用条件在keras中实现自定义丢失函数

Question

我需要一些keras损失功能的帮助.我一直在使用Tensorflow后端在keras上实现自定义丢失功能.

我已经在numpy中实现了自定义丢失函数,但如果它可以转换为keras损失函数则会很棒.loss函数采用数据帧和一系列用户id.如果user_id不同,则相同user_id的欧几里德距离为正和负.该函数返回数据帧的标量距离的总和.

def custom_loss_numpy (encodings, user_id):
# user_id: a pandas series of users
# encodings: a pandas dataframe of encodings

    batch_dist = 0

    for i in range(len(user_id)):
         first_row = encodings.iloc[i,:].values
         first_user = user_id[i]

         for j in range(i+1, len(user_id)):
              second_user = user_id[j]
              second_row = encodings.iloc[j,:].values

        # compute distance: if the users are same then Euclidean distance is positive otherwise negative.
            if first_user == second_user:
                tmp_dist = np.linalg.norm(first_row - second_row)
            else:
                tmp_dist = -np.linalg.norm(first_row - second_row)

            batch_dist += tmp_dist

    return batch_dist

我试图实现keras损失功能.我从y_true和y_pred张量对象中提取了numpy数组.

def custom_loss_keras(y_true, y_pred):
    # session of my program
    sess = tf_session.TF_Session().get()

    with sess.as_default():
        array_pred = y_pred.eval()
        print(array_pred)

但是我收到以下错误.

tensorflow.python.framework.errors_impl.InvalidArgumentError: You must feed a value for placeholder tensor 'dense_1_input' with dtype float and shape [?,102]
 [[Node: dense_1_input = Placeholder[dtype=DT_FLOAT, shape=[?,102], _device="/job:localhost/replica:0/task:0/device:CPU:0"]()]]

任何形式的帮助将非常感激.

Answer 1

首先,不可能在Keras损失函数中" 从y_true和中提取numpy数组y_pred".您必须使用Keras后端功能(或TF功能)来操作张量来计算损耗.

换句话说,最好不要使用if-else和循环来考虑计算损失的"矢量化"方法.

您的损失函数可以通过以下步骤计算:

1. 生成成对的欧几里德距离矩阵,在所有向量对之间encodings.
2. 生成一个矩阵,I其元素I_ij为1 if user_i == user_j和-1 if user_i != user_j.
3. 元素方式乘以两个矩阵,并总结元素以获得最终损失.

这是一个实现:

def custom_loss_keras(user_id, encodings):
    # calculate pairwise Euclidean distance matrix
    pairwise_diff = K.expand_dims(encodings, 0) - K.expand_dims(encodings, 1)
    pairwise_squared_distance = K.sum(K.square(pairwise_diff), axis=-1)

    # add a small number before taking K.sqrt for numerical safety
    # (K.sqrt(0) sometimes becomes nan)
    pairwise_distance = K.sqrt(pairwise_squared_distance + K.epsilon())

    # this will be a pairwise matrix of True and False, with shape (batch_size, batch_size)
    pairwise_equal = K.equal(K.expand_dims(user_id, 0), K.expand_dims(user_id, 1))

    # convert True and False to 1 and -1
    pos_neg = K.cast(pairwise_equal, K.floatx()) * 2 - 1

    # divide by 2 to match the output of `custom_loss_numpy`, but it's not really necessary
    return K.sum(pairwise_distance * pos_neg, axis=-1) / 2

我假设user_id在上面的代码中是整数.这里的技巧是K.expand_dims用于实现成对操作.乍一看可能有点难以理解,但它非常有用.

它应该给出与custom_loss_numpy(由于K.epsilon())会有一点点差异相同的损失值:

encodings = np.random.rand(32, 10)
user_id = np.random.randint(10, size=32)

print(K.eval(custom_loss_keras(K.variable(user_id), K.variable(encodings))).sum())
-478.4245

print(custom_loss_numpy(pd.DataFrame(encodings), pd.Series(user_id)))
-478.42953553795815

我在损失函数中犯了一个错误.

当此功能用于训练时,由于Keras自动变为y_true至少2D,因此参数user_id不再是1D张量.它的形状将是(batch_size, 1).

要使用此功能,必须删除额外的轴:

def custom_loss_keras(user_id, encodings):
    pairwise_diff = K.expand_dims(encodings, 0) - K.expand_dims(encodings, 1)
    pairwise_squared_distance = K.sum(K.square(pairwise_diff), axis=-1)
    pairwise_distance = K.sqrt(pairwise_squared_distance + K.epsilon())

    user_id = K.squeeze(user_id, axis=1)  # remove the axis added by Keras
    pairwise_equal = K.equal(K.expand_dims(user_id, 0), K.expand_dims(user_id, 1))

    pos_neg = K.cast(pairwise_equal, K.floatx()) * 2 - 1
    return K.sum(pairwise_distance * pos_neg, axis=-1) / 2

Answer 2

在 Keras 中实现参数化的自定义损失函数有两个步骤。首先，编写系数/度量的方法。其次，编写一个包装函数来按照 Keras 需要的方式格式化事物。

1. 对于像 DICE 这样的简单自定义损失函数，直接使用 Keras 后端而不是 tensorflow 实际上要干净一些。这是以这种方式实现的系数的示例：

   import keras.backend as K
   def dice_coef(y_true, y_pred, smooth, thresh):
       y_pred = y_pred > thresh
       y_true_f = K.flatten(y_true)
       y_pred_f = K.flatten(y_pred)
       intersection = K.sum(y_true_f * y_pred_f)
       return (2. * intersection + smooth) / (K.sum(y_true_f) + K.sum(y_pred_f) + smooth)
   

2. 现在是棘手的部分。Keras 损失函数只能(y_true, y_pred)作为参数。所以我们需要一个单独的函数来返回另一个函数：

   def dice_loss(smooth, thresh):
       def dice(y_true, y_pred)
           return -dice_coef(y_true, y_pred, smooth, thresh)
       return dice
   

最后，您可以在 Keras 中按如下方式使用它compile：

# build model 
model = my_model()
# get the loss function
model_dice = dice_loss(smooth=1e-5, thresh=0.5)
# compile model
model.compile(loss=model_dice)