Recall the mathematical notation:
$$L_1 = \left\{x^2 : x \in \{0\ldots 9\}\right\}$$$$L_2 = \left(1, 2, 4, 8,\ldots, 2^{12}\right)$$L12 = []
for x in range(10):
L12.append(x**2)
L12
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
L1 = [x**2 for x in range(10)] # range(n): returns an iterator over the numbers 0,...,n-1
L2 = [2**i for i in range(13)]
print (L1)
print (L2)
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81] [1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096]
List comprehension with conditions
$$M = \left\{x \mid x \in L_1 \text{ if } x \text{ is even}\right\}$$L3 = [x for x in L1 if x % 2 == 0]
print (L3)
[0, 4, 16, 36, 64]
L3 = [x**2 for x in range(10) if x % 2 == 0]
print (L3)
[0, 4, 16, 36, 64]
Nested use of link comprehension
[x for x in [x**2 for x in range(10)] if x % 2 == 0]
[0, 4, 16, 36, 64]
Use of list comprehension for string processing
words = 'The quick brown fox jumps over the lazy dog'.split()
print(words)
upper = [w.upper() for w in words]
print(upper)
upper_lower = [[w.upper(), w.lower()] for w in words]
print(upper_lower)
long_words = [x for x in words if len(x) > 3]
print(long_words)
['The', 'quick', 'brown', 'fox', 'jumps', 'over', 'the', 'lazy', 'dog'] ['THE', 'QUICK', 'BROWN', 'FOX', 'JUMPS', 'OVER', 'THE', 'LAZY', 'DOG'] [['THE', 'the'], ['QUICK', 'quick'], ['BROWN', 'brown'], ['FOX', 'fox'], ['JUMPS', 'jumps'], ['OVER', 'over'], ['THE', 'the'], ['LAZY', 'lazy'], ['DOG', 'dog']] ['quick', 'brown', 'jumps', 'over', 'lazy']
Use list comprehension for obtaining input
s = input('Give numbers separated by comma: ')
x = [int(n) for n in s.split(',')]
print(x)
Give numbers separated by comma: 1,2,3 [1, 2, 3]
Creating vectors and matrices
Create a vector of 10 zeros
z = [0 for i in range(10)]
print(z)
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
M = [[0 for i in range(10)] for j in range(10)]
M
[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
Set the diagonal to 1
for i in range(10): M[i][i] = 1
M
[[1, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 1]]
Create a list of 10 random integers in [0,99]
import random
R = [random.choice(range(100)) for i in range(10)]
print(R)
[7, 47, 56, 71, 80, 30, 44, 40, 32, 15]
Removing elements from a list
Removing elements from a list while you iterate through it can lead to problems. In the example below we wanted to remove all even numbers, but we failed to remove 4 and 8.
L = [1,2,4,5,6,8]
for x in L:
if x%2 == 0:
L.remove(x)
print(L)
[1, 4, 5]
Another way to do this using list comprehension:
We can create a new list with the entries we want to keep
L = [1,2,4,5,6,8]
L = [x for x in L if x%2 == 1] #creates a new list
print(L)
[1, 5]
Or create a list with the entries we want to remove
L = [1,2,4,5,6,8]
R =[y for y in L if y%2 == 0]
for x in R: L.remove(x)
print(L)
[1, 5]
L = [1,2,4,5,6,8]
R =[y for y in L if y%2 == 0]
L = [x for x in L if x not in R]
print(L)
[1, 5]
Using a dictionary in the list comprehension
D = {'A':1,'B':5,'C':4,'D':2}
print([x for x in D if D[x]>2])
['B', 'C']
We can create dictionaries in a similar way as with list comprehension
{str(i):i for i in [1,2,3,4,5]}
{'1': 1, '2': 2, '3': 3, '4': 4, '5': 5}
fruits = ['apple', 'mango', 'banana','cherry']
fl = {f:len(f) for f in fruits}
fl
{'apple': 5, 'mango': 5, 'banana': 6, 'cherry': 6}
f_dict = {f.capitalize():i for i,f in enumerate(fruits)}
print(f_dict)
{'Apple': 0, 'Mango': 1, 'Banana': 2, 'Cherry': 3}
{v:k for k,v in f_dict.items()}
{0: 'Apple', 1: 'Mango', 2: 'Banana', 3: 'Cherry'}
f_dict2 = {i:f.capitalize() for i,f in enumerate(fruits)}
print(f_dict2)
{0: 'Apple', 1: 'Mango', 2: 'Banana', 3: 'Cherry'}
Using the right data structure makes a big difference when handling large data. Dictionaries and sets have expected constant time for finding an element, while lists have linear complexity. Even logarithmic time is significantly faster than linear.
Example
Looking for 100K integers in a collection of 100K integers
import time
import random
L = [random.choice(range(10000000)) for i in range(100000)] # the integers we are searching in a list
S = set(L) # and in a set
Q = [random.choice(range(10000000)) for i in range(100000)] # the list of integers we will look up
#Searching the set
start = time.time()
[x for x in Q if x in S]
end = time.time()
print(end-start)
0.0139923095703125
#Searching the list
start = time.time()
[x for x in Q if x in L]
end = time.time()
print(end - start)
216.76644134521484
Example
You are given a graph in the form of a collection of edges, that is, pairs of vertices
How do you store it in order to be able to quickly answer if there is an edge (x,y) in the graph, and also to get all the neighbors of node?
Create a dictionary with nodes as the keys, and sets with neighboring nodes as values
E = [(1,2),(2,3),(2,5),(2,6),(2,7),(3,4),(3,5),(5,6),(5,7),(7,8),(8,9),(8,10)]
G = {}
for (x,y) in E:
if x not in G:
G[x] = set()
if y not in G:
G[y] = set()
G[x].add(y)
G[y].add(x)
G
{1: {2},
2: {1, 3, 5, 6, 7},
3: {2, 4, 5},
5: {2, 3, 6, 7},
6: {2, 5},
7: {2, 5, 8},
4: {3},
8: {7, 9, 10},
9: {8},
10: {8}}
G[2]
{1, 3, 5, 6, 7}
We will often need to work with randomness. A library for this that is part of the main Python distribution is the random library.
Useful functions:
How do I implement the following?
With probability 0.7 print 'A', with probability 0.2 print 'B', and with probability 0.1 do nothing.
import random
p = random.random()
if p <0.7: print('A')
elif p < 0.9: print('B')
From a list I want to sample k elements where k is a parameter. I want my samples for the different k's to have the property that smaller samples are subsets of bigger samples. That is, the sample of size k+1 will contain one more random element to the sample of size k.
L = [i for i in range(20)]
random.shuffle(L)
sample4 = L[:4]
sample5 = L[:5]
print(sample4)
print(sample5)
[2, 12, 16, 14] [2, 12, 16, 14, 10]