Just in case, if someone is looking for a similar problem...
Most solutions given here are one line and are quite readable and simple. Just wanted to add one more that does not need the use of lambda(I am assuming that you are trying to use lambda just for the sake of making it a one line code). Instead, you can use a simple list comprehension.
[print(i) for i in x]
BTW, the return values will be a list of Nones.
Just in case, if someone is looking for a similar problem...
Most solutions given here are one line and are quite readable and simple. Just wanted to add one more that does not need the use of lambda(I am assuming that you are trying to use lambda just for the sake of making it a one line code). Instead, you can use a simple list comprehension.
[print(i) for i in x]
BTW, the return values will be a list of Nones.
Since a for loop is a statement (as is print, in Python 2.x), you cannot include it in a lambda expression. Instead, you need to use the write method on sys.stdout along with the join method.
x = lambda x: sys.stdout.write("\n".join(x) + "\n")
python - Lambda in a loop - Stack Overflow
Issue with creating lambda functions using for loop
Lambda vs For each loops
I still don't understand the benefit of lambda functions
Videos
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)
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.
Hey all,
Can someone explain why when I run the below code:
arr = []
for n in range(1,5):
arr.append(lambda n: 5+n)
I can't run one of the functions in the array by simply calling arr[0]()with empty parenthesis?
it throws an error saying I need to pass one positional argument, n - but I'm passing it using for loop, am I not?
There is evidently something I'm missing.
What makes it worse to me is that ok, if I pass an argument, say I will call art[0](7) - that makes the n in my lambda and in for loop useless, because the value (of n) didn't get passed at all to that lambda n: definition...
Any help would be appreciated!