相关疑难解决方法(0)

import copy

a = "deepak"
b = 1, 2, 3, 4
c = [1, 2, 3, 4]
d = {1: 10, 2: 20, 3: 30}

a1 = copy.copy(a)
b1 = copy.copy(b)
c1 = copy.copy(c)
d1 = copy.copy(d)


print("immutable - id(a)==id(a1)", id(a) == id(a1))
print("immutable - id(b)==id(b1)", id(b) == id(b1))
print("mutable - id(c)==id(c1)", id(c) == id(c1))
print("mutable - id(d)==id(d1)", id(d) == id(d1))

我得到以下结果 -

immutable - id(a)==id(a1) True
immutable - id(b)==id(b1) True
mutable - id(c)==id(c1) False
mutable - id(d)==id(d1) False

如果我进行深度扫描 …

将数据从数组b复制到数组a的最快方法是什么,而不修改数组a的地址.我需要这个,因为外部库(PyFFTW)使用指向我的数组的指针,该指针无法更改.

例如:

a = numpy.empty(n, dtype=complex)
for i in xrange(a.size):
  a[i] = b[i]

有没有循环可以做到这一点？