Question

我正在使用带有以下图层的keras模型来预测输入标签（4个标签中）

embedding_layer = keras.layers.Embedding(MAX_NB_WORDS, 
                                         EMBEDDING_DIM, 
                                         weights=[embedding_matrix], 
                                         input_length=MAX_SEQUENCE_LENGTH,
                                         trainable=False)

sequence_input = keras.layers.Input(shape = (MAX_SEQUENCE_LENGTH,), 
                                    dtype = 'int32')

embedded_sequences = embedding_layer(sequence_input)

hidden_layer = keras.layers.Dense(50, activation='relu')(embedded_sequences)
flat = keras.layers.Flatten()(hidden_layer)
preds = keras.layers.Dense(4, activation='softmax')(flat)
model = keras.models.Model(sequence_input, preds)
model.compile(loss='categorical_crossentropy', optimizer='rmsprop', metrics=['acc'])
model.fit(X_train, Y_train, batch_size=32, epochs=100)

但是，softmax函数返回4的输出数（因为我有4个标签）

当我使用预测函数使用相同的模型获得预测的Y时，我得到的每个X的数组为4，而不是一个标签决定输入的标签。

model.predict(X_test, batch_size = None, verbose = 0, steps = None)

如何使keras模型的输出层或model.predict函数决定一个标签，而不是为每个标签输出权重？

Answer 1

以下是从概率向量中抽样的常用函数

@model Model.khachhang
    @using RaovatThuCung.Common

@{
    ViewBag.Title = "KetQua";
    Layout = "~/Views/Shared/_Layout.cshtml";
    var session = (UserLogin)Session[CommonConstants.USER_SESSION];
}

<h2>KetQua</h2>

@using (Html.BeginForm("KetQua","MuaVip",FormMethod.Post))
{
    @Html.AntiForgeryToken()
    
    <div class="form-horizontal">
        <h4>Thanh toán thành công</h4>
        <hr />
        @Html.ValidationSummary(true, "", new { @class = "text-danger" })
        @Html.Hidden("session.UserID",session.UserID)

        <div class="form-group">
            <div class="col-md-offset-2 col-md-10">
                <input type="submit" value="Kích hoạt vip" class="btn btn-default" />
            </div>
        </div>
    </div>
}

取自here。

温度参数决定概率权重之间的差异加权多少。温度为1正在考虑每个重量“按原样”，大于1的温度会减小重量之间的差异，小于1的温度会增加它们。

这是一个在3个标签上使用概率向量的例子：

def sample(preds, temperature=1.0):
    # helper function to sample an index from a probability array
    preds = np.asarray(preds).astype('float64')
    preds = np.log(preds) / temperature
    exp_preds = np.exp(preds)
    preds = exp_preds / np.sum(exp_preds)
    probas = np.random.multinomial(1, preds, 1)
    return np.argmax(probas)

要查看新的向量，请p = np.array([0.1, 0.7, 0.2]) # The first label has a probability of 10% of being chosen, the second 70%, the third 20% print(sample(p, 1)) # sample using the input probabilities, unchanged print(sample(p, 0.1)) # the new vector of probabilities from which to sample is [ 3.54012033e-09, 9.99996371e-01, 3.62508322e-06] print(sample(p, 10)) # the new vector of probabilities from which to sample is [ 0.30426696, 0.36962778, 0.32610526]返回sample。

在神经网络中使用softmax来确定输入的标签

1 个答案: