在numpy中使用二维索引的子样本1-D数组。

Background: The data I'm using is being extracted from a netCDF4 object, which creates a numpy masked array at initialization, but does not appear to support the numpy reshape() method, making it only possible to reshape after all the data has been copied = way too slow.

背景:我正在使用的数据是从netCDF4对象中提取的，它在初始化时创建了一个numpy屏蔽数组，但似乎没有支持numpy整形()方法，这使得在所有数据被复制后，只可能重新进行重构。

Question: How can I sub-sample a 1-D array, that is basically a flattened 2-D array, without reshaping it?

问:我怎样才能对一个一维数组进行子样本，它基本上是一个扁平的二维数组，而不需要重新构造它?

import numpy

a1 = np.array([[1,2,3,4],
               [11,22,33,44],
               [111,222,333,444],
               [1111,2222,3333,4444],
               [11111,22222,33333,44444]])

a2 = np.ravel(a1)

rows, cols = a1.shape

row1 = 1
row2 = 3

col1 = 1
col2 = 3

I would like to use a fast slicing method that doesn't require reshaping the 1-D array to a 2-D array.

我想使用一种快速切片方法，它不需要将一维数组转换为二维数组。

Desired Output:

期望的输出:

np.ravel(a1[row1:row2, col1:col2])

>> array([ 22,  33, 222, 333])

I got as far as getting the start and ending positions, but this just selects ALL data between these points (i.e. extra columns).

我得到了开始和结束的位置，但这只是选择了这些点之间的所有数据(即额外的列)。

idx_start = (row1 * cols) + col1
idx_end   = (row2 * cols) + col2

Update: I just tried Jaime's brilliant answer, but it appears that netCDF4 won't allow for 2-D indices.

更新:我刚刚尝试了Jaime的聪明答案，但似乎netCDF4不允许2d索引。

z = dataset.variables["z"][idx]
  File "netCDF4.pyx", line 2613, in netCDF4.Variable.__getitem__ (netCDF4.c:29583)
  File "/usr/local/lib/python2.7/dist-packages/netCDF4_utils.py", line 141, in _StartCountStride
    raise IndexError("Index cannot be multidimensional.")
IndexError: Index cannot be multidimensional.

3 个解决方案

#1

You can get what you want with a combination of np.ogrid and np.ravel_multi_index:

你可以用np来得到你想要的。ogrid np.ravel_multi_index:

>>> a1
array([    1,     2,     3,     4,    11,    22,    33,    44,   111,
         222,   333,   444,  1111,  2222,  3333,  4444, 11111, 22222,
       33333, 44444])
>>> idx = np.ravel_multi_index((np.ogrid[1:3,1:3]), (5, 4))
>>> a1[idx]
array([[ 22,  33],
       [222, 333]])

You could of course ravel this array to get a 1D return if that's what you are after. Notice also that this is a copy of your original data, not a view.

你当然可以拉这个数组得到1D返回如果这是你想要的。还要注意，这是原始数据的副本，而不是视图。

EDIT You can keep the same general approach, replacing np.ogrid with np.mgrid and reshaping it to get a flat return:

编辑您可以保持相同的通用方法，替换np。ogrid np。mgrid并对其进行重新调整以获得平坦的回报:

>>> idx = np.ravel_multi_index((np.mgrid[1:3,1:3].reshape(2, -1)), (5, 4))
>>> a1[idx]
array([ 22,  33, 222, 333])

#2

I came up with this, and though it doesn't copy ALL of the data, it is still copying data that I don't want into memory. This can probably be improved and I hope there is a better solution out there.

我想到了这个，虽然它没有复制所有的数据，但它仍然在复制我不想要的数据。这可能会得到改进，我希望有更好的解决方案。

zi = 0 
# Create zero array with the appropriate length for the data subset
z = np.zeros((col2 - col1) * (row2 - row1))
# Process number of rows for which data is being extracted
for i in range(row2 - row1):
    # Pull row, then desired elements of that row into buffer
    tmp = ((dataset.variables["z"][(i*cols):((i*cols)+cols)])[col1:col2])
    # Add each item in buffer sequentially to data array
    for j in tmp:
        z[zi] = j 
        # Keep a count of what index position the next data point goes to
        zi += 1

#3

Here a lean proposition

这里一个精益的命题

a1 = np.array([[1,2,3,4],
               [11,22,33,44],
               [111,222,333,444],
               [1111,2222,3333,4444],
               [11111,22222,33333,44444]])

row1 = 1; row2 = 3; ix = slice(row1,row2)
col1 = 1; col2 = 3; iy = slice(col1,col2)
n = (row2-row1)*(col2-col1)

print(a1[ix,iy]);    print()
print(a1[ix,iy].reshape(1,n))
.
[[ 22  33]
 [222 333]]

[[ 22  33 222 333]]

reshape in Python is not expensive, and slice is fast.

在Python中进行重构并不昂贵，而且切片速度很快。

#1