I'd like to assign a variable to the scope of a lambda that is called several times. Each time with a new instance of the variable. How do I do that?
f = lambda x: x + var.x - var.y
# Code needed here to prepare f with a new var
result = f(10)
In this case it's var I'd like to replace for each invocation without making it a second argument.
Variables undefined in the scope of a lambda are resolved from the calling scope at the point where it's called.
A slightly simpler example...
>>> y = 1
>>> f = lambda x: x + y
>>> f(1)
2
>>> y = 2
>>> f(1)
3
...so you just need to set var in the calling scope before calling your lambda, although this is more commonly used in cases where y is 'constant'.
A disassembly of that function reveals...
>>> import dis
>>> dis.dis(f)
1 0 LOAD_FAST 0 (x)
3 LOAD_GLOBAL 0 (y)
6 BINARY_ADD
7 RETURN_VALUE
If you want to bind y to an object at the point of defining the lambda (i.e. creating a closure), it's common to see this idiom...
>>> y = 1
>>> f = lambda x, y=y: x + y
>>> f(1)
2
>>> y = 2
>>> f(1)
2
...whereby changes to y after defining the lambda have no effect.
A disassembly of that function reveals...
>>> import dis
>>> dis.dis(f)
1 0 LOAD_FAST 0 (x)
3 LOAD_FAST 1 (y)
6 BINARY_ADD
7 RETURN_VALUE
f = functools.partial(lambda var, x: x + var.x - var.y, var) will give you a function (f) of one parameter (x) with var fixed at the value it was at the point of definition.
You cannot use the assignment statement in lambda expression, so you'll have to use a regular named function:
def f(x):
global var
var = x
Note the use of the "global" keyword. Without it, Python will assume you want to create a new "var" variable in the local scope of the function.
Make it another parameter:
f = lambda x,var: x + var.x - var.y
result = f(10, var)
Lambda or regular function, the only way you can change a variable in the scope of a function is via an argument.