[PPT] - Nested Lists and Dictionaries [Andersen, Gries, Lee, Marschner, Van PowerPoint Presentation

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Nested Lists and Dictionaries

Lecture 14

CS 1110:

Introduction to Computing Using Python

[Andersen, Gries, Lee, Marschner, Van Loan, White]

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Prelim = not Thursday or Friday
might be on weekend or next week
Dean of Faculty supports our request to extend

the drop deadline (but no official decision yet)

Announcements

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Tuesday and Thursday: Recursion.
Reading: 5.8-5.10

Next week: Recursion

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Announcements: Lab 8

Lab 8 has been released
Could be useful as extra for loop practice.
Prelim does not depend on any material

introduced in this lab.

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Nested Lists

Lists can hold any objects
Lists are objects
Therefore lists can hold other lists!

x = [1, [2, 1], [1, 4, [3, 1]], 5]

x[0] x[1][1] x[2][2][0] x[2][0] x[1] x[2] x[2][2]

b = [3, 1] c = [1, 4, b] a = [2, 1] x = [1, a, c, 5]

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Two Dimensional Lists

Table of Data Images

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5 4 7 3 4 8 9 7 5 1 2 3 4 1 2 9 6 7 8 0 0 1 2 3 1 4 2 3

Store them as lists of lists (row-major order) d = [[5,4,7,3],[4,8,9,7],[5,1,2,3],[4,1,2,9],[6,7,8,0]]

0 1 2 3 4 5 6 7 8 9 101112 1 2 3 4 5 6 7 8 9 10 11 12

Each row, col has a value Each row, col has an RGB value

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Overview of Two-Dimensional Lists

Access value at row 3, col 2:

d[3][2]

Assign value at row 3, col 2:

d[3][2] = 8

Number of rows of d:
len(d)
Number of cols in row r of d:
len(d[r])

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5 4 7 3 4 8 9 7 5 1 2 3 4 1 2 9 6 7 8 0 d 0 1 2 3 1 4 2 3

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How Multidimensional Lists are Stored

b = [[9, 6, 4], [5, 7, 7]]
b holds id of a one-dimensional list
Has len(b) elements
b[i] holds id of a one-dimensional list
Has len(b[i]) elements

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id2

9 6 4

id3

5 7 7

id1 id2 id3 id1 b

9 6 4 5 7 7

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1 1 2 1 2

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Ragged Lists: Rows w/ Different Length

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b = [[17,13,19],[28,95]]

id2

17 13 19

id3

28 95

id1 id1 b id2 id3

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1 1 2 1

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Slices and Multidimensional Lists

Only “top-level” list is copied.
Contents of the list are not altered
b = [[9, 6], [4, 5], [7, 7]]

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id2

9 6

id1 id2 id3 id1

b

id4 id3

4 5

id4

7 7

x = b[:2]

id5

x

id5 id2 id3

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1 1 1 1 1 2

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Slices and Multidimensional Lists

Create a nested list

>>> b = [[9,6],[4,5],[7,7]]

Get a slice

>>> x = b[:2]

Append to a row of x

>>> x[1].append(10)

What is now in x?

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A: [[9,6,10]] B: [[9,6],[4,5,10]] C: [[9,6],[4,5,10],[7,7]] D: [[9,6],[4,10],[7,7]] E: I don’t know

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Slices and Multidimensional Lists

Create a nested list

>>> b = [[9,6],[4,5],[7,7]]

Get a slice

>>> x = b[:2]

Append to a row of x

>>> x[1].append(10)

x now has nested list

[[9, 6], [4, 5, 10]]

What is now in b?

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A: [[9,6],[4,5],[7,7]] B: [[9,6],[4,5,10]] C: [[9,6],[4,5,10],[7,7]] D: [[9,6],[4,10],[7,7]] E: I don’t know

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Data Wrangling: Transpose

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1 2 3 4 5 6 1 3 5 2 4 6

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Data Wrangling: Transpose

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1 2 3 4 5 6 1 3 5 2 4 6

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Data Wrangling: Transpose

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1 2 3 4 5 6 1 3 5 2 4 6

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Data Wrangling: Transpose

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1 2 3 4 5 6 1 3 5 2 4 6

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Data Wrangling: Transpose

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1 2 3 4 5 6 1 3 5 2 4 6

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1 3 5 2 4 6

Data Wrangling: Transpose

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1 2 3 4 5 6

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1 3 5 2 4 6

Data Wrangling: Transpose

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1 2 3 4 5 6

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1 3 5 2 4 6

Data Wrangling: Transpose

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1 2 3 4 5 6

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1 3 5 2 4 6

Data Wrangling: Transpose

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1 2 3 4 5 6

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Data Wrangling: Transpose

def transpose(table): """Returns: copy of table with rows and columns swapped Precondition: table is a (non-ragged) 2d List""" numrows = len(table) numcols = len(table[0]) # All rows have same no. cols result = [] # Result accumulator for m in range(numcols): row = [] # Single row accumulator for n in range(numrows): row.append(table[n][m]) # Build up row result.append(row) # Add result to table return result

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1 2 3 4 5 6 1 3 5 2 4 6

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Data Wrangling

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1 2 3 1 2 3

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Dictionaries (Type dict)

Description

List of key-value pairs
Keys are unique
Values need not be
Example: net-ids
net-ids are unique (a key)
names need not be (values)
js1 is John Smith (class ’13)
js2 is John Smith (class ’16)

Python Syntax

Create with format:

{k1:v1, k2:v2, …}

Keys must be immutable
ints, floats, bools, strings
Not lists or custom objects
Values can be anything
Example:

d = {'ec1':'Ezra Cornell', 'ec2':'Ezra Cornell', 'ela63':'Erik Andersen'}

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Using Dictionaries (Type dict)

Access elts. like a list
d['ec1'] evaluates to 'Ezra'
But cannot slice ranges!

d = {'ec1':'Ezra','ec2':'Ezra', 'ela63':'Erik'}

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'ela63' id8 'Ezra' 'Ezra' 'Erik' dict 'ec2' 'ec1' id8 d

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Using Dictionaries (Type dict)

Dictionaries are mutable
Can reassign values
d['ec1'] = 'Ellis'

d = {'ec1':'Ezra','ec2':'Ezra', 'ela63':'Erik'}

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'ela63' id8 'Ezra' 'Ezra' 'Erik' dict 'ec2' 'ec1' id8 d

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Using Dictionaries (Type dict)

Dictionaries are mutable
Can reassign values
d['ec1'] = 'Ellis'

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id8 dict id8 d

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'ela63' 'Ezra' 'Ellis' 'Ezra' 'Erik' 'ec2' 'ec1' 

d = {'ec1':'Ezra','ec2':'Ezra', 'ela63':'Erik'}

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id8 dict

Using Dictionaries (Type dict)

Dictionaries are mutable
Can reassign values
d['ec1'] = 'Ellis'
Can add new keys
d['aa1'] = 'Allen'

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id8 d

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'ela63' 'Ezra' 'Ellis' 'Ezra' 'Erik' 'ec2' 'ec1' 

d = {'ec1':'Ezra','ec2':'Ezra', 'ela63':'Erik'}

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id8 dict

Using Dictionaries (Type dict)

Dictionaries are mutable
Can reassign values
d['ec1'] = 'Ellis'
Can add new keys
d['aa1'] = 'Allen'

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'aa1' 'Allen' id8 d

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'ela63' 'Ezra' 'Ellis' 'Ezra' 'Erik' 'ec2' 'ec1' 

d = {'ec1':'Ezra','ec2':'Ezra', 'ela63':'Erik','aa1':'Allen'}

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id8 dict

Using Dictionaries (Type dict)

Dictionaries are mutable
Can reassign values
d['ec1'] = 'Ellis'
Can add new keys
d['aa1'] = 'Allen'
Can delete keys
del d['ela63']

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'aa1' 'Allen' id8 d

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'ela63' 'Ezra' 'Ellis' 'Ezra' 'Erik' 'ec2' 'ec1' 

d = {'ec1':'Ezra','ec2':'Ezra', 'ela63':'Erik','aa1':'Allen'}

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id8 dict

Using Dictionaries (Type dict)

Dictionaries are mutable
Can reassign values
d['ec1'] = 'Ellis'
Can add new keys
d['aa1'] = 'Allen'
Can delete keys
del d['ela63']

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'aa1' 'Allen' id8 d

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'ela63' 'Ezra' 'Ellis' 'Ezra' 'Erik' 'ec2' 'ec1' 

 

Deleting key deletes both

d = {'ec1':'Ezra','ec2':'Ezra', 'aa1':'Allen'}

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New Way to do Test Cases

test_cases = {4:3, 5:3, 6:6, -2:0, 20:7} for item in test_cases: seq = orig[:] # this creates a copy of the list orig print "\tTesting input " + str(seq) + ", " + str(item)

utput = lab08.lesser_than(seq, item)

ct.assert_equals(test_cases[item], output) ct.assert_equals(seq, orig) # make sure argument list wasn't changed

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