访问多维数组的每个元素，而不需要知道各个维度。

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英文:

Access each element of a multidimensional array without knowing the individual dimensions

问题

import numpy as np
import random

myRand = np.random.rand(5, 6, 7)

这段代码适用于已知myRand具有3个维度的情况。如果我不知道呢？

mySum = 0.0
for i in range(myRand.shape[0]):
for j in range(myRand.shape[1]):
for k in range(myRand.shape[2]):
# 对myRand[i,j,k]执行某些操作

英文:

I have a multidimensional array but I won't know the number of dimensions or the size of each dimension. How can I generalize the code such that I can access each element of the array individually?

  1. import numpy as np
  2. import random
  3.     
  4. myRand = np.random.rand(5, 6, 7)
  5.     
  6. #print (myRand.shape[0])
  7. # This works great if I already know that myRand has 3 dimensions. What if I don't know that?
  8. mySum = 0.0
  9. for i in range(myRand.shape[0]):
  10.     for j in range(myRand.shape[1]):
  11.         for k in range(myRand.shape[2]):
  12. #           Do something with myRand[i,j,k]

答案1

得分: 1

如果您不需要在计算中保留每个维度内的索引，您可以只使用numpy.nditer。

  1. a = np.arange(8).reshape((2, 2, 2))
  2. a
  3. array([[[0, 1],
  4.         [2, 3]],
  6.        [[4, 5],
  7.         [6, 7]]])
  9. for i in np.nditer(a):
  10.     print(i)
  12. # 输出结果:
  13. 0
  14. 1
  15. 2
  16. 3
  17. 4
  18. 5
  19. 6
  20. 7

您通常不应该需要像这样使用for循环迭代数组。通常使用numpy的方法来执行您要进行的计算会更好。

英文:

If you do not need to retain the indices within each of the dimensions for calculation, you can just use numpy.nditer

  1. >>> a = np.arange(8).reshape((2, 2, 2))
  2. >>> a
  3. array([[[0, 1],
  4.         [2, 3]],
  6.        [[4, 5],
  7.         [6, 7]]])
  9. >>> for i in np.nditer(a):
  10. ...     print(i)
  11. ...
  12. 0
  13. 1
  14. 2
  15. 3
  16. 4
  17. 5
  18. 6
  19. 7

> You shouldn't really need to iterate over the array using a for-loop like this. There is usually a better way to perform whatever computation you are doing using numpy methods

答案2

得分: 1

你可以使用[itertools][1]来完成这个任务。

以下是生成索引的代码，通过这些索引你可以访问数组中的元素：

  1. import numpy as np
  2. import itertools
  4. v1 = np.random.randint(5, size=2)
  5. v2 = np.random.randint(5, size=(2, 4))
  6. v3 = np.random.randint(5, size=(2, 3, 2))
  8. # v1
  9. args1 = [list(range(e)) for e in list(v1.shape)]
  10. print(v1)
  11. for combination in itertools.product(*args1):
  12.     print(v1[combination])
  14. # v2
  15. args2 = [list(range(e)) for e in list(v2.shape)]
  16. print(v2)
  17. for combination in itertools.product(*args2):
  18.     print(v2[combination])
  20. # v3
  21. args3 = [list(range(e)) for e in list(v3.shape)]
  22. print(v3)
  23. for combination in itertools.product(*args3):
  24.     print(v3[combination])
  26. 在不同大小的简单数组上进行了测试，运行正常。
  28. [1]: https://docs.python.org/2/library/itertools.html
  30. <details>
  31. <summary>英文:</summary>
  33. You could use [`itertools`][1] to do so.
  35. the following piece of code will generate the indices which you will be able to access the array while obtaining them:
  37.     import numpy as np
  38.     import itertools
  39.     
  40.     v1 = np.random.randint(5, size=2)
  41.     v2 = np.random.randint(5, size=(2, 4))
  42.     v3 = np.random.randint(5, size=(2, 3, 2))
  43.     
  44.     # v1
  45.     args1 = [list(range(e)) for e in list(v1.shape)]
  46.     print(v1)
  47.     for combination in itertools.product(*args1):
  48.         print(v1[combination])
  49.     
  50.     # v2
  51.     args2 = [list(range(e)) for e in list(v2.shape)]
  52.     print(v2)
  53.     for combination in itertools.product(*args2):
  54.         print(v2[combination])
  55.         
  56.     # v3
  57.     args3 = [list(range(e)) for e in list(v3.shape)]
  58.     print(v3)
  59.     for combination in itertools.product(*args3):
  60.         print(v3[combination])
  62. Tested it on a simple arrays with different sizes and it works fine.
  67.   [1]: https://docs.python.org/2/library/itertools.html
  69. </details>

huangapple
  • 本文由 发表于 2020年1月6日 22:09:07
  • 转载请务必保留本文链接：https://go.coder-hub.com/59613577.html
  • numpy
  • python
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