initial commit

Author: dabeaz

src/1/calculating_with_dictionaries/example.py

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

src/1/determine_the_top_n_items_occurring_in_a_list/example.py

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

src/1/extracting_a_subset_of_a_dictionary/example.py

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

src/1/filtering_list_elements/example.py

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

src/1/finding_out_what_two_dictionaries_have_in_common/example.py

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

src/1/finding_the_largest_or_smallest_n_items/example.py

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

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

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

src/1/implementing_a_priority_queue/example.py

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

src/1/keeping_the_last_n_items/example.py

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

src/1/keeping_the_last_n_items/somefile.txt

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