Python Numpy 结构化数组（recarray）将值分配给切片

发布于 2024-09-06 03:24:26 字数 649 浏览 4 评论 0原文

以下示例显示了我想要执行的操作：

>>> test
rec.array([(0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0),
   (0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0)], 
  dtype=[('ifAction', '|i1'), ('ifDocu', '|i1'), ('ifComedy', '|i1')])

>>> test[['ifAction', 'ifDocu']][0]
(0, 0)

>>> test[['ifAction', 'ifDocu']][0] = (1,1)
>>> test[['ifAction', 'ifDocu']][0]
(0, 0)

因此，我想将值 (1,1) 分配给 test[['ifAction', 'ifDocu']][0]< /代码>。（最终，我想做一些类似 test[['ifAction', 'ifDocu']][0:10] = (1,1) 的事情，为 0 分配相同的值:10 我尝试了很多方法但没有成功，

谢谢。俊

原文

The following example shows what I want to do:

>>> test
rec.array([(0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0),
   (0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0)], 
  dtype=[('ifAction', '|i1'), ('ifDocu', '|i1'), ('ifComedy', '|i1')])

>>> test[['ifAction', 'ifDocu']][0]
(0, 0)

>>> test[['ifAction', 'ifDocu']][0] = (1,1)
>>> test[['ifAction', 'ifDocu']][0]
(0, 0)

So, I want to assign the values (1,1) to test[['ifAction', 'ifDocu']][0]. (Eventually, I want to do something like test[['ifAction', 'ifDocu']][0:10] = (1,1), assigning the same values for for 0:10. I have tried many ways but never succeeded. Is there any way to do this?

Thank you,
Joon

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不寐倦长更 2024-09-13 03:24:26

当您说 test['ifAction'] 时，您将看到数据视图。
当您说 test[['ifAction','ifDocu']] 时，您正在使用花式索引，从而获得数据的副本。该副本对您没有帮助，因为修改副本会使原始数据保持不变。

因此，解决这个问题的方法是分别为 test['ifAction'] 和 test['ifDocu'] 赋值：

test['ifAction'][0]=1
test['ifDocu'][0]=1

例如：

import numpy as np
test=np.rec.array([(0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0),
   (0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0)], 
  dtype=[('ifAction', '|i1'), ('ifDocu', '|i1'), ('ifComedy', '|i1')])

print(test[['ifAction','ifDocu']])
# [(0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0)]
test['ifAction'][0]=1
test['ifDocu'][0]=1

print(test[['ifAction','ifDocu']][0])
# (1, 1)
test['ifAction'][0:10]=1
test['ifDocu'][0:10]=1

print(test[['ifAction','ifDocu']])
# [(1, 1) (1, 1) (1, 1) (1, 1) (1, 1) (1, 1) (1, 1) (1, 1) (1, 1) (1, 1)]

为了更深入地了解底层，请参阅 Robert Kern 的这篇文章。

When you say test['ifAction'] you get a view of the data.
When you say test[['ifAction','ifDocu']] you are using fancy-indexing and thus get a copy of the data. The copy doesn't help you since modifying the copy leaves the original data unchanged.

So a way around this is to assign values to test['ifAction'] and test['ifDocu'] individually:

test['ifAction'][0]=1
test['ifDocu'][0]=1

For example:

import numpy as np
test=np.rec.array([(0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0),
   (0, 0, 0), (0, 0, 0), (0, 0, 0), (0, 0, 0)], 
  dtype=[('ifAction', '|i1'), ('ifDocu', '|i1'), ('ifComedy', '|i1')])

print(test[['ifAction','ifDocu']])
# [(0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0)]
test['ifAction'][0]=1
test['ifDocu'][0]=1

print(test[['ifAction','ifDocu']][0])
# (1, 1)
test['ifAction'][0:10]=1
test['ifDocu'][0:10]=1

print(test[['ifAction','ifDocu']])
# [(1, 1) (1, 1) (1, 1) (1, 1) (1, 1) (1, 1) (1, 1) (1, 1) (1, 1) (1, 1)]

For a deeper look under the hood, see this post by Robert Kern .

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