InvalidArgumentError 索引 [i,0] = x 不在 keras 中的 [0, x) 中

Question

我有使用keras 1.2和的代码tensorflow 1.1。我已经运行它但有错误

import numpy as np
import keras
from keras import backend as K
from keras import initializers
from keras.models import Sequential, Model, load_model, save_model
from keras.layers.core import Dense, Lambda, Activation
from keras.layers import Embedding, Input, Dense, Multiply, Reshape, Flatten
from keras.optimizers import Adagrad, Adam, SGD, RMSprop
from keras.regularizers import l2

from sklearn.metrics import average_precision_score
from sklearn.metrics import  auc

def init_normal(shape, name=None):

    return initializers.lecun_uniform(seed=None)

def get_model(num_a, num_b, num_c, dim, regs=[0,0,0]):

    a = Input(shape=(1,), dtype='int32', name = 'a')
    b = Input(shape=(1,), dtype='int32', name = 'b')
    c = Input(shape=(1,), dtype='int32', name = 'c')



    Embedding_a = Embedding(input_dim = num_a, output_dim = dim,
                              embeddings_initializer='uniform', W_regularizer = l2(regs[0]), input_length=1)
    Embedding_b = Embedding(input_dim = num_b, output_dim = dim,
                              embeddings_initializer='uniform', W_regularizer = l2(regs[1]), input_length=1)   
    Embedding_c = Embedding(input_dim = num_c, output_dim = dim,
                              embeddings_initializer='uniform', W_regularizer = l2(regs[2]), input_length=1)  


    a_latent = Flatten()(Embedding_a(a))
    b_latent = Flatten()(Embedding_b(b))
    c_latent = Flatten()(Embedding_c(c))


    predict_vector = Multiply()([a_latent, b_latent, b_latent])
    prediction = Dense(1, activation='sigmoid', init='lecun_uniform', name = 'prediction')(predict_vector)



    model = Model(input=[a, b, c], output=prediction)

    return model

def evaluate_model(model, test_pos, test_neg):

    global _model
    global _test_pos
    global _test_neg
    _model = model
    _test_pos = test_pos
    _test_neg = test_neg
    print(_test_neg)


    a, b, c, labels = [],[],[],[]
    for item in _test_pos:

        a.append(item[0])
        b.append(item[1])
        c.append(item[2])
        labels.append(1)

    for item in _test_neg:

        a.append(item[0])
        b.append(item[1])
        c.append(item[2])
        labels.append(0)

    a = np.array(a)
    b = np.array(b)
    c = np.array(c)


    predictions = _model.predict([a, b, c], 
                             batch_size=100, verbose=0)
    return average_precision_score(labels, predictions), auc(labels, predictions)

model = get_model(4, 8, 12, 2, [0,0,0])
model.compile(optimizer=Adam(lr=0.001), loss='binary_crossentropy')


pos_test = [[0, 0, 2], [4, 8, 8], [2, 5, 4], [0, 0, 0]]
neg_test = [[3, 3, 2], [2, 1, 8], [1, 4, 1], [3, 3, 12]]


aupr, auc = evaluate_model(model, pos_test, neg_test)
print(aupr, auc)

但是，它给了我错误：有什么方法可以解决吗？

InvalidArgumentError (see above for traceback): indices[1,0] = 4 is not in [0, 4)
     [[Node: embedding_4/embedding_lookup = Gather[Tindices=DT_INT32, Tparams=DT_FLOAT, _class=["loc:@embedding_4/embeddings"], validate_indices=true, _device="/job:localhost/replica:0/task:0/cpu:0"](embedding_4/embeddings/read, _recv_a_1_0)]]

Answer 1

问题是，您将嵌入定义input_dim为 4、8 和 12，而它应该是 5、9、13。因为input_dim在嵌入中应该是max_index + 1。Keras 文档中也明确提到了这一点：

词汇表的大小，即最大整数索引 + 1。

如何解决问题？

将get_model方法更改为：

model = get_model(5, 9, 13, 2, [0, 0, 0])

或者将数据索引更改为：

pos_test = [[0, 0, 2], [3, 7, 7], [2, 5, 4], [0, 0, 0]]
neg_test = [[3, 3, 2], [2, 1, 7], [1, 4, 1], [3, 3, 11]]