python函数在调用范围内修改变量

问题描述

谁能解释为什么fun1不会修改变量y的值，而fun2可以修改?我需要逐行修改一个数组，但是同时更新y并不是我想要的行为.

Could anyone explain why fun1 doesn't modify the value of the variable y, while fun2 does? I need to modify an array row by row, but updating y at the same time is not the behavior I'm looking for.

def fun1(x):
    x = 2*x
    return x


def fun2(x):
    for i in range(0, x.shape[0]):
        x[i, :] = 2*x[i, :]
    return x


y = np.random.uniform(0, 100, (10, 10))

z1 = fun1(y)
print(np.array(z1 == y).all())
# False

z2 = fun2(y)
print(np.array(z2 == y).all())
# True

推荐答案

修改您的函数以显示对象的id

Modifying your function to show the id of the objects

def fun1(x):
    print(id(x),id(y))
    x = 2*x
    print(id(x))
    return x

In [315]: y = np.arange(3)
In [316]: id(y)
Out[316]: 140296824014768
In [317]: z = fun1(y)
140296824014768 140296824014768
140296823720096
In [318]: id(z)
Out[318]: 140296823720096

因此，由y引用的数组将传递给函数，并且可以由x(参数变量)和y(外部变量)两者引用.但是赋值更改了x引用-该对象被传递回z. y不变.

So the array referenced by y is passed to the function, and can be referenced by both x (the argument variable) and y (the external variable). But the assignment changes the x reference - that object is passed back to z. y is unchanged.

def fun2(x):
    print(id(x), id(y))
    x[0] = 23
    print(id(x))
    return x

使用第二个功能，赋值会更改x的元素，但不会更改所引用对象的id. y，x和z都引用相同的数组.

With this 2nd function, the assignment changes an element of x, but doesn't change the id of the referenced object. y,x and z all reference the same array.

In [320]: y
Out[320]: array([0, 1, 2])
In [321]: id(y)
Out[321]: 140296824014768
In [322]: z = fun2(y)
140296824014768 140296824014768
140296824014768
In [323]: id(z)
Out[323]: 140296824014768
In [324]: z
Out[324]: array([23,  1,  2])
In [325]: y
Out[325]: array([23,  1,  2])

如果在将y复制到函数之前或在函数内部进行复制，则修改x不会修改y.

If we make a copy of y, either before passing it to the function, or inside the function, then modifying x will not modify y.

In [327]: y = np.arange(3)
In [328]: id(y)
Out[328]: 140296823645328
In [329]: z = fun2(y.copy())
140296823647968 140296823645328
140296823647968
In [330]: id(z)
Out[330]: 140296823647968
In [331]: z
Out[331]: array([23,  1,  2])
In [333]: y
Out[333]: array([0, 1, 2])

我们将数组传递给函数的事实并没有改变对副本的需求.即使只是在顶层执行操作，我们也会得到相同的行为.

The fact that we are passing the array to a function doesn't change the need for a copy. We'd get the same behavior even we just performed the action at the top level.

In [334]: y = np.arange(3)
In [335]: x = y.copy()
In [336]: x[:2]=22
In [337]: x
Out[337]: array([22, 22,  2])
In [338]: y
Out[338]: array([0, 1, 2])

---

如果对象是列表，我们将得到相同的行为:

---

We get the same behavior if the object is a list:

In [339]: yl = [1,2,3]
In [340]: fun1(yl)
140296925836360 ...
140296824729096
Out[340]: [1, 2, 3, 1, 2, 3]

In [341]: fun2(yl)
140296925836360 ...
140296925836360
Out[341]: [23, 2, 3]
In [343]: yl
Out[343]: [23, 2, 3]

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