在張量流中,我制作了一個將哈希作為輸入的常規網絡。作為一個例子,我使用了內置的 python hash()函數(是的,它在每個會話中都改變了鹽,但這是一個例子)代碼是這樣的:from time import timest = time()import tensorflow as tfprint(time() - st)import numpy as npimport chessimport atexitfrom numpy import shapedata = open("data.data", "r").readlines()[:10000]targets = open("targets.data", "r").readlines()[:10000]boards_data = []new_targets = []for i in data: boards_data.append(hash(i))for i in targets: new_targets.append(float(i))print(len(new_targets))print(len(boards_data))print(np.array(new_targets))print(np.array(boards_data))def create_model(): model = tf.keras.models.Sequential() model.add(tf.keras.layers.Reshape((1,1,1))) model.add(tf.keras.layers.Dense(1000, activation="tanh")) model.add(tf.keras.layers.Flatten()) model.add(tf.keras.layers.Dense(1, activation='tanh')) model.compile(loss="mse", optimizer="adam", metrics=['accuracy']) return modelmodel = create_model()model.fit(np.array(boards_data), np.array(new_targets), epochs=10)model.predict(np.array(hash("8/6P1/5k1K/6r1/8/8/8/8 b - - 0 83")))錯誤在預測中。我在如何修復 Tensorflow 中的“IndexError:列表索引超出范圍”中看到了 conv2d 示例 ,但事實并非如此......和回溯:Traceback (most recent call last): File "/Volumes/POOPOO USB/lichess-bot/engines/engine2/nn_evaluation/nn_evaluation2.py", line 36, in <module> model.predict(np.array(hash("8/6P1/5k1K/6r1/8/8/8/8 b - - 0 83"))) File "/Users/ofek/Library/Python/3.8/lib/python/site-packages/tensorflow/python/keras/engine/training.py", line 130, in _method_wrapper return method(self, *args, **kwargs) File "/Users/ofek/Library/Python/3.8/lib/python/site-packages/tensorflow/python/keras/engine/training.py", line 1569, in predict data_handler = data_adapter.DataHandler( File "/Users/ofek/Library/Python/3.8/lib/python/site-packages/tensorflow/python/keras/engine/data_adapter.py", line 1105, in __init__ self._adapter = adapter_cls(
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TA貢獻1789條經驗 獲得超10個贊
問題是您正在從哈希值創建一個 0d numpy 字符串。預測只能在至少具有一維的數組上運行。您可以檢查您的散列值是否為 0d:
print(np.array(hash("8/6P1/5k1K/6r1/8/8/8/8 b - - 0 83")).shape) # outputs: ()
與將哈希值放入列表相比:
print(np.array([hash("8/6P1/5k1K/6r1/8/8/8/8 b - - 0 83")]).shape) # outputs: (1,)
第二個np.array
預測運行沒有錯誤。
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