ex39a.py

# Exercise 39: Dictionaries, Oh Lovely Dictionaries
# http://learnpythonthehardway.org/book/ex39.html

# create a mapping of state to abbreviation
states = {
    'Oregon': 'OR',
    'Florida': 'FL',
    'California': 'CA',
    'New York': 'NY',
    'Michigan': 'MI'
}

# create a basic set of states and some cities in them
cities = {
    'CA': 'San Francisco',
    'MI': 'Detroit',
    'FL': 'Jacks onville'
}

# add some more cities
cities['NY'] = 'New York'
cities['OR'] = 'Portland'

# print out some cities
print '-' * 10
print "NY State has: ", cities['NY']
print "OR State has: ", cities['OR']

# accessing values in dictionary via keys: dict['key'] returns 'value'
print '-' * 10
print "Michigan's abbreviation is: ", states['Michigan']
print "Florida's abbreviation is: ", states['Florida']

# Because value in one dict can be key in another, nested accessing is possible
# dict1[dict2['key2']] equals:
# value2 = dict2['key2']
# value1 = dict1['value2']
print '-' * 10
print "Michigan has: ", cities[states['Michigan']]
print "Florida has: ", cities[states['Florida']]

# print every state abbreviation
print '-' * 10
for state, abbrev in states.items():  # iterators = keys in dictionary; items() returns dict contents
    print "%s means %s" % (abbrev, state)

# print every city in state
print '-' * 10
for abbrev, city in cities.items():
    print "%s has the city %s" % (abbrev, city)

# now do both at the same time
print '-' * 10
for state, abbrev in states.items():
    print "State %s means %s which contains %s." % (abbrev, state, cities[abbrev])

print '-' * 10
# safely get a abbreviation by state that might not be there
state = states.get('Texas')

if not state:
    print "Sorry, no Texas."

# get a city with a default value
city = cities.get('TX', 'Does Not Exist')
print "The city for the state 'TX' is: %s" % city