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Lambda function error while using python3

I have a piece of code that works in Python 2.7 but not in Python3.7. Here I am trying to sort by values of a lambda function.
def get_nearest_available_slot(self):
"""Method to find nearest availability slot in parking
"""
available_slots = filter(lambda x: x.availability, self.slots.values())
if not available_slots:
return None
return sorted(available_slots, key=lambda x: x.slotNum)[0]
The error I get is:
File "/home/xyz/Desktop/parking-lot/parking-lot-1.4.2/parking_lot/bin/source/parking.py", line 45, in get_nearest_available_slot
return sorted(available_slots, key=lambda x: x.slotNum)[0]
IndexError: list index out of range
What am I doing wrong here?

The answer is simple: it's because of how filter works.
In Python 2, filter is eagerly evaluated, which means that once you call it, it returns a list:
filter(lambda x: x % 2 == 0, [1, 2, 3])
Output:
[2]
Conversely, in Python 3, filter is lazily evaluated; it produces an object you can iterate over once, or an iterator:
<filter at 0x110848f98>
In Python 2, the line if not available_slots stops execution if the result of filter is empty, since an empty list evaluates to False.
However, in Python 3, filter returns an iterator, which always evaluates to True, since you cannot tell if an iterator has been exhausted without trying to get the next element, and an iterator has no length. See this for more information.
Because of this, a case exists where an empty iterator gets passed to sorted, producing another empty list. You cannot access the element at position 0 of an empty list, so you get an IndexError.
To fix this, I suggest evaluating the condition strictly. You could do something like this, replacing sorted with min, since we only need one value:
def get_nearest_available_slot(self):
"""Method to find nearest availability slot in parking
"""
available_slots = [slot for slot in self.slots.values() if slot.availability]
if available_slots:
return min(available_slots, key=lambda x: x.slotNum)
else:
return None

Recursion can't append list

Here I use a simplified code to demonstrate what I've encountered:
def recTry(inList):
print(inList)
if len(inList) < 10:
recTry(inList.append(1))
I hope with recursion it'll grow like [1], [1, 1], [1, 1, 1] ...
But actually run that will yield this:
>>> recTry([1])
[1]
None
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 4, in recTry
File "<stdin>", line 3, in recTry
TypeError: object of type 'NoneType' has no len()
inList become None since the second recursion.
So How can I code to make the inList appended with each recursion?

inList.append(1) returns None so you pass None while recursion
what you can do is:
def recTry(inList):
print(inList)
if len(inList) < 10:
inList.append(1)
recTry(inList)
else:
return inList

you try to call function with result of inList.append(1) but it is None
try to:
if len(inList) < 10:
inList.append(1)
recTry(inList)

It is a convention in Python that methods that mutate sequences return None. Thus the value your passing in first returns nothing which is your problem.

On the 4th line of your function,
recTry(inList.append(1))
inList.append(1) returns None after appending 1 to the existing list. This None is what you are passing to recTry method. You can try this instead to make the inList appended with each recursion:
inList.append(1)
recTry(inList)

What do you call the item of list when used as an iterator in a for loop?

I'm not sure how you name the n in the following for loop. Is there are a term for it?
for n in [1,2,3,4,5]:
print i
And, am I correct that the list itself is the iterator of the for loop ?

While n is called a loop variable the list is absolutely not an iterator. It is iterable object, i.e. and iterable, but it is not an iterator. An iterable may be an iterator itself, but not always. That is to say, iterators are iterable, but not all iterables are iterators. In the case of a list it is simply an iterable.
It is an iterable because it implements an __iter__ method, which returns an iterator:
From the Python Glossary an iterable is:
An object capable of returning its members one at a time. Examples of
iterables include all sequence types (such as list, str, and tuple)
and some non-sequence types like dict, file objects, and objects of
any classes you define with an __iter__() or __getitem__() method.
Iterables can be used in a for loop and in many other places where a
sequence is needed (zip(), map(), ...). When an iterable object is
passed as an argument to the built-in function iter(), it returns an
iterator for the object. This iterator is good for one pass over the
set of values. When using iterables, it is usually not necessary to
call iter() or deal with iterator objects yourself. The for statement
does that automatically for you, creating a temporary unnamed variable
to hold the iterator for the duration of the loop.
So, observe:
>>> x = [1,2,3]
>>> iterator = iter(x)
>>> type(iterator)
<class 'list_iterator'>
>>> next(iterator)
1
>>> next(iterator)
2
>>> next(iterator)
3
>>> next(iterator)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
It is illuminating to understand that a for-loop in Python such as the following:
for n in some_iterable:
# do something
is equivalent to:
iterator = iter(some_iterable)
while True:
try:
n = next(iterator)
# do something
except StopIteration as e:
break
Iterators, which are returned by a call to an object's __iter__ method, also implement the __iter__ method (usually returning themselves) but they also implement a __next__ method. Thus, an easy way to check if something is an iterable is to see if it implements a next method
>>> next(x)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'list' object is not an iterator
Again, from the Python Glossary, an iterator is:
An object representing a stream of data. Repeated calls to the
iterator’s __next__() method (or passing it to the built-in function
next()) return successive items in the stream. When no more data are
available a StopIteration exception is raised instead. At this point,
the iterator object is exhausted and any further calls to its
__next__() method just raise StopIteration again. Iterators are required to have an __iter__() method that returns the iterator object
itself so every iterator is also iterable and may be used in most
places where other iterables are accepted. One notable exception is
code which attempts multiple iteration passes. A container object
(such as a list) produces a fresh new iterator each time you pass it
to the iter() function or use it in a for loop. Attempting this with
an iterator will just return the same exhausted iterator object used
in the previous iteration pass, making it appear like an empty
container.
I've illustrated the bevahior of an iterator with the next function above, so now I want to concentrate on the bolded portion.
Basically, an iterator can be used in the place of an iterable because iterators are always iterable. However, an iterator is good for only a single pass. So, if I use a non-iterator iterable, like a list, I can do stuff like this:
>>> my_list = ['a','b','c']
>>> for c in my_list:
... print(c)
...
a
b
c
And this:
>>> for c1 in my_list:
... for c2 in my_list:
... print(c1,c2)
...
a a
a b
a c
b a
b b
b c
c a
c b
c c
>>>
An iterator behaves almost in the same way, so I can still do this:
>>> it = iter(my_list)
>>> for c in it:
... print(c)
...
a
b
c
>>>
However, iterators do not support multiple iteration (well, you can make your an iterator that does, but generally they do not):
>>> it = iter(my_list)
>>> for c1 in it:
... for c2 in it:
... print(c1,c2)
...
a b
a c
Why is that? Well, recall what is happening with the iterator protocol which is used by a for loop under the hood, and consider the following:
>>> my_list = ['a','b','c','d','e','f','g']
>>> iterator = iter(my_list)
>>> iterator_of_iterator = iter(iterator)
>>> next(iterator)
'a'
>>> next(iterator)
'b'
>>> next(iterator_of_iterator)
'c'
>>> next(iterator_of_iterator)
'd'
>>> next(iterator)
'e'
>>> next(iterator_of_iterator)
'f'
>>> next(iterator)
'g'
>>> next(iterator)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
>>> next(iterator_of_iterator)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
>>>
When I used iter() on an iterator, it returned itself!
>>> id(iterator)
139788446566216
>>> id(iterator_of_iterator)
139788446566216

The example you gave is an "iterator-based for-loop"
n is called the loop variable.
The role that list plays is more troublesome to name.
Indeed, after an interesting conversation with #juanpa.arrivillaga I've concluded that there simply isn't a "clearly correct formal name", nor a commonly used name, for that syntactic element.
That being said, I do think that if you referred to it in context in a sentence as "the loop iterator" everyone would know what you meant.
In doing so, you take the risk of confusing yourself or someone else with the fact that the syntactic element in that position is not in fact an iterator, its a collection or (loosely, but from the definition in the referenced article) an "iterable of some sort".
I suspect that one reason why there isn't a name for this is that we hardly ever have to refer to it in a sentence. Another is that they types of element that can appear in that position vary widely, so it is hard to safely cover them all with a label.

What is the purpose of __iter__ returning the iterator object itself?

I don't understand exactly why the __iter__ special method just returns the object it's called on (if it's called on an iterator). Is it essentially just a flag indicating that the object is an iterator?
EDIT: Actually, I discovered that "This is required to allow both containers and iterators to be used with the for and in statements." https://docs.python.org/3/library/stdtypes.html#iterator.iter
Alright, here's how I understand it: When writing a for loop, you're allowed to specify either an iterable or an iterator to loop over. But Python ultimately needs an iterator for the loop, so it calls the __iter__ method on whatever it's given. If it's been given an iterable, the __iter__ method will produce an iterator, and if it's been given an iterator, the __iter__ method will likewise produce an iterator (the original object given).

When you loop over something using for x in something, then the loop actually calls iter(something) first, so it has something to work with. In general, the for loop is approximately equivalent to something like this:
something_iterator = iter(something)
while True:
try:
x = next(something_iterator)
# loop body
except StopIteration:
break
So as you already figured out yourself, in order to be able to loop over an iterator, i.e. when something is already an iterator, iterators should always return themselves when calling iter() on them. So this basically makes sure that iterators are also iterable.

This depends what object you call iter on. If an object is already an iterator, then there is no operation required to convert it to an iterator, because it already is one. But if the object is not an iterator, but is iterable, then an iterator is constructed from the object.
A good example of this is the list object:
>>> x = [1, 2, 3]
>>> iter(x) == x
False
>>> iter(x)
<list_iterator object at 0x7fccadc5feb8>
>>> x
[1, 2, 3]
Lists are iterable, but they are not themselves iterators. The result of list.__iter__ is not the original list.

In Python when ever you try to use loops, or try to iterate over any object like below..
Lets try to understand for list object..
>>> l = [1, 2, 3] # Defined list l
If we iterate over the above list..
>>> for i in l:
... print i
...
1
2
3
When you try to do this iteration over list l, Python for loop checks for l.__iter__() which intern return an iterator object.
>>> for i in l.__iter__():
... print i
...
1
2
3
To understand this more, lets customize the list and create anew list class..
>>> class ListOverride(list):
... def __iter__(self):
... raise TypeError('Not iterable')
...
Here I've created ListOverride class which intern inherited from list and overrided list.__iter__ method to raise TypeError.
>>> ll = ListOverride([1, 2, 3])
>>> ll
[1, 2, 3]
And i've created anew list using ListOverride class, and since it's list object it should iterate in the same way as list does.
>>> for i in ll:
... print i
...
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in __iter__
TypeError: Not iterable
If we try to iterate over ListOverride object ll, we'll endup getting NotIterable exception..

How to use python generator expressions to create a oneliner to run a function multiple times and get a list output

I am wondering if there is there is a simple Pythonic way (maybe using generators) to run a function over each item in a list and result in a list of returns?
Example:
def square_it(x):
return x*x
x_set = [0,1,2,3,4]
squared_set = square_it(x for x in x_set)
I notice that when I do a line by line debug on this, the object that gets passed into the function is a generator.
Because of this, I get an error:
TypeError: unsupported operand type(s) for *: 'generator' and 'generator'
I understand that this generator expression created a generator to be passed into the function, but I am wondering if there is a cool way to accomplish running the function multiple times only by specifying an iterable as the argument? (without modifying the function to expect an iterable).
It seems to me that this ability would be really useful to cut down on lines of code because you would not need to create a loop to fun the function and a variable to save the output in a list.
Thanks!

You want a list comprehension:
squared_set = [square_it(x) for x in x_set]

There's a builtin function, map(), for this common problem.
>>> map(square_it, x_set)
[0,1,4,9,16] # On Python 3, a generator is returned.
Alternatively, one can use a generator expression, which is memory-efficient but lazy (meaning the values will not be computed now, only when needed):
>>> (square_it(x) for x in x_set)
<generator object <genexpr> at ...>
Similarly, one can also use a list comprehension, which computes all the values upon creation, returning a list.
Additionally, here's a comparison of generator expressions and list comprehensions.

You want to call the square_it function inside the generator, not on the generator.
squared_set = (square_it(x) for x in x_set)

As the other answers have suggested, I think it is best (most "pythonic") to call your function explicitly on each element, using a list or generator comprehension.
To actually answer the question though, you can wrap your function that operates over scalers with a function that sniffs the input, and has different behavior depending on what it sees. For example:
>>> import types
>>> def scaler_over_generator(f):
... def wrapper(x):
... if isinstance(x, types.GeneratorType):
... return [f(i) for i in x]
... return f(x)
... return wrapper
>>> def square_it(x):
... return x * x
>>> square_it_maybe_over = scaler_over_generator(square_it)
>>> square_it_maybe_over(10)
100
>>> square_it_maybe_over(x for x in range(5))
[0, 1, 4, 9, 16]
I wouldn't use this idiom in my code, but it is possible to do.
You could also code it up with a decorator, like so:
>>> #scaler_over_generator
... def square_it(x):
... return x * x
>>> square_it(x for x in range(5))
[0, 1, 4, 9, 16]
If you didn't want/need a handle to the original function.

Note that there is a difference between list comprehension returning a list
squared_set = [square_it(x) for x in x_set]
and returning a generator that you can iterate over it:
squared_set = (square_it(x) for x in x_set)