Conditional expressions
=======================

.. code:: ipython3

    age = 28
    min_age = 18

**Write an expression which prints ‘You are not allowed in’ if
``age < min_age``:**

.. code:: ipython3

    if age < min_age:
        print('You are not allowed in')


**Copy your code from above to the cell below, then do the following:**
- add a second ``print()`` statement to be executed if
``age < min_age``, which uses string formatting to print ‘Come back in
``<x>`` years’, with an appropriate value for ``x`` - remember f-strings
and the ``.format()`` method - add an ``else`` statement, such that
‘Welcome, ``<name>``’ is printed if ``age >= min_age``.

.. code:: ipython3

    name = 'Dave'

.. code:: ipython3

    if age < min_age:
        print('You are not allowed in')
        print(f'Come back in {min_age - age} years')
        
    else:
        print(f'Welcome, {name}')



.. parsed-literal::

    Welcome, Dave


.. code:: ipython3

    adult = True
    scary = False
    cool = False
    rich = True

**Write a boolean expression which assigns ``True`` to a variable called
``allowed`` if the following conditions are satisfied:**

-  ``adult`` must be ``True``
-  ``scary`` must **not** be ``True``
   and
-  either ``cool`` or ``rich`` must be ``True``

.. code:: ipython3

    allowed = adult and not scary and (cool or rich)
    allowed




.. parsed-literal::

    True



Iteration
---------

.. code:: ipython3

    ages = [17, 22, 14, 55, 23, 19, 16, 21, 33, 42, 25]
    names = ['John', 'Mandy', 'Siri', 'Robert', 'Saul', 'Paulina', 'Fay', 'Mahmood', 'Ben', 'Angela', 'Vicky']
    details = tuple(zip(names, ages))

You don’t need to worry about the code above, but can see that it has
been used to create a list of tuples:

.. code:: ipython3

    details




.. parsed-literal::

    (('John', 17),
     ('Mandy', 22),
     ('Siri', 14),
     ('Robert', 55),
     ('Saul', 23),
     ('Paulina', 19),
     ('Fay', 16),
     ('Mahmood', 21),
     ('Ben', 33),
     ('Angela', 42),
     ('Vicky', 25))



**Use a ``for`` loop to iterate through each tuple in ``details``. For
any element which has a number ``>= min_age``, ``.append()`` the name to
the ``inside`` list; otherwise, add the name to ``refused``:**

.. code:: ipython3

    inside = []
    refused = []

.. code:: ipython3

    for person in details:
        if person[1] >= min_age:
            inside.append(person[0])
        else:
            refused.append(person[0])
            
    print(inside)
    print(refused)


.. parsed-literal::

    ['Mandy', 'Robert', 'Saul', 'Paulina', 'Mahmood', 'Ben', 'Angela', 'Vicky']
    ['John', 'Siri', 'Fay']


**Calculate the percentage of all people who are ``inside`` and print
‘``x``\ % of people are inside’:** - use the built-in ``round()``
function to display the value to one decimal place

.. code:: ipython3

    percent = (len(inside) / (len(inside) + len(refused)) * 100)
    print(f'{round(percent, 1)}% of people are inside')



.. parsed-literal::

    72.7% of people are inside


Run the cell below to ``.clear()`` the ``inside`` and ``outside`` lists:
- Alternatively, assign an empty list to each variable

.. code:: ipython3

    inside.clear() # or inside = []
    refused.clear() # or refused = []

.. code:: ipython3

    capacity = 5
    reserves = []

**Copy your ``for`` loop code from above and update it so that if the
number is ``>=min_age`` but ``len(inside) > capacity``, ``.append()``
the name to the ``reserves`` list:**

.. code:: ipython3

    for person in details:
        if person[1] >= min_age:
            
            if len(inside) >= capacity:            
                reserves.append(person[0])
            
            else:    
                inside.append(person[0])
            
        else:
            refused.append(person[0])
            
    print(inside)
    print(reserves)
    print(refused)


.. parsed-literal::

    ['Mandy', 'Robert', 'Saul', 'Paulina', 'Mahmood']
    ['Ben', 'Angela', 'Vicky']
    ['John', 'Siri', 'Fay']


.. code:: ipython3

    birthdays = ['Saul', 'Mandy']

**For each of the people listed in ``birthdays``, find out their ``age``
from ``details`` and ``print()`` a string in the format ‘``<name>`` is
``<age>`` today!’:**

.. code:: ipython3

    for person in birthdays:
        for entry in details:
            if entry[0] == person:
                print(f'{person} is {entry[1]} today!')



.. parsed-literal::

    Saul is 23 today!
    Mandy is 22 today!


**Can you do the previous task without using ``details``?** - Try using
``.index()`` with ``names`` and ``ages`` defined earlier in the notebook
(run the cell below to see these again)

.. code:: ipython3

    print(names)
    print(ages)


.. parsed-literal::

    ['John', 'Mandy', 'Siri', 'Robert', 'Saul', 'Paulina', 'Fay', 'Mahmood', 'Ben', 'Angela', 'Vicky']
    [17, 22, 14, 55, 23, 19, 16, 21, 33, 42, 25]


.. code:: ipython3

    for person in birthdays:
        idx = names.index(person)
        age = ages[idx]
        print(f'{person} is {age} today!')



.. parsed-literal::

    Saul is 23 today!
    Mandy is 22 today!