问题描述
问题
选择非连续索引的 Numpy 元组/数组索引的 Tensorflow 等价物是什么?使用 numpy,可以使用元组或数组选择多行/多列。
a = np.arange(12).reshape(3,4)
print(a)
print(a[
(0,2),# select row 0 and 2
1 # select col 0
])
---
[[ 0 1 2 3] # a[0][1] -> 1
[ 4 5 6 7]
[ 8 9 10 11]] # a[2][1] -> 9
[1 9]
看着 Multi-axis indexing 但似乎没有等效的方法。
通过传递多个索引来索引更高等级的张量。
使用元组或数组会导致 ypeError: Only integers,slices (`:`),ellipsis (`...`),tf.newaxis (`None`) and scalar tf.int32/tf.int64 tensors are valid indices,got (0,2,5)
。
from tensorflow.keras.layers.experimental.preprocessing import TextVectorization
training_data = np.array([["This is the 1st sample."],["And here's the 2nd sample."]])
vectorizer = TextVectorization(output_mode="int")
vectorizer.adapt(training_data)
word_indices = vectorizer(training_data)
word_indices = tf.cast(word_indices,dtype=tf.int8)
print(f"word vocabulary:{vectorizer.get_vocabulary()}\n")
print(f"word indices:\n{word_indices}\n")
index_to_word = tf.reshape(tf.constant(vectorizer.get_vocabulary()),(-1,1))
print(f"index_to_word:\n{index_to_word}\n")
# Numpy tuple indexing
print(f"indices to words:{words.numpy()[(0,5),::]}")
# What is TF equivalent indexing?
print(f"indices to words:{words[(0,::]}") # <--- cannot use tuple/array indexing
结果:
word vocabulary:['','[UNK]','the','sample','this','is','heres','and','2nd','1st']
word indices:
[[4 5 2 9 3]
[7 6 2 8 3]]
index_to_word:
[[b'']
[b'[UNK]']
[b'the']
[b'sample']
[b'this']
[b'is']
[b'heres']
[b'and']
[b'2nd']
[b'1st']]
indices to words:[[b'']
[b'the']
[b'is']]
TypeError: Only integers,5)
Tensorflow 中有哪些索引可以选择不连续的多个索引?
解决方法
您可以使用 tf.gather
。
>>> tf.gather(words,[0,2,5])
<tf.Tensor: shape=(3,1),dtype=string,numpy=
array([[b''],[b'the'],[b'is']],dtype=object)>
在指南中阅读更多内容:Introduction to tensor slicing