Storing Multiple Values in Lists

Overview

Teaching: 30 min
Exercises: 15 min

Questions

• How can I store many values together?

Objectives

• Explain what a list is.

• Create and index lists of simple values.

• Change the values of individual elements

• Append values to an existing list

• Reorder and slice list elements

• Create and manipulate nested lists

• motivation:
  • lots of things to keep in lists
  • built-in in Python standard library, i.e. no need for numpy
  • many things are too trivial for numpy arrays

Python lists

• built-in so no need to load anything
• put values into square brackets, comma-separated

odds = [1, 3, 5, 7]
print('odds are:', odds)

odds are: [1, 3, 5, 7]

• access via indices, same as with arrays

print('first element:', odds[0])
print('last element:', odds[3])
print('"-1" element:', odds[-1])

first element: 1
last element: 7
"-1" element: 7

• negative indices
  • -1 last element
  • -2 second to last, etc.

• odds[3] and odds[-1] point to the same element here.

• important difference:
  • values in lists can be changed
  • characters in strings cannot be changed
  • terminology: mutable vs immutable

names = ['Curie', 'Darwing', 'Turing']  # typo in Darwin's name
print('names is originally:', names)
names[1] = 'Darwin'  # correct the name
print('final value of names:', names)

names is originally: ['Curie', 'Darwing', 'Turing']
final value of names: ['Curie', 'Darwin', 'Turing']

works, but:

name = 'Darwin'
name[0] = 'd'

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-8-220df48aeb2e> in <module>()
      1 name = 'Darwin'
----> 2 name[0] = 'd'

TypeError: 'str' object does not support item assignment

does not.

Ch-Ch-Ch-Ch-Changes

Be careful when modifying data in-place. If two variables refer to the same list, and you modify the list value, it will change for both variables!

salsa = ['peppers', 'onions', 'cilantro', 'tomatoes']
my_salsa = salsa        # <-- my_salsa and salsa point to the *same* list data in memory
salsa[0] = 'hot peppers'
print('Ingredients in my salsa:', my_salsa)

Ingredients in my salsa: ['hot peppers', 'onions', 'cilantro', 'tomatoes']

If you want variables with mutable values to be independent, you must make a copy of the value when you assign it.

salsa = ['peppers', 'onions', 'cilantro', 'tomatoes']
my_salsa = list(salsa)        # <-- makes a *copy* of the list
salsa[0] = 'hot peppers'
print('Ingredients in my salsa:', my_salsa)

Ingredients in my salsa: ['peppers', 'onions', 'cilantro', 'tomatoes']

Because of pitfalls like this, code which modifies data in place can be more difficult to understand. However, it is often far more efficient to modify a large data structure in place than to create a modified copy for every small change. You should consider both of these aspects when writing your code.

Nested Lists

Since a list can contain any Python variables, it can even contain other lists.

For example, we could represent the products in the shelves of a small grocery shop:

x = [['pepper', 'zucchini', 'onion'],
     ['cabbage', 'lettuce', 'garlic'],
     ['apple', 'pear', 'banana']]

Heterogeneous Lists

Lists in Python can contain elements of different types. Example:

sample_ages = [10, 12.5, 'Unknown']

There are many ways to change the contents of lists besides assigning new values to individual elements:

odds.append(11)
print('odds after adding a value:', odds)

odds after adding a value: [1, 3, 5, 7, 11]

removed_element = odds.pop(0)
print('odds after removing the first element:', odds)
print('removed_element:', removed_element)

odds after removing the first element: [3, 5, 7, 11]
removed_element: 1

odds.reverse()
print('odds after reversing:', odds)

odds after reversing: [11, 7, 5, 3]

EXERCISE: Overloading (more modifying)

+ usually means addition, but when used on strings or lists, it means “concatenate”. Given that, what do you think the multiplication operator * does on lists? In particular, what will be the output of the following code?

counts = [2, 4, 6, 8, 10]
repeats = counts * 2
print(repeats)

1. [2, 4, 6, 8, 10, 2, 4, 6, 8, 10]
2. [4, 8, 12, 16, 20]
3. [[2, 4, 6, 8, 10],[2, 4, 6, 8, 10]]
4. [2, 4, 6, 8, 10, 4, 8, 12, 16, 20]

The technical term for this is operator overloading: a single operator, like + or *, can do different things depending on what it’s applied to.

Solution

The multiplication operator * used on a list replicates elements of the list and concatenates them together:

[2, 4, 6, 8, 10, 2, 4, 6, 8, 10]

It’s equivalent to:

counts + counts

While modifying in place, it is useful to remember that Python treats lists in a slightly counter-intuitive way.

As we saw earlier, when we modified the salsa list item in-place, if we make a list, (attempt to) copy it and then modify this list, we can cause all sorts of trouble. This also applies to modifying the list using the above functions:

odds = [3, 5, 7]
primes = odds
primes.append(2)
print('primes:', primes)
print('odds:', odds)

primes: [3, 5, 7, 2]
odds: [3, 5, 7, 2]

This is because Python stores a list in memory, and then can use multiple names to refer to the same list.

• for simple lists: use list() to create a copy

odds = [3, 5, 7]
primes = list(odds)
primes.append(2)
print('primes:', primes)
print('odds:', odds)

primes: [3, 5, 7, 2]
odds: [3, 5, 7]

• the return of the slicer :D
• as with arrays, we can specify ranges to access values

binomial_name = 'Drosophila melanogaster'
group = binomial_name[0:10]
print('group:', group)

species = binomial_name[11:23]
print('species:', species)

chromosomes = ['X', 'Y', '2', '3', '4']
autosomes = chromosomes[2:5]
print('autosomes:', autosomes)

last = chromosomes[-1]
print('last:', last)

group: Drosophila
species: melanogaster
autosomes: ['2', '3', '4']
last: 4

INFO: Slicing From the End

Use slicing to access only the last four characters of a string or entries of a list.

string_for_slicing = 'Observation date: 02-Feb-2013'
list_for_slicing = [['fluorine', 'F'],
                    ['chlorine', 'Cl'],
                    ['bromine', 'Br'],
                    ['iodine', 'I'],
                    ['astatine', 'At']]

'2013'
[['chlorine', 'Cl'], ['bromine', 'Br'], ['iodine', 'I'], ['astatine', 'At']]

Would your solution work regardless of whether you knew beforehand the length of the string or list (e.g. if you wanted to apply the solution to a set of lists of different lengths)? If not, try to change your approach to make it more robust.

Hint: Remember that indices can be negative as well as positive

Solution

Use negative indices to count elements from the end of a container (such as list or string):

string_for_slicing[-4:]
list_for_slicing[-4:]

EXERCISE: Non-Continuous Slices

So far we’ve seen how to use slicing to take single blocks of successive entries from a sequence. But what if we want to take a subset of entries that aren’t next to each other in the sequence?

You can achieve this by providing a third argument to the range within the brackets, called the step size. The example below shows how you can take every third entry in a list:

primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]
subset = primes[0:12:3]
print('subset', subset)

subset [2, 7, 17, 29]

Notice that the slice taken begins with the first entry in the range, followed by entries taken at equally-spaced intervals (the steps) thereafter. If you wanted to begin the subset with the third entry, you would need to specify that as the starting point of the sliced range:

primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]
subset = primes[2:12:3]
print('subset', subset)

subset [5, 13, 23, 37]

Use the step size argument to create a new string that contains only every other character in the string “In an octopus’s garden in the shade”. Start with creating a variable to hold the string:

beatles = "In an octopus's garden in the shade"

What slice of beatles will produce the following output (i.e., the first character, third character, and every other character through the end of the string)?

I notpssgre ntesae

Solution

To obtain every other character you need to provide a slice with the step size of 2:

beatles[0:35:2]

You can also leave out the beginning and end of the slice to take the whole string and provide only the step argument to go every second element:

beatles[::2]

If you want to take a slice from the beginning of a sequence, you can omit the first index in the range:

date = 'Monday 4 January 2016'
day = date[0:6]
print('Using 0 to begin range:', day)
day = date[:6]
print('Omitting beginning index:', day)

Using 0 to begin range: Monday
Omitting beginning index: Monday

And similarly, you can omit the ending index in the range to take a slice to the very end of the sequence:

months = ['jan', 'feb', 'mar', 'apr', 'may', 'jun', 'jul', 'aug', 'sep', 'oct', 'nov', 'dec']
sond = months[8:12]
print('With known last position:', sond)
sond = months[8:len(months)]
print('Using len() to get last entry:', sond)
sond = months[8:]
print('Omitting ending index:', sond)

With known last position: ['sep', 'oct', 'nov', 'dec']
Using len() to get last entry: ['sep', 'oct', 'nov', 'dec']
Omitting ending index: ['sep', 'oct', 'nov', 'dec']

Key Points

• [value1, value2, value3, ...] creates a list.

• Lists can contain any Python object, including lists (i.e., list of lists).

• Lists are indexed and sliced with square brackets (e.g., list[0] and list[2:9]), in the same way as strings and arrays.

• Lists are mutable (i.e., their values can be changed in place).

• Strings are immutable (i.e., the characters in them cannot be changed).