Numpy 如何迭代三维张量

我有一个张量说：

y_true = np.array([[[1.], [0.], [3.]], [[5.], [0.], [0.]]])

我想在y_true上迭代访问所有独立值。我想在java中执行以下操作：

for(i=0;i<y_true.length;i++){
    arr2 = y_true[i];
    for(j=0;j<arr2.length;j++){
        print(arr2[j][0])
    }
}. 

---

for（i=0；i您是否正在寻找使用
[：，：，0]
进行切片

>>> y_true[:,:,0]
array([[1., 0., 3.],
       [5., 0., 0.]])

有两种情况：

你知道等级（维度）在示例中创建的numpy数组的
y\u-true
中，数组的秩为3，您可以检查
y\u-true.shape
属性，该属性应为您提供
y\u-true
的每个维度的精确大小，然后您可以为秩为
y\u-true
的循环编写尽可能多的for，并分别输出每个元素，例如：

import numpy as np

y_true = np.array([[[1.], [0.], [3.]], [[5.], [0.], [0.]]])
dims = y_true.shape
for i in range(dims[0]):
    for j in range(dims[1]):
        for k in range(dims[2]):
            print("Element of np array with indices {} is equal to {}".format([i, j, k], y_true[i, j, k]))

import numpy as np    


def recursively_print_elems(np_arr, idx, pos):
    if pos >= len(np_arr.shape):
        print("Element of np array with indeces {} is equal to: {}".format(idx, np_arr[tuple(idx)]))
        return
    for i in range(np_arr.shape[pos]):
        idx[pos] = i
        recursively_print_elems(np_arr, idx, pos + 1)


def print_elems(np_arr):
    idx = [0] * len(np_arr.shape)
    recursively_print_elems(np_arr, idx, 0)


y_true = np.array([[[1.], [0.], [3.]], [[5.], [0.], [0.]]])
print_elems(y_true)


如果不知道要打印的张量的秩，则可以编写递归函数来打印所有元素，例如：

import numpy as np

y_true = np.array([[[1.], [0.], [3.]], [[5.], [0.], [0.]]])
dims = y_true.shape
for i in range(dims[0]):
    for j in range(dims[1]):
        for k in range(dims[2]):
            print("Element of np array with indices {} is equal to {}".format([i, j, k], y_true[i, j, k]))

import numpy as np    


def recursively_print_elems(np_arr, idx, pos):
    if pos >= len(np_arr.shape):
        print("Element of np array with indeces {} is equal to: {}".format(idx, np_arr[tuple(idx)]))
        return
    for i in range(np_arr.shape[pos]):
        idx[pos] = i
        recursively_print_elems(np_arr, idx, pos + 1)


def print_elems(np_arr):
    idx = [0] * len(np_arr.shape)
    recursively_print_elems(np_arr, idx, 0)


y_true = np.array([[[1.], [0.], [3.]], [[5.], [0.], [0.]]])
print_elems(y_true)


第二种方法更一般，它适用于任何维张量

您的阵列：

In [19]: y_true
Out[19]: 
array([[[1.],
        [0.],
        [3.]],

       [[5.],
        [0.],
        [0.]]])
In [20]: y_true.shape
Out[20]: (2, 3, 1)

使用大小为1的最后一个维度，我们可以重塑它

In [21]: y_true.reshape(2,3)
Out[21]: 
array([[1., 0., 3.],
       [5., 0., 0.]])

在该索引上选择也可以

但是，您可以通过展开/展平按顺序访问所有值：

In [22]: y_true.ravel()
Out[22]: array([1., 0., 3., 5., 0., 0.])

或者获取1迭代器：

In [23]: yiter = y_true.flat
In [24]: yiter?
Type:            flatiter
String form:     <numpy.flatiter object at 0x1fdd200>
Length:          6
File:            ~/.local/lib/python3.6/site-packages/numpy/__init__.py
Docstring:       <no docstring>
Class docstring:
Flat iterator object to iterate over arrays.

A `flatiter` iterator is returned by ``x.flat`` for any array `x`.
It allows iterating over the array as if it were a 1-D array,
either in a for-loop or by calling its `next` method.
...

ndenumerate
使用此平面迭代器，并返回坐标和值：

In [26]: list(np.ndenumerate(y_true))
Out[26]: 
[((0, 0, 0), 1.0),
 ((0, 1, 0), 0.0),
 ((0, 2, 0), 3.0),
 ((1, 0, 0), 5.0),
 ((1, 1, 0), 0.0),
 ((1, 2, 0), 0.0)]

这方面的一个变体是
ndix
：

In [27]: indexs = np.ndindex(y_true.shape)
In [28]: for ijk in indexs:
    ...:     print(ijk, y_true[ijk])
    ...:     
(0, 0, 0) 1.0
(0, 1, 0) 0.0
(0, 2, 0) 3.0
(1, 0, 0) 5.0
(1, 1, 0) 0.0
(1, 2, 0) 0.0

但是如果可能的话，最好是对整个数组进行操作，而不是进行迭代。这些整个数组操作在编译代码中进行迭代。
您是否尝试过编写与java代码极其相似的python等效代码？