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‎Software Architecture with Python/4Performance/Listings.txt‎

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1. common_items.py
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2. plot.py
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3. primes.py
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4. mem_profile_example.py
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5. sub_string.py
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6. objgraph_example.py
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7. rotate.py
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8. sub_string2.py
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9. defaultdict_example.py
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10. ordered.py
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11. hound_words.py
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12. chainmap_example.py
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13. named.py
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14. bloomtest.py

‎Software Architecture with Python/4Performance/big_o.py‎

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# Extra Code
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"""
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Big-O notation image generator. Code which generated the Big-O
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growth graph (order) in the book.
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"""
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import numpy as np
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import math
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import matplotlib.pyplot as plt
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from functools import reduce
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def f(t):
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return np.exp(-t) * np.cos(2*np.pi*t)
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def k(t):
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return [0]
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def l(t):
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return list(map(math.log, t))
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def x(t):
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return list(map(lambda x: x, t))
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def xl(t):
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return list(map(lambda x: math.log(x)*x, t))
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def sq(t):
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return list(map(lambda x: x*x, t))
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def p2(t):
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return list(map(lambda x: pow(2,x), t))
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def fact(n):
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return reduce(lambda x,y : x*y, range(1, n+1))
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def facn(t):
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return list(map(lambda x: fact(x), t))
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t = range(1, 101)
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plt.axis([0, 50, 0, 50])
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line1=plt.plot(t, [1]*100, '#00ee00')
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print(line1)
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plt.text(45, 1.5, r'O(1)')
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line2=plt.plot(t, l(t), color='#00bb00')
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plt.text(40, 4.8, r'O(log(n))')
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line3=plt.plot(t, x(t), color='#008800')
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plt.text(42, 40, r'O((n))')
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line4=plt.plot(t, xl(t), color='#eeee00')
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plt.text(18, 45, r'O(nlog(n))')
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line5=plt.plot(t, sq(t), color='#440000')
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plt.text(8, 44, r'O(n^2)')
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line6=plt.plot(t, p2(t), color='#bb0000')
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plt.text(2, 35, r'O(2^n)')
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line7=plt.plot(t, facn(t), color='#ff0000')
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plt.text(0, 45, r'O(n!)')
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plt.show()

‎Software Architecture with Python/4Performance/bloomtest.py‎

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# Code Listing #12
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"""
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Example of using bloom filter
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NOTE: This works only with Python2.x
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"""
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from pybloom import BloomFilter
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import requests
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# Uncomment for profiling with line profiler or memory profiler
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# @profile
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def hound():
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f=BloomFilter(capacity=100000, error_rate=0.01)
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text=requests.get('https://www.gutenberg.org/files/2852/2852-0.txt').text
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for word in text.split():
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word = word.lower().strip()
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f.add(word)
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print len(f)
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print len(text.split())
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for w in ('holmes','watson','hound','moor','queen'):
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print 'Found',w,w in f
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if __name__ == "__main__":
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hound()
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‎Software Architecture with Python/4Performance/chainmap_example.py‎

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# Code Listing #10
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"""
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Example of using ChainMap
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"""
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from collections import ChainMap
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d1 = {i:i for i in range(100)}
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d2 = {i:i*i for i in range(100)}
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c = ChainMap(d1, d2)
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# Older value still accessible
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print(c[5])
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print(c.maps[0][5])
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# Updating d1 with d2
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d1.update(d2)
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print(d1)
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# Olde value got updated
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print(c[5])
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print(c.maps[0][5])

‎Software Architecture with Python/4Performance/defaultdict_example.py‎

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# Code Listing #7
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"""
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Examples of using defaultdict
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"""
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from collections import defaultdict
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counts = {}
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text="""Python is an interpreted language. Python is an object-oriented language.
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Python is easy to learn. Python is an open source language. """
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word="Python"
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# Implementations with simple dictionary
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for word in text.split():
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word = word.lower().strip()
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try:
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counts[word] += 1
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except KeyError:
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counts[word] = 1
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print("Counts of word",word,'=>',counts[word])
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cities = ['Jakarta','Delhi','Newyork','Bonn','Kolkata','Bangalore','Seoul']
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cities_len = {}
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for city in cities:
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clen = len(city)
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# First create entry
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if clen not in cities_len:
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cities_len[clen] = []
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cities_len[clen].append(city)
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print('Cities grouped by length=>',cities_len)
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# Implementation using default dict
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# 1. Counts
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counts = defaultdict(int)
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for word in text.split():
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word = word.lower().strip()
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# Value is set to 0 and incremented by 1 in one go
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counts[word] += 1
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print("Counts of word",word,'=>',counts[word])
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# 2. Cities grouped by length
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cities = ['Jakarta','Delhi','Newyork','Bonn','Kolkata','Bangalore','Seoul']
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cities_len = defaultdict(list)
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for city in cities:
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# Empty list is created as value and appended to in one go
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cities_len[len(city)].append(city)
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print('Cities grouped by length=>',cities_len)

‎Software Architecture with Python/4Performance/hound_words.py‎

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# Code Listing #9
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"""
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Return top 10 most common words from the online text of the "The Hound of Baskervilles"
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"""
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import requests, operator
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from collections import defaultdict, Counter
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print('Using defaultdict')
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text=requests.get('https://www.gutenberg.org/files/2852/2852-0.txt').text
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freq=defaultdict(int)
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for word in text.split():
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if len(word.strip())==0: continue
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freq[word.lower()] += 1
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print(sorted(freq.items(), key=operator.itemgetter(1), reverse=True)[:10])
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print('Using Counter')
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freq=Counter(filter(None, map(lambda x:x.lower().strip(), text.split())))
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print(freq.most_common(10))

‎Software Architecture with Python/4Performance/mem_profile_example.py‎

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# Code Listing #4
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"""
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Memory profiler example
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"""
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@profile
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def squares(n):
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return [x*x for x in range(1, n+1)]
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squares(1000000)

‎Software Architecture with Python/4Performance/named.py‎

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# Code Listing #11
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"""
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Example of using namedtuple
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"""
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from collections import namedtuple
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Employee = namedtuple('Employee', 'name, age, gender, title, department')
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print(Employee)
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# Create an employee
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jack = Employee('Jack',25,'M','Programmer','Engineering')
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print(jack)
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for field in jack:
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print(field)
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# This will raise an error
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# jack.age=32
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# This works fine
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print(jack._replace(age=32))
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‎Software Architecture with Python/4Performance/objgraph_example.py‎

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# Code Listing #6
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"""
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Objgraph example code.
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Make sure you install graphviz and xdot before running this.
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"""
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import objgraph
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class MyRefClass(object):
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pass
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ref=MyRefClass()
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class C(object):pass
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c_objects=[]
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for i in range(100):
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c=C()
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c.ref=ref
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c_objects.append(c)
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import pdb; pdb.set_trace()
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# Run this code in pdb prompt, you just need to press 'c'
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objgraph.show_backrefs(ref, max_depth=2, too_many=2, filename='refs.png')
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‎Software Architecture with Python/4Performance/ordered.py‎

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# Code Listing #8
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"""
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Examples of using OrderedDict
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"""
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from collections import OrderedDict
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cities = ['Jakarta','Delhi','Newyork','Bonn','Kolkata','Bangalore','Seoul']
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# Dictionary
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cities_dict = dict.fromkeys(cities)
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print(cities_dict)
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# Ordered dictionary
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ocities_dict = OrderedDict.fromkeys(cities)
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print(ocities_dict)
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# Dropping duplicates while preserving order with Ordered dictionary
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cities = ['Jakarta','Delhi','Newyork','Bonn','Kolkata','Bangalore','Bonn','Seoul','Delhi','Jakarta','Mumbai']
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cities_odict = OrderedDict.fromkeys(cities)
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print(cities_odict.keys())
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print(cities_odict.popitem())
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print(cities_odict.popitem(last=False))
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# class LRU
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class LRU(OrderedDict):
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""" Least recently used cache (LRU) dictionary
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using OrderedDict
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"""
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def __init__(self, size=10):
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self.size = size
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def set(self, key):
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# If key is there delete and reinsert so
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# it moves to end.
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if key in self:
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del self[key]
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self[key] = 1
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if len(self)>self.size:
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# Pop from left
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self.popitem(last=False)
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if __name__ == "__main__":
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d=LRU(size=5)
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d.set('bangalore')
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d.set('chennai')
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d.set('mumbai')
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# Bangalore is set again - moves to right
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d.set('bangalore')
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d.set('kolkata')
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d.set('delhi')
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# Chennai is set again - so moves to right
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d.set('chennai')
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print('Length=>',len(d))
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print(d)
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# Kochi is appended and mumbai drops off
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d.set('kochi')
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print(d)

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