如何在 tensorflow 2.0 中重用变量？

Question

When using tensorflow 2.0, I find something weird about tf.Variable? There are two cases bellow.

The first one

x1 = tf.Variable(12., name='x')
x2 = tf.Variable(12., name='x')
print(x1 is x2)
x1.assign(1.)
print(x1)
print(x2)

The output is

False
<tf.Variable 'x:0' shape=() dtype=float32, numpy=1.0>
<tf.Variable 'x:0' shape=() dtype=float32, numpy=12.0>

which means variables with the same name don't share the same memory.

The second one

x = tf.Variable(12., name='x')
print(x)
y = x.assign(5.)
print(y)
print(x is y)

x.assign(3.)
print(x)
print(y)

The output is

<tf.Variable 'x:0' shape=() dtype=float32, numpy=12.0>
<tf.Variable 'UnreadVariable' shape=() dtype=float32, numpy=5.0>
False
<tf.Variable 'x:0' shape=() dtype=float32, numpy=3.0>
<tf.Variable 'UnreadVariable' shape=() dtype=float32, numpy=3.0>

The result is unexpected, variables x and y with different names share the same memory, but id(x) is not equal to id(y) .

Therefore, the name of variable can't distinguish whether variables are identical(share the same memory). And how can I reuse variables in tensorflow 2.0, like with tf.variable_scope("scope", reuse=True) tf.get_variable(...) in tensorflow 1.0?

Answer 1

Quoted from your question:

The result is unexpected, variables x and y with different names share the same memory, but id(x) is not equal to id(y).

No, this is incorrect. From the docs of tf.Variable.assign , where read_value is default to True :

read_value: if True, will return something which evaluates to the new value of the variable; if False will return the assign op.

Here "something" should be a new operation, which isn't x , but is evaluated to the value of x .

To reuse x , just access x :

y = x
print(y is x) # True

Finally, regarding:

which means variables with the same name don't share the same memory.

Therefore, the name of variable can't distinguish whether [...]

You have to create different(thus distinguishable) name s yourself, regarding your first example. You might want to take a look at the comments of this accepted answer https://stackoverflow.com/a/73024334/5290519