Keras自定义指标迭代 [英] Keras custom metric iteration
问题描述
我对Keras相当陌生,我正在尝试定义自己的指标.它计算出一致性指数,作为对回归问题的一种度量.
I'm pretty new to Keras and I'm trying to define my own metric. It calculates concordance index which is a measure for regression problems.
def cindex_score(y_true, y_pred):
sum = 0
pair = 0
for i in range(1, len(y_true)):
for j in range(0, i):
if i is not j:
if(y_true[i] > y_true[j]):
pair +=1
sum += 1* (y_pred[i] > y_pred[j]) + 0.5 * (y_pred[i] == y_pred[j])
if pair is not 0:
return sum/pair
else:
return 0
def baseline_model(hidden_neurons, inputdim):
model = Sequential()
model.add(Dense(hidden_neurons, input_dim=inputdim, init='normal', activation='relu'))
model.add(Dense(hidden_neurons, init='normal', activation='relu'))
model.add(Dense(1, init='normal')) #output layer
model.compile(loss='mean_squared_error', optimizer='adam', metrics=[cindex_score])
return model
def run_model(P_train, Y_train, P_test, model):
history = model.fit(numpy.array(P_train), numpy.array(Y_train), batch_size=50, nb_epoch=200)
plotLoss(history)
return model.predict(P_test)
baseline_model,run_model和cindex_score函数位于one.py中,以下函数位于two.py中,我将其称为模型,
baseline_model, run_model and cindex_score functions are in one.py and the following function is in two.py where I called the model,
def experiment():
hidden_neurons = 250
dmodel=baseline_model(hidden_neurons, train_pair.shape[1])
predicted_Y = run_model(train_pair,train_Y, test_pair, dmodel)
但是出现以下错误,类型为'Tensor'的对象没有len()".它也不适用于shape属性.
But I get the following error, "object of type 'Tensor' has no len()". It does not work with shape attribute as well.
例如,y_true表示为Tensor("dense_4_target:0",shape =(?,?),dtype = float32),其形状为Tensor("strided_slice:0",shape =(),dtype = int32 ).
For instance, y_true is represented as Tensor("dense_4_target:0", shape=(?, ?), dtype=float32) and its shape is Tensor("strided_slice:0", shape=(), dtype=int32).
请问如何在Tensor对象中进行迭代?
Could you please help me about how to iterate within a Tensor object?
最好
推荐答案
如果您习惯使用tensorflow
,则可以尝试使用以下代码:
If you are comfortable using tensorflow
, then you can try using this code instead:
def cindex_score(y_true, y_pred):
g = tf.subtract(tf.expand_dims(y_pred, -1), y_pred)
g = tf.cast(g == 0.0, tf.float32) * 0.5 + tf.cast(g > 0.0, tf.float32)
f = tf.subtract(tf.expand_dims(y_true, -1), y_true) > 0.0
f = tf.matrix_band_part(tf.cast(f, tf.float32), -1, 0)
g = tf.reduce_sum(tf.multiply(g, f))
f = tf.reduce_sum(f)
return tf.where(tf.equal(g, 0), 0.0, g/f)
以下代码可验证这两种方法是否等效:
Here is some code that verifies that both approaches are equivalent:
def _ref(J, K):
_sum = 0
_pair = 0
for _i in range(1, len(J)):
for _j in range(0, _i):
if _i is not _j:
if(J[_i] > J[_j]):
_pair +=1
_sum += 1* (K[_i] > K[_j]) + 0.5 * (K[_i] == K[_j])
return 0 if _pair == 0 else _sum / _pair
def _raw(J, K):
g = tf.subtract(tf.expand_dims(K, -1), K)
g = tf.cast(g == 0.0, tf.float32) * 0.5 + tf.cast(g > 0.0, tf.float32)
f = tf.subtract(tf.expand_dims(J, -1), J) > 0.0
f = tf.matrix_band_part(tf.cast(f, tf.float32), -1, 0)
g = tf.reduce_sum(tf.multiply(g, f))
f = tf.reduce_sum(f)
return tf.where(tf.equal(g, 0), 0.0, g/f)
for _ in range(100):
with tf.Session() as sess:
inputs = [tf.placeholder(dtype=tf.float32),
tf.placeholder(dtype=tf.float32)]
D = np.random.randint(low=10, high=1000)
data = [np.random.rand(D), np.random.rand(D)]
r1 = sess.run(_raw(inputs[0], inputs[1]),
feed_dict={x: y for x, y in zip(inputs, data)})
r2 = _ref(data[0], data[1])
assert np.isclose(r1, r2)
请注意,这仅适用于一维张量(在keras中很少出现这种情况).
Please note that this only works for 1D-tensors (rarely a case you will have in keras).
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