List Methods

Lists are (Mutable) Objects

Lists are objects, and they have a slew of methods that you can call on them:

• lists are mutable …
• many methods actually change the list in place! Let's see how this differs from strings →

>>> s = 'hello'
>>> s.upper()
'HELLO'
>>> print(s)
hello

>>> numbers = [5, 3, 4]
>>> numbers.sort()
>>> print(numbers)
[3, 4, 5]

List Methods Continued

So…. why does this matter?

• most list methods modify the list in place
• most list methods don't return a value!
• that means…
  • if you're getting None…
  • you're probably assigning the return result of a list method to a variable

Some methods that you can call on lists

li = [1, 2, 3]

• li.append(4) # appends element to end of list
• li.extend([3, 4]) # appends all items of one list to the other…
• li.insert(2, "asdf") # insert object before index
• li.remove(4) # removes first occurrence of…
• li.pop() # returns and removes the last element
• li.sort() # sorts list in place
• li.count(1) # returns number of occurrences of arg in list
• li.index(1) # returns the index of the element supplied

Adding Elements

• append(object) - append object to end of list, even another list!
• extend(iterable) - appends all items of one iterable (list, string, etc.) to the original list
• insert(index, object) - insert object before index

Usage notes:

• use append when you want the whole object added to the end of a list →
• use extend when you want every element in another list added as an individual element to the original →
• use insert when you want to add an element to an arbitrary place in the original list →

Removing Elements

• remove(object) - removes first occurrence of object in list
• pop() - returns and removes the last element, takes an optional argument to specify index

Usage notes:

• use remove for finding and removing an element →
• use pop when you want to delete an element at a specific index (by default, the last element)… you also get the element removed as a return value! →

Miscellaneous Methods

• sort() - sorts a list in place →
• count(object) - counts the number of occurrences of object in the original list →
• index(object) - returns the index of the object supplied; causes an error if the object doesn't exist →

Some Exercises

• make the last element in a list the first element
• filter a list of strings
• average

Make the First Element Last

• define a function called last_to_first
• it should take a list as an argument
• it should return a new list with the last element as the first element… and every element shifted up one
  • last_to_first([1, 2, 3, 4] →
• if there's one element or less in the list, return the list as is
• use at least 3 assertions to test your last_to_first function
• try with just slicing and addition
• or try using pop and insert

Make the First Element Last Solution v1

def last_to_first(items):
	""" makes last element in list first, shifts every other element up one """
	if len(items) <= 1:
		return items
	else: 
		return [items[-1]] + items[0:-1]

assert [1] == last_to_first([1]), 'test that one element returns same list'
assert [] == last_to_first([]), 'test that empty list returns empty list'
assert [4, 1, 2, 3] == last_to_first([1, 2, 3, 4]), 'test that all elements shifted by one'

Make the First Element Last Solution v2

def last_to_first(items):
	""" makes last element in list first, shifts every other element up one """
	new_items = items[:]
	if len(items) > 1:
		new_items.insert(0, new_items.pop())
	return new_items


assert [1] == last_to_first([1]), 'test that one element returns same list'
assert [] == last_to_first([]), 'test that empty list returns empty list'
assert [4, 1, 2, 3] == last_to_first([1, 2, 3, 4]), 'test that all elements shifted by one'