dot product between all embeddings
    # shape (batch_size, batch_size)
    dot_product = tf.matmul(embeddings, tf.transpose(embeddings))

    # Get squared L2 norm for each embedding. We can just take the diagonal of `dot_product`.
    # This also provides more numerical stability (the diagonal of the result will be exactly 0).
    # shape (batch_size,)
    square_norm = tf.diag_part(dot_product)

    # Compute the pairwise distance matrix as we have:
    # ||a - b||^2 = ||a||^2  - 2 <a, b> + ||b||^2
    # shape (batch_size, batch_size)
    distances = tf.expand_dims