Lambda in a loop [duplicate]

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Lambda in a loop [duplicate]

Considering the following code snippet:

# directorys == {'login': <object at ...>, 'home': <object at ...>}
for d in directorys:
    self.command["cd " + d] = (lambda : self.root.change_directory(d))

I expect to create a dictionary of two function as following :

# Expected :
self.command == {
    "cd login": lambda: self.root.change_directory("login"),
    "cd home": lambda: self.root.change_directory("home")
}

but it looks like the two lambda function generated are exactly the same :

# Result :
self.command == {
    "cd login": lambda: self.root.change_directory("login"),
    "cd home": lambda: self.root.change_directory("login")   # <- Why login ?
}

I really don’t understand why. Do you have any suggestions ?

Answer #1:

You need to bind d for each function created. One way to do that is to pass it as a parameter with a default value:

lambda d=d: self.root.change_directory(d)

Now the d inside the function uses the parameter, even though it has the same name, and the default value for that is evaluated when the function is created. To help you see this:

lambda bound_d=d: self.root.change_directory(bound_d)

Remember how default values work, such as for mutable objects like lists and dicts, because you are binding an object.

This idiom of parameters with default values is common enough, but may fail if you introspect function parameters and determine what to do based on their presence. You can avoid the parameter with another closure:

(lambda d=d: lambda: self.root.change_directory(d))()
# or
(lambda d: lambda: self.root.change_directory(d))(d)
Answered By: Roger Pate

Answer #2:

This is due to the point at which d is being bound. The lambda functions all point at the variable d rather than the current value of it, so when you update d in the next iteration, this update is seen across all your functions.

For a simpler example:

funcs = []
for x in [1,2,3]:
  funcs.append(lambda: x)
for f in funcs:
  print f()
# output:
3
3
3

You can get around this by adding an additional function, like so:

def makeFunc(x):
  return lambda: x
funcs = []
for x in [1,2,3]:
  funcs.append(makeFunc(x))
for f in funcs:
  print f()
# output:
1
2
3

You can also fix the scoping inside the lambda expression

lambda bound_x=x: bound_x

However in general this is not good practice as you have changed the signature of your function.

Answered By: robbie_c

Answer #3:

Alternatively, instead of lambda, you can use functools.partial which, in my opinion, has a cleaner syntax.

Instead of:

for d in directorys:
    self.command["cd " + d] = (lambda d=d: self.root.change_directory(d))

it will be:

for d in directorys:
    self.command["cd " + d] = partial(self.root.change_directory, d)

Or, here is another simple example:

numbers = [1, 2, 3]
lambdas = [lambda: print(number)
           for number in numbers]
lambdas_with_binding = [lambda number=number: print(number)
                        for number in numbers]
partials = [partial(print, number)
            for number in numbers]
for function in lambdas:
    function()
# 3 3 3
for function in lambdas_with_binding:
    function()
# 1 2 3
for function in partials:
    function()
# 1 2 3
Answered By: Georgy

Answer #4:

I met the same problem. The selected solution helped me a lot, but I consider necessary to add a precision to make functional the code of the question: define the lambda function outside of the loop. By the way, default value is not necessary.

foo = lambda d: lambda : self.root.change_directory(d)
for d in directorys:
    self.command["cd " + d] = (foo(d))
The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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