Today: new data type dict and the dict-count algorithm

Dict - Hash Table - Fast

• Python "dict" type
• A key/value "dictionary"
• In CS generally known as a "hash table"
  Sounds like a real hacker thing
  CS106B!
• Dict is a bit advanced, compared to basic string/list
• Defining features of a dict:
  1. Can get/set a value under a key
  2. Can get/set operations are fast

For more details sees the chapter in the Guide: Dict

Dict Story Arc

• Have: have some big data set, data is not organized, perhaps random order
• Dict:
  Pick out data items we want
  Store each item under a key in the dict
• Done: now the data is organized by key
• The dict is fast doing get/set by key, its defining superpower
• Job interview pattern:
  Interview question has some messed up data
  Best answer inevitably uses a dict to organize the data
  Because the dict is powerful and fast...
  Interviewers cannot resist using it

Dict Restaurant Order Story

Suppose you are out ordering dinner at a restaurant, and the order is proceeding in a chaotic way like this

Alice: I'd like to start with a cup of gazpacho
Bob:   I like beignets for dessert
Alice: Then a ceaser salad
Zoe:   I'll have lasagna
Bob:   Actually two orders of beignets
Alice: Then I'll have tacos
Bob:   And a hot dog
...

People mention the parts of their order piece by piece in no organized order - fine. However, what is needed for the kitchen is to organize each order by person:

Alice: gazpacho, ceasar, tacos
Bob: hot dog, two orders of beignets
...

This is what the dictionary does - data comes in randomly and the dict can organize it by a chosen part of the data, here the name.

Dict Basics

• Organize data around keys
• 1. Set - for a key, store one associated value
  In drawing with an arrow: key -> value
• 2. Get - look up the value for any key

Dict-1 - Set key:value into Dict

• Create empty dict: d = {}
• Set: d[key] = value
• e.g. d['a'] = 'alpha'
• Set creates that key entry in dict if needed
• Overwrites any previous value for that key
  i.e. Each key has one value

>>> d = {}             # Start with empty dict {}
>>> d['a'] = 'alpha'   # Set key/value
>>> d['g'] = 'gamma'
>>> d['b'] = 'beta'
>>> # Now we have built the picture above
>>> # Python can input/output a dict using
>>> # the literal { .. } syntax.
>>> d
{'a': 'alpha', 'g': 'gamma', 'b': 'beta'}
>>>

Dict-2 - Get value out of Dict

• Get a value out of a dict by key
• Get: d[key] - returns the value for that key
• e.g. d['a'] returns 'alpha'
• Note Left/Right:
  On left of = - setting
  On right side - getting
• Handy: +=
  Does a get/set series on the value
  This will be a handy pattern
• e.g. d['a'] += '!!!'
  Equivalent to: d['a'] = d['a'] + '!!!'
  Adds '!!!' to end of that value

>>> s = d['g']         # Get by key
>>> s
'gamma'
>>> d['b']
'beta'
>>> d['a'] = 'apple'   # Overwrite 'a' key
>>> d['a']
'apple'
>>>
>>> # += modify value
>>> d['a'] += '!!!'
>>> d['a']
'apple!!!'
>>>
>>> d
{'a': 'apple!!!', 'g': 'gamma', 'b': 'beta'}
>>>

Dict-3 - Get Error / "in" Test

• There is one big catch
• Problem: get d[key] only works if the key is in the dict
  If the key is not in the dict, get d[key] is an error
• Solution: use in to check if a key is in the dict
• Habit: when you see d[key] .. think if that key is good
• Pattern: before using a key to get a value, in-check the key
  Sort of the "guard" pattern again
• Note: the in check is for keys, not values
  All dict logic is on the key, the value is just stored

>>> # Can initialize dict with literal
>>> d = {'a': 'alpha', 'g': 'gamma', 'b': 'beta'}
>>>
>>> val = d['x']         # Key not in -> Error
Error:KeyError('x',)
>>>
>>> 'a' in d             # "in" key tests
True
>>> 'x' in d
False
>>> 
>>> # Guard pattern (else ..)
>>> if 'x' in d:
      val = d['x']
>>>
>>> # "in" uses key, not value, so this does not work:
>>> 'alpha' in d 
False
>>>

Dict = Memory, Meals Examples

At a high level, the dict is like memory - code can store a piece of information at one time, retrieve it later.

Meals problems: use dict to remember what food was eaten under the keys 'breakfast', 'lunch', 'dinner'. For example meals['breakfast'] = 'apple'`

>>> meals = {}
>>> meals['breakfast'] = 'apple'
>>> meals['lunch'] = 'donut'
>>>
>>> # time passes, other lines run
>>>
>>> # what was lunch again?
>>> meals['lunch']
'donut'
>>> 
>>> # did I have breakfast and dinner yet?
>>> 'breakfast' in meals
True
>>> 'dinner' in meals
False
>>>

Basic Dict Code Examples - Meals

Look at the dict1 "meals" exercises on the experimental server

> dict1 meals exercises

With the "meals" examples, the keys are 'breakfast', 'lunch', 'dinner' and the values are like 'hot dot' and 'bagel'. A key like 'breakfast' may or may not be in the dict, so need to "in" check first.

• bad_start() - check for bad breakfast - return True if no breakfast or if it is 'candy'
• candyish() - check for candy lunch or dinner
• enkale() - if 'candy' for dinner, change it to 'kale'

Inevitable case to think bout: dict[key]

The code can only get the value for a key, if the key is in the dict. Otherwise it's an error. Therefore, need to structure the code with an "in" guard check or something to make sure the key is in the dict before trying to get its value.

1. bad_start()

> bad_start()

bad_start(meals): Given a "meals" dict which contains key/value pairs like 'lunch' -> 'hot dog'. The possible keys are 'breakfast', 'lunch', 'dinner'. Return True if there is no 'breakfast' key in meals, or the value for 'breakfast' is 'candy'. Otherwise return False.

bad_start() Solution Code

Question: is the meals['breakfast'] == 'candy' line safe? Yes. The if-statement guards the [ ].

def bad_start(meals):
    if 'breakfast' not in meals:
        return True
    if meals['breakfast'] == 'candy':
        return True
    return False
    # Can be written with "or" / short-circuiting
    # if 'breakfast' not in meals or meals['breakfast'] == 'candy':

2. enkale()

> enkale()

enkale(meals): Given a "meals" dict which contains key/value pairs like 'lunch' -> 'hot dog'. The possible keys are 'breakfast', 'lunch', 'dinner'. If the key 'dinner' is in the dict with the value 'candy', change the value to 'kale'. Otherwise leave the dict unchanged. Return the dict in all cases.

enkale() Solution Code

Demo: work out the code, see key error

Cannot access meals['dinner'] in the case that dinner is not in the dict, so need logic to avoid that case.

def enkale(meals):
    if 'dinner' in meals and meals['dinner'] == 'candy':
        meals['dinner'] = 'kale'
    return meals

Typical pattern: "in" check guards the meals['dinner'] access, since the short-circuit and only proceeds when the first test is True. Could write it out in this longer form with two if-statements which is ok — works exactly the same as the above and/short-circuit form:

def enkale(meals):
    if 'dinner' in meals:
        if meals['dinner'] == 'candy':
            meals['dinner'] = 'kale'
    return meals

Exercise: is_boring()

> is_boring()

is_boring(meals): Given a "meals" dict. We'll say the meals dict is boring if lunch and dinner are both present and are the same food. Return True if the meals dict is boring, False otherwise.

Dict Observations

Dict Random Order

• Note: the order of the keys in the dict is kind of random
  It is the order they were added
  Simplest to think of it as random
• Key type is typically a str or int (immutable)
• Value type can be anything (str, list, ...)
• The get/set/in logic is on the keys
• The values are just dumb payload - stored, not used by in/[]

"in" Guard Pattern

• Very often see "in" checks just before key access
• Accessing meals['dinner'] = an error if dinner not in the dict
• Therefore: check dinner in meals first
• Only access meals['dinner'] when the key is present
• The and on the following line does this, only proceeding when in is True:
  if 'dinner' in meals and meals['dinner'] == 'candy':
• Aka "short circuiting" of boolean expressions

Key and Value - Different Roles

• Note that get/set/in are all by key
• The key is the control, value is just dumb payload
• Could say key/value are asymmetric, having specialized roles
• YES set by key: d['a'] = 'alpha'
• YES get by key: d['a'] -> 'alpha'
• YES in check of key: 'a' in d -> True
• NO, in check by value: 'alpha' in d
  Does not work
• get/set/in all work with key

Dict vs. List - Keys

• Dict and List both remember things
• What's the difference?
• Keys!
• The "keys" for list are always index numbers
  0, 1, 2, 3, ... len-1
• The "keys" for dict, you choose!
  Any string or int etc. value is a key
  Can set the values in any order
• Can install anything, make that key on the fly...
• e.g. d['dinner'] = 'radish'

Dict-Count Algorithm

• Extremely important dict algorithm pattern
• (Read: we'll use it a lot)
• A "counts" dict:
  We have some big data set
  Store a key for each distinct value in the data
  The value for each key is count of occurrences of that key in the data
• e.g. strs: 'a', 'b', 'a', 'c', 'b'
• Compute output "counts" dict: {'a': 2, 'b': 2, 'c': 1}

Dict Count Code Examples

> dict2 Count exercises

Dict-Count Algorithm Steps

• 1. Start with empty dict counts = {}
• 2. For each str, test: not seen before?
• 3. Not seen before: store key = str, value = 1
• 4. Seen before: key = str, value = value + 1

Dict-Count abacb

Go through these strs
strs = ['a', 'b', 'a',  'c',  'b']

Sketch out counts dict here:

Counts dict ends up as {'a': 2, 'b': 2, 'c': 1}:

• strs: 'a', 'b', 'a', 'c', 'b'
• Each distinct str is a key in the dict
• The value for each key is the number of times it is seen
• Algorithm: loop through all s, update dict with counts as we go
• Each s: seen this before or not?

1. str-count1() - if/else

> str_count1()

str_count1 demo, canonical dict-count algorithm

• Central test of this algorithm: not seen before?
• if/else solution
• Test: not seen before?
  not seen before: counts[s] = 1
  seen before: counts[s] += 1
• This if/else approach is fine, but we'll see another way below
• Demo: write code on board, then fix in next step

str_count1() Solution

def str_count1(strs):
    counts = {}
    for s in strs:
        # s not seen before?
        if s not in counts:
            counts[s] = 1   # first time
        else:
            counts[s] +=1   # every later time
    return counts

2. str-count2() - Unified/Invariant Version, no else

> str_count2()

• A slight unified/invariant improvement on the above code
• Same central test: not seen s before?
• If not seen before - set to zero - aka "fix" dict for that s
  counts[s] = 0
• With fix done, following line works for all cases:
• counts[s] += 1
• "Unified" - 1 line works for all cases ("invariant")
• I have a very slight preference this version
  It's one fewer lines and does not use else
• All counting goes through that one += 1 line

Standard Dict-Count Code - Unified/Invariant Version

def str_count2(strs):
    counts = {}
    for s in strs:
        # fix counts/s if not seen before
        if s not in counts:
            counts[s] = 0
        # Unified: now s is in counts one way or
        # another, so this works for all cases:
        counts[s] += 1
    return counts

Int Count - Exercise

> int_count()

Apply the dict-count algorithm to a list of int values, return a counts dict, counting how many times each int value appears in the list.

Char Count - Exercise

> char_count()

Apply the dict-count algorithm to chars in a string. Build a counts dict of how many times each char, converted to lowercase, appears in a string so 'Coffee' returns {'c': 1, 'o': 1, 'f': 2, 'e': 2}.