Question

我正在尝试修改numpy数组＆＃34;就地＆＃34;。我有兴趣在适当的位置重新排列数组（而不是返回：重新排列的数组版本）。

以下是一个示例代码：

  from numpy import *

  def modar(arr):
    arr=arr[[1,0]] # comment & uncomment this line to get different behaviour
    arr[:,:]=0 
    print "greetings inside modar:"
    print arr

  def test2():
    arr=array([[4,5,6],[1,2,3]])
    print "array before modding"
    print arr
    print
    modar(arr)
    print
    print "array now"
    print arr

  test2()

分配ar = arr [[1,0]]打破了＆＃34; arr＆＃34;的对应关系。将原始数组传递给函数＆＃34; modar＆＃34;。您可以通过注释/取消注释该行来确认这一点。当然，这会发生，因为必须创建一个新数组。

如何告诉python新数组仍然对应于＆＃34; arr＆＃34;？

简单地说，我如何制作＆＃34; modar＆＃34;重新排列阵列＆＃34;就地＆＃34;？

好的..我修改了那段代码并替换了＆＃34; modarr＆＃34;由：

def modar(arr):
  # arr=arr[[1,0]] # comment & uncomment this line to get different behaviour
  # arr[:,:]=0 
  arr2=arr[[1,0]]
  arr=arr2
  print "greetings inside modar:"
  print arr

例程＆＃34; test2＆＃34;仍然从＆＃34; modar＆＃34;。

获得一个未经修改的数组

Answer 1

“对于索引数组的所有情况，返回的是一个副本原始数据，而不是切片的视图。“

http://docs.scipy.org/doc/numpy/user/basics.indexing.html

Answer 2

在这种情况下，你可以这样做：

  arr2 = arr[[1, 0]]
  arr[...] = arr2[...]

其中临时数组arr2用于存储花式索引结果。最后一行将数据从arr2复制到原始数组，保留引用。

注意：请务必在您的操作中arr2具有arr相同的形状，以避免产生奇怪的结果......

Answer 3

这是一个附加游戏的解决方案。基本上和Saullo一样。

from numpy import *

def modar1(arr):
  # arr=arr[[1,0]] # (a)
  arr[:,:]=arr[[1,0]][:,:] # (b)
  print "greetings inside modar:"
  print arr
  # (a) arr is now referring to a new array .. python does not know if it 
  # has the same type / size as the original parameter array 
  # and therefore "arr" does not point to the original parameter array anymore. DOES NOT WORK.
  #
  # (b) explicit copy of each element.  WORKS.

def modar2(arr):
  arr2=arr.copy()
  arr2=arr2[[1,0]]
  # arr=arr2 # (a)
  arr[:,:]=arr2[:,:] # (b)
  print "greetings inside modar:"
  print arr
  # (a) same problem as in modar1
  # .. it seems that *any* reference "arr=.." will point "arr" to something else as than original parameter array
  # and "in-place" modification does not work. DOES NOT WORK
  #
  # (b) does an explicit copying of each array element.  WORKS
  #

def modar3(arr):
  arr2=arr.copy()
  arr2=arr2[[1,0]]
  for i in range(arr.shape[0]):
    arr[i]=arr2[i]
  print "greetings inside modar:"
  print arr
  # this works, as there is no reference "arr=", i.e. to the whole array

def test2():
  #
  # the goal:
  # give an array "arr" to a routine "modar"
  # After calling that routine, "arr" should appear re-arranged
  #
  arr=array([[4,5,6],[1,2,3]])
  print "array before modding"
  print arr
  print
  modar1(arr) # OK
  # modar2(arr) # OK
  # modar3(arr) # OK
  print
  print "array now"
  print arr

test2()

重新安排numpy数组到位

3 个答案: