Tuples and more

Question 1

What makes a tuple different from a list?

Answer 1

A tuple is immutable

Tuples are unchangeable and always in the same order

Question 2

What do you need to add if you only have one item in a tuple?

Answer 2

A coma

tuple = (‘one’,)

Question 3

What constructor do you use to make something into a tuple?

Answer 3

tuple()

Question 4

What is an array?

Answer 4

linear data structure
There are 4 array like structures in python
Must be same data type to be an array technically.

Arrays are indexed

Question 5

Show a way to change a tuple item for:

x = (‘apple’, ‘banana’, ‘cherry’)

Answer 5

y = list(x)
y[1] = ‘kiwi’
x = tuple(y)

Question 6

Show a way to add orange to the below
x = (‘apple’, ‘banana’, ‘cherry’)

Answer 6

y = list(x)
y.append(‘orange’)
x = tuple(y)

Question 7

Add two tuples together

Answer 7

thistuple = (“apple”, “banana”, “cherry”)
y = (“orange”,)
thistuple += y

Question 8

What does packing mean
What does unpacking mean

ie: packing a tuple

Answer 8

assigning values to it

extracting values back into variables.
So if you have the list fruits (like below) and you put it’s contents into separate variables (like below) then you just unpacked the list.

fruits = (“apple”, “banana”, “cherry”)

(green, yellow, red) = fruits

print(green)
print(yellow)
print(red)

Question 9

Using the below tuple, place it’s elements into 3 variables, the last variable should contain the remaining elements in a list

tup = (‘apple’, ‘blueberry’, ‘grape’, ‘banana’)

Answer 9

(red, blue, *pellow) = tup

Question 10

Using the below tuple, put the first and last element into a respective variable, throw the ones in the middle into their own variable

tup = (‘apple’, ‘blueberry’, ‘kiwi’, ‘grape’, ‘banana’)

Answer 10

(red, *fruit, yellow) = tup

Question 11

Double all elements of the below tuple:

tup = (‘apple’, ‘blueberry’, ‘kiwi’, ‘grape’, ‘banana’)

Answer 11

mytup = tup * 2

Question 12

Using the below tuple, show how many times ‘apple’ was used.

Show what index apple is in

tup = (‘apple’, ‘blueberry’, ‘kiwi’, ‘grape’, ‘apple’)

Answer 12

tup.count(‘apple’)

tup.index(‘apple’)

Question 13

Name the four principles of a {set}

Answer 13

Unordered
Unchangeable
- you can’t change items, but you can add and remove.
Unindexed
No duplicates

Question 14

What elements are considered exactly the same when placed in a set?

Answer 14

True/1
False/0
So if they’re both in a set together one will be deleted

Question 15

Using the below set, add ‘orange’ to it
thisset = {“apple”, “banana”, “cherry”}

We don’t want ‘orange’ anymore… take it out NOW

Answer 15

thisset.add(‘orange’)

thisset.remove(‘orange’)
thisset.discard(‘orange’)
thisset.pop() <- might get it, might not

Question 16

Add all the items from ‘tropical’ into ‘thisset’.

ALSO, what would it output if ‘tropical’ was a list?

thisset = {“apple”, “banana”, “cherry”}
tropical = {“pineapple”, “mango”, “papaya”}

Answer 16

thisset.update(tropical)

It would still produce a set

Question 17

The set below is trash. Remove its contents

thisset = {“apple”, “banana”, “cherry”}

In fact I don’t want to see this set EVER again. Delete it…

Answer 17

thisset.clear()

del thisset

Question 18

We’re going to be using two sets here and we have four different things we want to accomplish:

Join the sets together into set3, show how you would do more than one

Add set2 to set1

create a new set that only contains items that match from set1 and set2

Update set1 to only contain duplicates from it and set2

update set1 to only contain elements that aren’t present in set2

in a new set, keep only elements not present in both sets
Return a new set that contains only items from the first set that aren’t present in the second

update set1 to keep items that are not present in both set
set1 = {“a”, “b”, “c”}
set2 = {1, 2, 3}

Answer 18

set3 = set1.union(set2, set1)
or
set3 = set1 | set2

set1.update(set2)

set3 = set1.intersection(set2)
or
set3 = set1 & set2

set1.intersection_update(set2)

set3 = set1.difference(set2)
or
set3 = set1 - set2

set1.difference_update(set2)

set3 = set1.symmetric_difference()
or
set3 = set1 ^ set2

set1.symmetric_difference_update(set2)

Question 19

Create a dictionary, and then pull the value for one of the keys. There’s two ways to get a value.

Answer 19

thisdict = {
“brand”: “Ford”,
“model”: “Mustang”,
“year”: 1964
}
x = thisdict[“model”]

x = thisdict.get(‘model)

Question 20

get all the keys for thisdict

Next, get all the values

Answer 20

x = thisdict.keys()

x = thisdict.values()

Question 21

Add a new key and value of
color = white to the below

car = {
“brand”: “Ford”,
“model”: “Mustang”,
“year”: 1964
}

Answer 21

car[‘color’] = ‘white’

or
car.update({‘color’: ‘white’})

Question 22

Get the below dictionary in the form of a list of tuples. Each tuple will have a key and value

car = {
“brand”: “Ford”,
“model”: “Mustang”,
“year”: 1964
}

Answer 22

x = car.items()

Question 23

Using an if statement, see if the key ‘model’ exists in the below dictionary

car = {
“brand”: “Ford”,
“model”: “Mustang”,
“year”: 1964
}

Answer 23

if ‘model’ in car:
print(‘yea’)

Question 24

Change the value of the year below in two different ways

thisdict = {
“brand”: “Ford”,
“model”: “Mustang”,
“year”: 1964
}

Answer 24

thisdict[‘year’] = 2018

or

thisdict.update({‘year’: 2020})

or if you had dictionaries nested:
thisdict[client1].update({‘year’:2020})

Question 25

If you didn’t want the ‘model’ key:value in the below dictionary, what would you do?

How do you remove a random item

remove the dictionaries contents

thisdict = {
“brand”: “Ford”,
“model”: “Mustang”,
“year”: 1964
}

Answer 25

thisdict.pop(‘model’)

or

del thisdict[‘model’]

thisdict.popitem()

thisdict.clear()

Question 26

Loop through the keys of a dictionary

Now do it for the values

Loop through both keys and values

Answer 26

for x in thisdict:
print(x)
or
for x in thisdict.keys():
print(x)

for x in thisdict:
print(thisdict[x])
or
for x in thisdict.values():
print(x)

for x, y in thisdict.items():
print(x, y)

Question 27

Copying a dictionary doesn’t work because the second dictionary will only be a reference to the first and if the first changes those changes apply to the new one.

how would you do a copy properly?

Answer 27

newdict = thisdict.copy()

or

newdict = dict(thisdict)

Question 28

What is a nested dictionary?

If you have three dictionaries, and wanted to combine them into one, how would you do this

get the name of the child2 from your new aggregated dictionary

child1 = {
“name” : “Emil”,
“year” : 2004
}
child2 = {
“name” : “Tobias”,
“year” : 2007
}
child3 = {
“name” : “Linus”,
“year” : 2011
}

Answer 28

child1 = {
“name” : “Emil”,
“year” : 2004
}
child2 = {
“name” : “Tobias”,
“year” : 2007
}
child3 = {
“name” : “Linus”,
“year” : 2011
}

myfamily = {
“child1” : child1,
“child2” : child2,
“child3” : child3
}

print(myfamily[‘child2’][‘name’])