如何使python lambda表达式的索引静态

Question

Say I have these lines of code

index = 1
test = lambda t : t[index]+1
index = 0
print(test([5, 0]))

The result is 6, I would expect the result to be 1. How can I make the index inside the lambda expression static without writing t[1]+1 , ie using a variable?

Answer 1

Python's closures are late binding . This means that the values of variables used in closures are looked up at the time the function is called.

To avoid the late binding effect you can use a lambda with a default arg:

index = 1
test = lambda t, index=index: t[index]+1  # binds index at definition time
index = 0
print(test([5, 0]))  # 1

Answer 2

index is neither an argument nor a local variable in test() so it is indeed resolved as aa nonlocal (+> it's looked up in the enclosing scopes).

The simple solution is to make index an argument of test with a default value capturing the value of index at definition time:

index = 1
test = lambda t, _index=index: t[_index]+1
index = 0
print(test([5, 0])) 

Answer 3

You can use a lambda to return a lambda :

test = (lambda index: lambda t: t[index] + 1)(index)

This way the index variable is local to the returned lambda function.

Answer 4

You can use operator.itemgetter to create a callable object which extracts the n th item from a list:

from operator import itemgetter

index = 1
get_val = itemgetter(index)
test = lambda t: get_val(t) +1
index = 0

print(test([5, 0]))  # 1

But there's no reason here for a lambda statement, you can define a function explicitly:

def test(t):
    return get_val(t) + 1