Practical 3: Foundations (Part 2)

Getting to grips with Dictionaries, LOLs and DOLs

In this notebook we are exploring basic (in the sense of fundamental) data structures so that you understand both how to manage more complex types of data and are prepared for what we will encounter when we start using pandas to perform data analysis. To achieve that, you will need to be ‘fluent’ in “nested” lists and dictionaries; we will focus primarily on lists-of-lists and dictionaries-of-lists, but note that file formats like JSON can be understood as dictionaries-of-dictionaries-of-lists-of-… so this is just a taster of real-world data structures.

Tip

You should download this notebook and then save it to your own copy of the repository. Follow the process used last week (i.e. git add ..., git commit -m "...", git push) right away and then do this again at the end of the class and you’ll have a record of everything you did.

From Lists to Data (Little Steps)

We’re going to start off using lists and dictionaries that we define right at the start of the ‘program’, but the real value of these data structures comes when we build a list or dictionary from data such as a file or a web page… and that’s what we’re going to do below!

First, here’s a reminder of some useful methods (i.e. functions) that apply to lists which we covered in the lecture and practical in Week 2:

Method	Action
`list.count(x)`	Return the number of times x appears in the list
`list.insert(i, x)`	Insert value `x` at a given position `i`
`list.pop([i])`	Remove and return the value at position `i` (`i` is optional)
`list.remove(x)`	Remove the first element from the list whose value is `x`
`list.reverse()`	Reverse the elements of the list in place
`list.sort()`	Sort the items of the list in place
`list.index(x)`	Find the first occurence of `x` in the list
`list[x:y]`	Slice the list from index `x` to `y-1`

This should all be revision… because it’s how we finished things up last week. But I want to go over it briefly again because we’re going to build on it this week.

Hint

As before, ?? will highlight where one or more bit of code are missing and need to be filled in…

List Refresher

Difficulty: Low.

To complete these tasks, all of the methods that you need are listed above, so this is about testing yourself on your understanding both of how to read the help and how to index elements in a list.

The next line creates a list of city names (each element is a string):

cities = ["Bristol", "London", "Manchester", "Edinburgh", "Belfast", "York"]

List Arithmetic

Replace the ?? so that it prints Belfast.

Question

print(cities[?? + 2])

Negative List Arithmetic

Use a negative index to print Belfast:

Question

print(cities[??])

Finding a Position in a List

Replace the ?? so that it prints the index for Manchester in the list.

Question

print("The position of Manchester in the list is: " + str( ?? ))

Looking Across Lists

Connections

This section draws on the LOLs lecture and you will also find Code Camp’s Loops session useful here.

Difficulty: Medium.

Notice that the list of temperatures below is the same length as the list of cities, that’s because these are (roughly) the average temperatures for each city.

cities = ["Bristol", "London", "Manchester", "Edinburgh", "Belfast", "York"]
temperatures = [15.6, 16.5, 13.4, 14.0, 15.2, 14.8]

Lateral Thinking

Given what you know about cities and temperatures, how do you print:

“The average temperature in Manchester is 13.4 degrees”

But you have to do this without doing any of the following:

Using a list index directly (i.e. cities[2] and temperatures[2]) or
Hard-coding the name of the city?

To put it another way, neither of these solutions is the answer:

print("The average temperature in Manchester is " + str(temperatures[2]) + " degrees.")
# ...OR...
city=2
print("The average temperature in " + cities[city] + " is " + str(temperatures[city]) + " degrees.")

Tip

You will need to combine some of the ideas above and also think about the fact that the list index is that we need is the same in both lists… Also, remember that you’ll need to wrap a str(...) around your temperature to make it into a string.

Question

city="Manchester" # Use this to get the solution...

# This way is perfectly fine
print("The average temperature in " + ?? + " is " + str(??))

# This way is more Python 3 and a bit easier to read
print(f"The average temperature in {??} is {??}")

Double-Checking Your Solution

You’ll know that you got the ‘right’ answer to the question above if you can copy+paste your code and change only one thing in order to print out: “The average temperature in Belfast is 15.2 degrees”

Question

city="Belfast"
print(??)

Loops

Now use a for loop over the cities to print out the average temperature in each city:

Question

for c in cities:
    ??

The output should be:

The average temperature in Bristol is 15.6
The average temperature in London is 16.5
The average temperature in Manchester is 13.4
The average temperature in Edinburgh is 14.0
The average temperature in Belfast is 15.2
The average temperature in York is 14.8

Dictionaries

Connections

This section draws on the Dictionaries lecture and Code Camp Dictonaries session.

Difficulty: Low.

Remember that dictionaries (a.k.a. dicts) are like lists in that they are data structures containing multiple elements. A key difference between dictionaries and lists is that while elements in lists are ordered, dicts (in most programming languages, though not Python) are unordered. This means that whereas for lists we use integers as indexes to access elements, in dictonaries we use ‘keys’ (which can multiple different types; strings, integers, etc.). Consequently, an important concept for dicts is that of key-value pairs.

Creating an Atlas

The code below creates an Atlas using a dictionary. The dictionary key is a city name, and the value is the latitude, longitude, and main airport code.

cities = {
    'San Francisco': [37.77, -122.43, 'SFO'],
    'London': [51.51, -0.08, 'LDN'],
    'Paris': [48.86,2.29, 'PAR'],
    'Beijing': [39.92,116.40 ,'BEI'],
}

Adding to a Dict

Add a record to the dictionary for Chennai (data here)

Question

??

Accessing a Dict

In one line of code, print out the airport code for Chennai (MAA):

Question

??

Dealing With Errors

Check you understand the difference between the following two blocks of code by running them.

try:
    print(cities['Berlin'])
except KeyError as e:
    print("Error found")
    print(e)

Error found
'Berlin'

try:
    print(cities.get('Berlin','Not Found'))
except KeyError as e:
    print("Error found")
    print(e)

Not Found

Notice that trying to access a non-existent element of a dict triggers a KeyError, while asking the dict to get the same element does not, it simply returns None. Can you think why, depending on the situtation, either of these might be the ‘correct’ answer?

Thinking Data

This section makes use of both the Dictionaries lecture and the DOLs to Data lecture.

Tip

In this section you’ll need to look up (i.e. Google) and make use of a few new functions that apply to dictionaries: <dictionary>.items(), <dictionary>.keys(). Remember: if in doubt, add print(...) statements to see what is going on!

Iterating over a Dict

Adapting the code below, print out the city name and airport code for every city in our Atlas.

Question

for c in cities.keys():
    print(??)

The output should look something like this:

San Francisco -> SFO
London -> LDN
Paris -> PAR
Beijing -> BEI
Chennai -> MAA

More Complex Dicts

How would your code need to change to produce the same output from this data structure:

cities = {
    'San Francisco': {
        'lat': 37.77, 
        'lon': -122.43,
        'airport': 'SFO'},
    'London': {
        'lat': 51.51, 
        'lon': -0.08, 
        'airport': 'LDN'},
    'Paris': {
        'lat': 48.86,
        'lon': 2.29, 
        'airport': 'PAR'},
    'Beijing': {
        'lat': 39.92,
        'lon': 116.40,
        'airport': 'BEI'},
    'Chennai': { 
        'lat': 13.08, 
        'lon': 80.28, 
        'airport': 'MAA'}
}

Question

for c in cities.keys():
    print(??)

More Dictionary Action!

And how would it need to change to print out the name and latitude of every city?

Question

for c in cities.keys():
    print(??)

The output should be something like this:

San Francisco is at latitude 37.77
London is at latitude 51.51
Paris is at latitude 48.86
Beijing is at latitude 39.92
Chennai is at latitude 13.08

And Another Way to Use a Dict

Now produce the same output using this new data structure:

cities_alt = [
    {'name':     'San Francisco',
     'position': [37.77, -122.43],
     'airport':  'SFO'},
    {'name':     'London',
     'position': [51.51, -0.08],
     'airport':  'LDN'},
    {'name':     'Paris',
     'position': [48.86, 2.29],
     'airport':  'PAR'},
    {'name':     'Beijing',
     'position': [39.92, 116.40],
     'airport':  'BEI'},
    {'name':     'Chennai', 
     'position': [13.08, 80.28],
      'airport': 'MAA'}
]

Question

for c in cities_alt:
    print(??)

The output should be something like this:

San Francisco is at latitude 37.77
London is at latitude 51.51
Paris is at latitude 48.86
Beijing is at latitude 39.92
Chennai is at latitude 13.08

Think Data!

What are some of the main differences that you can think of between cities and cities_alt as data? There is no right answer.

Tip

I just want you to think about these as data! If you were trying to use cities and cities_alt as data what differences would you find when accessing one or more ‘records’?

Point 1 here.
Point 2 here.
Point 3 here.

Add to Git/GitHub

Now follow the same process that you used last week to ensure that your edited notebook is updated in Git and then synchronised with GitHub.