## 问题：切片NumPy 2d数组，或者如何从nxn数组（n> m）中提取mxm子矩阵？

``````from numpy import *
x = range(16)
x = reshape(x,(4,4))

print x
[[ 0  1  2  3]
[ 4  5  6  7]
[ 8  9 10 11]
[12 13 14 15]]``````

``````In [33]: x[0:2,0:2]
Out[33]:
array([[0, 1],
[4, 5]])

In [34]: x[2:,2:]
Out[34]:
array([[10, 11],
[14, 15]])``````

``````In [35]: x[[1,3],[1,3]]
Out[35]: array([ 5, 15])``````

``````    In [61]: x[[1,3]][:,[1,3]]
Out[61]:
array([[ 5,  7],
[13, 15]])``````

I want to slice a NumPy nxn array. I want to extract an arbitrary selection of m rows and columns of that array (i.e. without any pattern in the numbers of rows/columns), making it a new, mxm array. For this example let us say the array is 4×4 and I want to extract a 2×2 array from it.

Here is our array:

``````from numpy import *
x = range(16)
x = reshape(x,(4,4))

print x
[[ 0  1  2  3]
[ 4  5  6  7]
[ 8  9 10 11]
[12 13 14 15]]
``````

The line and columns to remove are the same. The easiest case is when I want to extract a 2×2 submatrix that is at the beginning or at the end, i.e. :

``````In [33]: x[0:2,0:2]
Out[33]:
array([[0, 1],
[4, 5]])

In [34]: x[2:,2:]
Out[34]:
array([[10, 11],
[14, 15]])
``````

But what if I need to remove another mixture of rows/columns? What if I need to remove the first and third lines/rows, thus extracting the submatrix `[[5,7],[13,15]]`? There can be any composition of rows/lines. I read somewhere that I just need to index my array using arrays/lists of indices for both rows and columns, but that doesn’t seem to work:

``````In [35]: x[[1,3],[1,3]]
Out[35]: array([ 5, 15])
``````

I found one way, which is:

``````    In [61]: x[[1,3]][:,[1,3]]
Out[61]:
array([[ 5,  7],
[13, 15]])
``````

First issue with this is that it is hardly readable, although I can live with that. If someone has a better solution, I’d certainly like to hear it.

Other thing is I read on a forum that indexing arrays with arrays forces NumPy to make a copy of the desired array, thus when treating with large arrays this could become a problem. Why is that so / how does this mechanism work?

## 回答 0

As Sven mentioned, `x[[[0],[2]],[1,3]]` will give back the 0 and 2 rows that match with the 1 and 3 columns while `x[[0,2],[1,3]]` will return the values x[0,1] and x[2,3] in an array.

There is a helpful function for doing the first example I gave, `numpy.ix_`. You can do the same thing as my first example with `x[numpy.ix_([0,2],[1,3])]`. This can save you from having to enter in all of those extra brackets.

## 回答 1

NumPy通过引入步幅来解决此问题。在计算要访问的内存偏移量时`x[i,j]`，实际计算的是`i*x.strides[0]+j*x.strides[1]`（并且这已经包括int大小的因数）：

``````x.strides
(16, 4)``````

`y`像上面那样提取时，NumPy不会创建新的缓冲区，但是创建一个引用相同缓冲区的新数组对象（否则`y`将等于`x`。）然后，新数组对象将具有不同的形状，`x`并且可能以不同的开头偏移到缓冲区中，但将与`x`（至少在这种情况下）共享跨步：

``````y.shape
(2,2)
y.strides
(16, 4)``````

``x[[[1],[3]],[1,3]]``

``x[1::2, 1::2]``

To answer this question, we have to look at how indexing a multidimensional array works in Numpy. Let’s first say you have the array `x` from your question. The buffer assigned to `x` will contain 16 ascending integers from 0 to 15. If you access one element, say `x[i,j]`, NumPy has to figure out the memory location of this element relative to the beginning of the buffer. This is done by calculating in effect `i*x.shape[1]+j` (and multiplying with the size of an int to get an actual memory offset).

If you extract a subarray by basic slicing like `y = x[0:2,0:2]`, the resulting object will share the underlying buffer with `x`. But what happens if you acces `y[i,j]`? NumPy can’t use `i*y.shape[1]+j` to calculate the offset into the array, because the data belonging to `y` is not consecutive in memory.

NumPy solves this problem by introducing strides. When calculating the memory offset for accessing `x[i,j]`, what is actually calculated is `i*x.strides[0]+j*x.strides[1]` (and this already includes the factor for the size of an int):

``````x.strides
(16, 4)
``````

When `y` is extracted like above, NumPy does not create a new buffer, but it does create a new array object referencing the same buffer (otherwise `y` would just be equal to `x`.) The new array object will have a different shape then `x` and maybe a different starting offset into the buffer, but will share the strides with `x` (in this case at least):

``````y.shape
(2,2)
y.strides
(16, 4)
``````

This way, computing the memory offset for `y[i,j]` will yield the correct result.

But what should NumPy do for something like `z=x[[1,3]]`? The strides mechanism won’t allow correct indexing if the original buffer is used for `z`. NumPy theoretically could add some more sophisticated mechanism than the strides, but this would make element access relatively expensive, somehow defying the whole idea of an array. In addition, a view wouldn’t be a really lightweight object anymore.

This is covered in depth in the NumPy documentation on indexing.

Oh, and nearly forgot about your actual question: Here is how to make the indexing with multiple lists work as expected:

``````x[[[1],[3]],[1,3]]
``````

This is because the index arrays are broadcasted to a common shape. Of course, for this particular example, you can also make do with basic slicing:

``````x[1::2, 1::2]
``````

## 回答 2

``a[[1,3],:][:,[1,3]]``

I don’t think that `x[[1,3]][:,[1,3]]` is hardly readable. If you want to be more clear on your intent, you can do:

``````a[[1,3],:][:,[1,3]]
``````

I am not an expert in slicing but typically, if you try to slice into an array and the values are continuous, you get back a view where the stride value is changed.

e.g. In your inputs 33 and 34, although you get a 2×2 array, the stride is 4. Thus, when you index the next row, the pointer moves to the correct position in memory.

Clearly, this mechanism doesn’t carry well into the case of an array of indices. Hence, numpy will have to make the copy. After all, many other matrix math function relies on size, stride and continuous memory allocation.

## 回答 3

``````In [49]: x=np.arange(16).reshape((4,4))
In [50]: x[1:4:2,1:4:2]
Out[50]:
array([[ 5,  7],
[13, 15]])``````

``````In [51]: y=x[1:4:2,1:4:2]

In [52]: y[0,0]=100

In [53]: x   # <---- Notice x[1,1] has changed
Out[53]:
array([[  0,   1,   2,   3],
[  4, 100,   6,   7],
[  8,   9,  10,  11],
[ 12,  13,  14,  15]])``````

`z=x[(1,3),:][:,(1,3)]`使用高级索引并因此返回副本：

``````In [58]: x=np.arange(16).reshape((4,4))
In [59]: z=x[(1,3),:][:,(1,3)]

In [60]: z
Out[60]:
array([[ 5,  7],
[13, 15]])

In [61]: z[0,0]=0``````

``````In [62]: x
Out[62]:
array([[ 0,  1,  2,  3],
[ 4,  5,  6,  7],
[ 8,  9, 10, 11],
[12, 13, 14, 15]])``````

If you want to skip every other row and every other column, then you can do it with basic slicing:

``````In [49]: x=np.arange(16).reshape((4,4))
In [50]: x[1:4:2,1:4:2]
Out[50]:
array([[ 5,  7],
[13, 15]])
``````

This returns a view, not a copy of your array.

``````In [51]: y=x[1:4:2,1:4:2]

In [52]: y[0,0]=100

In [53]: x   # <---- Notice x[1,1] has changed
Out[53]:
array([[  0,   1,   2,   3],
[  4, 100,   6,   7],
[  8,   9,  10,  11],
[ 12,  13,  14,  15]])
``````

while `z=x[(1,3),:][:,(1,3)]` uses advanced indexing and thus returns a copy:

``````In [58]: x=np.arange(16).reshape((4,4))
In [59]: z=x[(1,3),:][:,(1,3)]

In [60]: z
Out[60]:
array([[ 5,  7],
[13, 15]])

In [61]: z[0,0]=0
``````

Note that `x` is unchanged:

``````In [62]: x
Out[62]:
array([[ 0,  1,  2,  3],
[ 4,  5,  6,  7],
[ 8,  9, 10, 11],
[12, 13, 14, 15]])
``````

If you wish to select arbitrary rows and columns, then you can’t use basic slicing. You’ll have to use advanced indexing, using something like `x[rows,:][:,columns]`, where `rows` and `columns` are sequences. This of course is going to give you a copy, not a view, of your original array. This is as one should expect, since a numpy array uses contiguous memory (with constant strides), and there would be no way to generate a view with arbitrary rows and columns (since that would require non-constant strides).

## 回答 4

``````>>> x[1:4:2, 1:4:2]
array([[ 5,  7],
[13, 15]])``````

With numpy, you can pass a slice for each component of the index – so, your `x[0:2,0:2]` example above works.

If you just want to evenly skip columns or rows, you can pass slices with three components (i.e. start, stop, step).

``````>>> x[1:4:2, 1:4:2]
array([[ 5,  7],
[13, 15]])
``````

Which is basically: slice in the first dimension, with start at index 1, stop when index is equal or greater than 4, and add 2 to the index in each pass. The same for the second dimension. Again: this only works for constant steps.

The syntax you got to do something quite different internally – what `x[[1,3]][:,[1,3]]` actually does is create a new array including only rows 1 and 3 from the original array (done with the `x[[1,3]]` part), and then re-slice that – creating a third array – including only columns 1 and 3 of the previous array.

## 回答 5

``````columns_to_keep = [1,3]
rows_to_keep = [1,3]``````

``x[np.ix_(rows_to_keep, columns_to_keep)] ``

``````array([[ 5,  7],
[13, 15]])``````

I have a similar question here: Writting in sub-ndarray of a ndarray in the most pythonian way. Python 2 .

Following the solution of previous post for your case the solution looks like:

``````columns_to_keep = [1,3]
rows_to_keep = [1,3]
``````

An using ix_:

``````x[np.ix_(rows_to_keep, columns_to_keep)]
``````

Which is:

``````array([[ 5,  7],
[13, 15]])
``````

## 回答 6

`````` x=np.arange(16).reshape((4,4))
x[range(1,3), :][:,range(1,3)] ``````

I’m not sure how efficient this is but you can use range() to slice in both axis

`````` x=np.arange(16).reshape((4,4))
x[range(1,3), :][:,range(1,3)]
``````