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调用 numpy 数组时,“[()]”是什么意思?

[英]What does "[()]" mean when called upon a numpy array?

I just came across this piece of code:我刚刚遇到了这段代码:

x = np.load(lc_path, allow_pickle=True)[()]

And I've never seen this pattern before: [()] .我以前从未见过这种模式: [()] What does it do and why is this syntacticly correct?它有什么作用,为什么这在语法上是正确的?


a = np.load(lc_path, allow_pickle=True)
>>> array({'train_ppls': [1158.359413193576, 400.54333992093854, ...],
 'val_ppls': [493.0056070137404, 326.53203520368623, ...],
 'train_losses': [340.40905952453613, 675.6475067138672, ...],
 'val_losses': [217.46258735656738, 438.86770486831665, ...],
 'times': [19.488852977752686, 20.147733449935913, ...]}, dtype=object)

So I guess a is a dict wrapped in an array for some reason by the person who saved it所以我猜a是保存它的人出于某种原因包装在数组中的dict

It a way (the only way) of indexing a 0d array:这是索引 0d 数组的一种方式(唯一方式):

In [475]: x=np.array(21)                                                                       
In [476]: x                                                                                    
Out[476]: array(21)
In [477]: x.shape                                                                              
Out[477]: ()
In [478]: x[()]                                                                                
Out[478]: 21

In effect it pulls the element out of the array.实际上,它将元素从数组中拉出。 item() is another way: item()是另一种方式:

In [479]: x.item()                                                                             
Out[479]: 21
In [480]: x.ndim                                                                               
Out[480]: 0

In

x = np.load(lc_path, allow_pickle=True)[()]

most likely the np.save was given a non-array;很可能np.save被赋予了一个非数组; and wrapped in a 0d object dtype array to save it.并包装在一个 0d 对象 dtype 数组中以保存它。 This is a way of recovering that object.这是恢复该对象的一种方式。

In [481]: np.save('test.npy', {'a':1})                                                         
In [482]: x = np.load('test.npy', allow_pickle=True)                                           
In [483]: x                                                                                    
Out[483]: array({'a': 1}, dtype=object)
In [484]: x.ndim                                                                               
Out[484]: 0
In [485]: x[()]                                                                                
Out[485]: {'a': 1}

In general when we index a nd array, eg x[1,2] we are really doing x[(1,2)] , that is, using a tuple that corresponds to the number of dimensions.一般来说,当我们索引一个 nd 数组时,例如x[1,2]我们实际上是在做x[(1,2)] ,也就是说,使用对应于维数的元组。 If x is 0d, the only tuple that works is an empty one, () .如果x是 0d,则唯一有效的元组是空元组()

That's indexing the array with a tuple of 0 indices.那是用 0 个索引的元组对数组进行索引。 For most arrays, this just produces a view of the whole array, but for a 0-dimensional array, it extracts the array's single element as a scalar.对于大多数数组,这只会生成整个数组的视图,但对于 0 维数组,它将数组的单个元素提取为标量。

In this case, it looks like someone made the weird choice to dump a non-NumPy object to an array with numpy.save , resulting in NumPy saving a 0-dimensional array of object dtype wrapping the original object.在这种情况下,看起来有人做出了奇怪的选择,将非 NumPy 对象转储到numpy.save数组,导致 NumPy 保存了一个 0 维object numpy.save数组,该数组包含原始对象。 The use of allow_pickle=True and the empty tuple index extracts the object from the 0-dimensional array.使用allow_pickle=True和空元组索引从 0 维数组中提取对象。

They probably should have picked something other than numpy.save to save this object.他们可能应该选择numpy.save以外的其他东西来保存这个对象。

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