The collection or compound data types str, list and tuple have a sequential structure. You may access any element of the object by its index in the range [0, len(object)-1].
In many cases you probably do not know the meaning of the value with a given index. Does alist[2] give the name or age or street address of an individual?
In Python, a Dictionary is another data collection object that has a built-in mapping type.
A map is an unordered, associative pairing process.
For dictionaries, the mapping, or pairing, is between a key (any immutable object) and a value (any Python data object)
Constructing a dictionary - one way.
d = {} # define 'd' to be a dictionary indicated by the braces
d['Mark'] = 21 # the first key-value pair has the key 'Mark' and value 21
print(d)Activity 1: Write a function called make_dictionary(l) that takes the list l of tuples to generate a dictionary using the names as key words and the ages as values. Use the list below as the argument.
a_list = [('Mark', 21), ('Angelo', 41), ('Wilson', 72), ('Roberta', 61), ('Xie', 19)] The value item in a dictionary can be a dictionary (just like a list may have an element that is a list). Copy and paste the line below into an active code window or pythontutor.com.
d = {'first': 'Bob', 'last': 'White', 'age': 32, 'bday': {'day': 3, 'month': 5, 'year': 1997}}Activity 2: Write a function that takes a string, such as the one below, and returns a dictionary with the structure shown in the line above. The five-digit number at the beginning of the string is the person’s ID number.
info = "03459 Bob White 32 05/03/1997" Activity 3: Modify the code written in the previous activity so that it creates a dictionary of dictionaries. The dictionary entries in the main dictionary uses the person’s five-digit number at the beginning of the string as the person’s dictionary key.
md = {03459 : {'first': 'Bob', 'last': 'White', 'age': 32, 'bday': {'day': 3, 'month': 5, 'year': 1997}}}Delete a key-value pair:
d = {'first': 'Bob', 'last': 'White', 'age': 32, 'bday': {'day': 3, 'month': 5, 'year': 1997}}
del d['age']
print(d) Changing a dictionary value for a given key:
d = {'first': 'Bob', 'last': 'White', 'age': 32, 'bday': {'day': 3, 'month': 5, 'year': 1997}}
d['age'] = 33
print(d)
d['age'] += 1
print(d)keys() - returns a view of a dictionary’s keys, that can be iterated over or converted to a list using the list() function
values() - returns a view of a dictionary’s values, that can be iterated over or converted to a list using the list() function
items() - returns a view of a dictionary’s key-value pairs, that can be iterated over or converted to a list of tuples using the list() function
get(key) - returns the value associated with the parameter key, none otherwise
get(key,_alt_) - returns the value associated with the parameter key, alt otherwise
Use the dictionary below to experiment with the various dictionary methods listed above.
d = {'first': 'Bob', 'last': 'White', 'age': 32, 'bday': {'day': 3, 'month': 5, 'year': 1997}} Activity 4: Using a list comprehension, generate a list of random elements of integers or characters. Then use a dictionary to determine the count of each element in the list.
Dictionaries, like lists, are mutable, so you need to be aware of aliasing.
d1 = {'first': 'Bob', 'last': 'White', 'age': 32, 'bday': {'day': 3, 'month': 5, 'year': 1997}}
d2 = d1
d2['age'] = 33
print(d1['age']) Use the copy() method to avoid unintended results
d1 = {'first': 'Bob', 'last': 'White', 'age': 32, 'bday': {'day': 3, 'month': 5, 'year': 1997}}
d2 = d1.copy()
d2['age'] = 33
print(d1['age'])
d2['bday']['day'] = 10
print(d1['bday]['day'])In Python, a matrix may be thought of as a list of sub-lists, where each sub-list is a row of the matrix, and the items of each sub-list are column entries of the given row. Example are shown below:
m1 = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
m2 = [[1, 2, 3, 4], [5, 6, 7, 8]]Matrices are mutable because they are list of lists, and list are mutable.
m1[2][0] = 1
print(m1) There are common applications that use matrices, and many time the matrix is sparse in that it has few non-zero entries.
m3 = [[0, 3, 0], [1, 0, 0], [0, 0, -2]] A dictionary can provide an efficient way to represent a sparse matrix.
d_m3 = {(0, 1): 3, (1, 0): 1, (2, 2): -2}
m3 = [[0]*3]*3
for akey in d_m3.keys():
m3[akey[0]][akey[1]] = d_m3[akey] Another try…
d_m3 = {(0, 1): 3, (1, 0): 1, (2, 2): -2}
m3 = [[0]*3, [0]*3, [0]*3]
for akey in d_m3.keys():
m3[akey[0]][akey[1]] = d_m3[akey] Coding solution for Assignment 28:
def sparse_fill(d, r, c):
m = []
ec = 0
for row in range(r):
m.append([0]*c)
for akey in d:
if akey[0] >= r and akey[1] < c:
ec = 1
elif akey[0] < r and akey[1] >= c:
ec = 2
elif akey[0] >= r and akey[1] >= c:
ec = 3
else:
m[akey[0]][akey[1]] = d[akey]
return m, ec