如何返回更改后的 Python numpy 数组大小参考参数

Question

这是一个 Python 函数参数传递问题。 我想要一个 Python 函数来调整作为函数引用参数之一的 numpy 数组的大小。

传递的数组的内容似乎在函数内部和外部发生了变化。不知何故，数组对象的更新大小/形状没有从函数中导出，即使我认为 Python 通过引用传递参数。我是 Python 编程的新手，并希望通过引用来更新对象的所有方面。我是否需要明确“导出”更改？

#!/opt/local/bin/python2.7

# Function Test returning changed array
import numpy

def adjust( a1, a2 ) :
  " Adjust passed arrays (my final function will choose which one to adjust from content) "
  print str(a1.shape) + " At start inside function"
  a1[-1,0] = 99
  a1 = numpy.delete(a1, -1, 0)
  print str(a1.shape) + " After delete inside function"
  return None


d1 = numpy.array( [ [ 1,  2,  3],
                    [11, 12, 13],
                    [21, 22, 23],
                    [31, 32, 33]  ] )
d2 = numpy.array( [ [ 9,  8,  7],
                    [19, 18, 17]  ] )

print str(d1.shape) + " At start"
# Let us delete the last row
d1 = numpy.delete(d1, -1, 0)
print str(d1.shape) + " After delete"
# Worked as expected

# So far so good, now do it by object reference parameters in a function......
adjust( d1, d2 )
print d1
print str(d1.shape) + " After function delete return"
# Reference fails to update object properties

不知何故，引用的数组对象没有更新它的大小/形状属性。返回的数组中应该只有 2 行。

(4, 3) At start
(3, 3) After delete
(3, 3) At start inside function
(2, 3) After delete inside function
[[ 1  2  3]
 [11 12 13]
 [99 22 23]]
(3, 3) After function delete return

所以主线/全局代码按预期工作，该函数无法调整大小，但末尾现在删除的行显示了更新的数据。 记住最终函数会选择几个参数中的哪一个来调整，我如何从函数中完全导出参数的更改形状/大小？

Answer 1

https://docs.scipy.org/doc/numpy-1.15.1/reference/generated/numpy.delete.html

返回：“删除了 obj 指定的元素的 arr 副本。请注意，删除不会就地发生。如果 axis 为 None，out 是扁平化数组。”

这意味着delete 不会修改原始数组——它只是制作一个副本。将此与元素分配 a1[-1,0] = 99 进行对比，后者确实修改了数组。我不相信 numpy 出于性能原因允许动态数组调整大小。

至于您的困惑，参数是通过引用传递的（因此a1 的数组最初与d1 的数组相同）。然而，a1 = numpy.delete(a1, -1, 0) 的赋值是重新绑定名称a1，而不是修改它指向的数组，所以d1 没有改变。

如果这没有意义，您应该阅读更多关于变量如何在 Python https://mathieularose.com/python-variables/ 中工作的信息，但基本上，名称（eq a1、d1）是字典键，而对名称的赋值是更改与该键关联的值，这不会影响与该值关联的任何其他键。

Answer 2

另一种解决方案（比 numpy 调用效率低）.....

我确信我在搜索答案时遇到了一些“就地删除”示例，但我当时并不完全理解这可能是我所需要的。 诀窍是确保本地名称（形式参数）不会通过函数中的“=”重新分配给新引用，因此通过这种方式，所有更改都对与传递参数相同的引用（名称）进行（实际参数）到函数。

鉴于名称在赋值 (=) 上的行为方式，Python 的引用调用可能更像是值指针调用。

# Warning Unsafe - array parameter resized in place
# Insure array is not in use on calling
# Requires array to be in C order
def inplace_row_delete( a, r, c=1 ) :
  " Dangerously and inefficiently delete a row in-place on a numpy array (no range checking) "
  # a  reference to the numpy array to have a row deleted from
  #    This routine takes care not to re-assign this in effect "local" name
  #    so as to change the called array in place.
  # r  row to delete in numpy array a
  # c  Nr of rows from r (inclusive) to delete (default 1)

  rows = a.shape[0]
  if (r < 0):
    # support -ve indexing back from end
    r = rows + r
  # Move all the elements after row r, forward one (or c) place(s)
  for index in range(r, rows - c):
    a[index] = a[index + c]

  # Now make the array smaller, disposing of the last (now repeated) element(s)
  # tuples are immutable, but we need to reduce the first element (for the rows in
  # the array) by 1 (c), but keep however many others there are, the same.
  sh    =  list(a.shape)
  sh[0] -= c
  sh    =  tuple(sh)
  # This numpy re-size happens inplace
  # REQUIRES C order
  a.resize(sh, refcheck=False)
  return None