Re-arranging numpy array in place

问题

I am trying to modify a numpy array "in-place". I am interested in re-arranging the array in-place (instead of return:ing a re-arranged version of the array).

Here is an example code:

  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()

The assignment ar=arr[[1,0]] breaks the correspondence of "arr" to the original array passed to the function "modar". You can confirm this by commenting/uncommenting that line.. this happens, of course, as a new array has to be created.

How can I tell python that the new array still corresponds to "arr"?

Simply, how can I make "modar" to rearrange the array "in-place"?

Ok.. I modified that code and replaced "modarr" by:

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

The routine "test2" still gets an unmodified array from "modar".

回答1:

In this case you could do:

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

where the temporary array arr2 is used to store the fancy indexing result. The last line copies the data from arr2 to the original array, keeping the reference.

Note: be sure in your operations that arr2 has the same shape of arr in order to avoid strange results...

回答2:

"For all cases of index arrays, what is returned is a copy of the original data, not a view as one gets for slices."

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

回答3:

Here is a solution with additional playing around. Basically the same as Saullo's.

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()

来源：https://stackoverflow.com/questions/26257067/re-arranging-numpy-array-in-place

标签

python

arrays

numpy

in-place