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‎src/1/calculating_with_dictionaries/example.py

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# example.py
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#
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# Example of calculating with dictionaries
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prices = {
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'ACME': 45.23,
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'AAPL': 612.78,
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'IBM': 205.55,
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'HPQ': 37.20,
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'FB': 10.75
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}
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# Find min and max price
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min_price = min(zip(prices.values(), prices.keys()))
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max_price = max(zip(prices.values(), prices.keys()))
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print('min price:', min_price)
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print('max price:', max_price)
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print('sorted prices:')
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prices_sorted = sorted(zip(prices.values(), prices.keys()))
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for price, name in prices_sorted:
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print(' ', name, price)
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‎src/1/determine_the_top_n_items_occurring_in_a_list/example.py

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# example.py
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#
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# Determine the most common words in a list
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words = [
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'look', 'into', 'my', 'eyes', 'look', 'into', 'my', 'eyes',
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'the', 'eyes', 'the', 'eyes', 'the', 'eyes', 'not', 'around', 'the',
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'eyes', "don't", 'look', 'around', 'the', 'eyes', 'look', 'into',
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'my', 'eyes', "you're", 'under'
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]
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from collections import Counter
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word_counts = Counter(words)
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top_three = word_counts.most_common(3)
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print(top_three)
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# outputs [('eyes', 8), ('the', 5), ('look', 4)]
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# Example of merging in more words
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morewords = ['why','are','you','not','looking','in','my','eyes']
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word_counts.update(morewords)
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print(word_counts.most_common(3))
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‎src/1/extracting_a_subset_of_a_dictionary/example.py

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# example of extracting a subset from a dictionary
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from pprint import pprint
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prices = {
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'ACME': 45.23,
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'AAPL': 612.78,
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'IBM': 205.55,
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'HPQ': 37.20,
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'FB': 10.75
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}
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# Make a dictionary of all prices over 200
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p1 = { key:value for key, value in prices.items() if value > 200 }
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print("All prices over 200")
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pprint(p1)
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# Make a dictionary of tech stocks
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tech_names = { 'AAPL', 'IBM', 'HPQ', 'MSFT' }
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p2 = { key:value for key,value in prices.items() if key in tech_names }
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print("All techs")
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pprint(p2)

‎src/1/filtering_list_elements/example.py

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# Examples of different ways to filter data
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mylist = [1, 4, -5, 10, -7, 2, 3, -1]
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# All positive values
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pos = [n for n in mylist if n > 0]
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print(pos)
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# All negative values
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neg = [n for n in mylist if n < 0]
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print(neg)
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# Negative values clipped to 0
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neg_clip = [n if n > 0 else 0 for n in mylist]
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print(neg_clip)
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# Positive values clipped to 0
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pos_clip = [n if n < 0 else 0 for n in mylist]
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print(pos_clip)
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# Compressing example
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addresses = [
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'5412 N CLARK',
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'5148 N CLARK',
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'5800 E 58TH',
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'2122 N CLARK',
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'5645 N RAVENSWOOD',
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'1060 W ADDISON',
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'4801 N BROADWAY',
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'1039 W GRANVILLE',
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]
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counts = [ 0, 3, 10, 4, 1, 7, 6, 1]
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from itertools import compress
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more5 = [ n > 5 for n in counts ]
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a = list(compress(addresses, more5))
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print(a)
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‎src/1/finding_out_what_two_dictionaries_have_in_common/example.py

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# example.py
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#
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# Find out what two dictionaries have in common
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a = {
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'x' : 1,
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'y' : 2,
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'z' : 3
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}
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b = {
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'w' : 10,
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'x' : 11,
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'y' : 2
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}
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print('Common keys:', a.keys() & b.keys())
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print('Keys in a not in b:', a.keys() - b.keys())
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print('(key,value) pairs in common:', a.items() & b.items())
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‎src/1/finding_the_largest_or_smallest_n_items/example.py

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# example.py
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#
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# Example of using heapq to find the N smallest or largest items
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import heapq
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portfolio = [
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{'name': 'IBM', 'shares': 100, 'price': 91.1},
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{'name': 'AAPL', 'shares': 50, 'price': 543.22},
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{'name': 'FB', 'shares': 200, 'price': 21.09},
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{'name': 'HPQ', 'shares': 35, 'price': 31.75},
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{'name': 'YHOO', 'shares': 45, 'price': 16.35},
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{'name': 'ACME', 'shares': 75, 'price': 115.65}
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]
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cheap = heapq.nsmallest(3, portfolio, key=lambda s: s['price'])
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expensive = heapq.nlargest(3, portfolio, key=lambda s: s['price'])
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print(cheap)
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print(expensive)

‎src/1/grouping-records-together-based-on-a-field/grouping.py

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rows = [
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{'address': '5412 N CLARK', 'date': '07/01/2012'},
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{'address': '5148 N CLARK', 'date': '07/04/2012'},
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{'address': '5800 E 58TH', 'date': '07/02/2012'},
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{'address': '2122 N CLARK', 'date': '07/03/2012'},
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{'address': '5645 N RAVENSWOOD', 'date': '07/02/2012'},
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{'address': '1060 W ADDISON', 'date': '07/02/2012'},
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{'address': '4801 N BROADWAY', 'date': '07/01/2012'},
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{'address': '1039 W GRANVILLE', 'date': '07/04/2012'},
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]
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from itertools import groupby
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rows.sort(key=lambda r: r['date'])
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for date, items in groupby(rows, key=lambda r: r['date']):
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print(date)
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for i in items:
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print(' ', i)
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# Example of building a multidict
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from collections import defaultdict
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rows_by_date = defaultdict(list)
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for row in rows:
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rows_by_date[row['date']].append(row)
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for r in rows_by_date['07/01/2012']:
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print(r)
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‎src/1/implementing_a_priority_queue/example.py

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# example.py
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#
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# Example of a priority queue
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import heapq
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class PriorityQueue:
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def __init__(self):
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self._queue = []
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self._index = 0
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def push(self, item, priority):
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heapq.heappush(self._queue, (-priority, self._index, item))
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self._index += 1
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def pop(self):
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return heapq.heappop(self._queue)[-1]
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# Example use
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class Item:
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def __init__(self, name):
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self.name = name
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def __repr__(self):
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return 'Item({!r})'.format(self.name)
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q = PriorityQueue()
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q.push(Item('foo'), 1)
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q.push(Item('bar'), 5)
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q.push(Item('spam'), 4)
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q.push(Item('grok'), 1)
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print("Should be bar:", q.pop())
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print("Should be spam:", q.pop())
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print("Should be foo:", q.pop())
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print("Should be grok:", q.pop())

‎src/1/keeping_the_last_n_items/example.py

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from collections import deque
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def search(lines, pattern, history=5):
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previous_lines = deque(maxlen=history)
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for line in lines:
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if pattern in line:
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yield line, previous_lines
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previous_lines.append(line)
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# Example use on a file
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if __name__ == '__main__':
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with open('somefile.txt') as f:
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for line, prevlines in search(f, 'python', 5):
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for pline in prevlines:
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print(pline, end='')
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print(line, end='')
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print('-'*20)

‎src/1/keeping_the_last_n_items/somefile.txt

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=== Keeping the Last N Items
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==== Problem
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You want to keep a limited history of the last few items seen
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during iteration or during some other kind of processing.
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==== Solution
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Keeping a limited history is a perfect use for a `collections.deque`.
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For example, the following code performs a simple text match on a
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sequence of lines and prints the matching line along with the previous
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N lines of context when found:
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[source,python]
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----
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from collections import deque
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def search(lines, pattern, history=5):
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previous_lines = deque(maxlen=history)
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for line in lines:
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if pattern in line:
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for pline in previous_lines:
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print(lline, end='')
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print(line, end='')
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print()
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previous_lines.append(line)
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# Example use on a file
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if __name__ == '__main__':
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with open('somefile.txt') as f:
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search(f, 'python', 5)
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----
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==== Discussion
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Using `deque(maxlen=N)` creates a fixed size queue. When new items
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are added and the queue is full, the oldest item is automatically
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removed. For example:
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[source,pycon]
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----
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>>> q = deque(maxlen=3)
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>>> q.append(1)
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>>> q.append(2)
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>>> q.append(3)
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>>> q
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deque([1, 2, 3], maxlen=3)
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>>> q.append(4)
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>>> q
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deque([2, 3, 4], maxlen=3)
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>>> q.append(5)
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>>> q
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deque([3, 4, 5], maxlen=3)
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----
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Although you could manually perform such operations on a list (e.g.,
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appending, deleting, etc.), the queue solution is far more elegant and
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runs a lot faster.
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More generally, a `deque` can be used whenever you need a simple queue
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structure. If you don't give it a maximum size, you get an unbounded
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queue that lets you append and pop items on either end. For example:
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[source,pycon]
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----
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>>> q = deque()
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>>> q.append(1)
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>>> q.append(2)
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>>> q.append(3)
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>>> q
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deque([1, 2, 3])
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>>> q.appendleft(4)
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>>> q
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deque([4, 1, 2, 3])
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>>> q.pop()
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3
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>>> q
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deque([4, 1, 2])
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>>> q.popleft()
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4
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----
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Adding or popping items from either end of a queue has O(1)
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complexity. This is unlike a list where inserting or removing
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items from the front of the list is O(N).

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