Python lambda's binding to local values

Question

The following code spits out 1 twice, but I expect to see 0 and then 1.

def pv(v) :
  print v

x = []
for v in range(2)         


        

Answer 1

The lambda's closure holds a reference to the variable being used, not its value, so if the value of the variable later changes, the value in the closure also changes. That is, the closure variable's value is resolved when the function is called, not when it is created. (Python's behavior here is not unusual in the functional programming world, for what it's worth.)

There are two solutions:

1. Use a default argument, binding the current value of the variable to a local name at the time of definition. lambda v=v: pv(v)

2. Use a double-lambda and immediately call the first one. (lambda v: lambda: pv(v))(v)

Answer 2

Change x.append(lambda : pv(v)) to x.append(lambda v=v: pv(v)).

You expect "python lambdas to bind to the reference a local variable is pointing to, behind the scene", but that is not how Python works. Python looks up the variable name at the time the function is called, not when it is created. Using a default argument works because default arguments are evaluated when the function is created, not when it is called.

This is not something special about lambdas. Consider:

x = "before foo defined"
def foo():
    print x
x = "after foo was defined"
foo()

prints

after foo was defined