Question

I’m having an issue considering the built-in Python List-methods.

As I learned Python, I always thought Python mutators, as any value class mutators should do, returned the new variable it created.

Take this example:

a = range(5)
# will give [0, 1, 2, 3, 4]
b = a.remove(1)
# as I learned it, b should now be [0, 2, 3, 4]
# what actually happens:
# a = [0, 2, 3, 4]
# b = None

The main problem with this list mutator not returning a new list, is that you cannot to multiple mutations subsequently.
Say I want a list ranging from 0 to 5, without the 2 and the 3.
Mutators returning new variables should be able to do it like this:

a = range(5).remove(2).remove(3)

This sadly isn’t possible, as range(5).remove(2) = None.

Now, is there a way to actually do multiple mutations on lists like I wanna do in my example? I think even PHP allows these types of subsequent mutations with Strings.

I also can’t find a good reference on all the built-in Python functions. If anyone can find the actual definition (with return values) of all the list mutator methods, please let me know. All I can find is this page: http://docs.python.org/tutorial/datastructures.html

Answer 1

Rather than both mutating and returning objects, the Python library chooses to have just one way of using the result of a mutator. From import this:

There should be one– and preferably only one –obvious way to do it.

Having said that, the more usual Python style for what you want to do is using list comprehensions or generator expressions:

[x for x in range(5) if x != 2 and x != 3]

You can also chain these together:

>>> [x for x in (x for x in range(5) if x != 2) if x != 3]
[0, 1, 4]

The above generator expression has the added advantage that it runs in O(n) time because Python only iterates over the range() once. For large generator expressions, and even for infinite generator expressions, this is advantageous.