重新安排numpy阵列

Question

我想修改一个numpy数组“in-place”。我有兴趣重新安排阵列（而不是返回：重新排列的数组版本）。

下面是一个例子代码：

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() 

AR = ARR分配[[1,0]]打破的 “ARR” 原始数组传递给函数 “modar” 的对应关系。您可以通过评论/取消注释该行来确认这一点..当然，这发生了，因为必须创建一个新的数组。

我该如何告诉python新数组仍然对应于“arr”？

简单地说，我怎样才能让“modar”重新排列“in-place”数组？

确定..我修改由该代码和替换 “modarr”：

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 “test2的” 仍然得到从 “modar” 未修饰的阵列。

Answer 1

在这种情况下，你可以这样做：

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

在临时数组arr2用于存储花哨的索引结果。最后一行将数据从arr2复制到原始数组，保留参考。

注意：一定要在你的操作是arr2拥有的，以避免奇怪的结果arr相同的形状......

Answer 2

“对于数组索引的所有情况下，什么是返回是一个副本原始数据的 ，而不是一个获取切片的视图。“

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

Answer 3

这里是额外玩弄的解决方案。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()