当我试图用tf.data构建一个输入管道时,我得到了一个非常奇怪的错误。我正在把我的参考图像和我的画组合成一个元组。然后我把它加到清单上。这应该能行,
但是现在我发现了一个奇怪的错误:
train_dataset = train_dataset.map(load_image_train, num_parallel_calls=tf.data.AUTOTUNE)这是我的代码:
@tf.function()
def load_image_train(a_training_datapoint):
print(type(a_training_datapoint))
print("here 1")
real_image_path, drawing_path = zip(*a_training_datapoint)
print("here 2")
real_image = convert_images_to_tensor(real_image_path)
print("here 3")
drawing_image = convert_images_to_tensor(drawing_path)
real_image, drawing_image = random_jitter(real_image, drawing_image)
real_image, drawing_image = normalize(real_image, drawing_image)
return real_image, drawing_image然后我有了这个:
test_dataset_list = []
for data in test_set:
test_dataset_list.append(zip(data.reference_image, data.drawing))
print(test_dataset_list)这里是唯一一个打印出来的。
所以它似乎不喜欢我如何解开我的元组,但我相信我是正确的。
它还说:
当我打印出a_test_datapoint的类型时
发布于 2022-10-04 04:18:46
这个Zip方法不是压缩为文件加密,但它是引用映射,您不能读取它,因为一些硬件尝试,但对象仍然存在,如果不是,它不能还原-引用。
与您处理路由路径或模拟问题的问题相同,我作为游戏示例,查找下一个操作,并存在于解决方案中。
服务器-硬件安全任务,压缩Fn的一个角色
试一试看看工作日志。
示例:用于重叠工作输入文本序列的Zip。
import tensorflow as tf
import tensorflow_text as tft
input_word = tf.constant(' \'Cause it\'s easy as an ice cream sundae Slipping outta your hand into the dirt Easy as an ice cream sundae Every dancer gets a little hurt Easy as an ice cream sundae Slipping outta your hand into the dirt Easy as an ice cream sundae Every dancer gets a little hurt Easy as an ice cream sundae Oh, easy as an ice cream sundae ' )
print( 'input_word: ' + str(input_word) )
print( " " )
dataset = tf.data.Dataset.from_tensors( tf.strings.bytes_split(input_word ) )
window_size = 6
dataset = dataset.map(lambda x: tft.sliding_window(x, width=window_size, axis=0) ).flat_map(tf.data.Dataset.from_tensor_slices)
print( dataset )
test_dataset_list = []
icount = 0
for data in dataset:
test_dataset_list.append(zip(tf.constant(data, name=str(icount)), tf.constant(data, name=str(icount + 1))))
icount = icount + 1
print(test_dataset_list)

https://stackoverflow.com/questions/73942922
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