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Fixup and simplify docstrings and doctests.
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1 changed files with 14 additions and 16 deletions
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@ -173,7 +173,7 @@ class Counter(dict):
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of elements to their counts.
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>>> c = Counter() # a new, empty counter
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>>> c = Counter('hocus pocus') # a new counter from an iterable
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>>> c = Counter('gallahad') # a new counter from an iterable
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>>> c = Counter({'a': 4, 'b': 2}) # a new counter from a mapping
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'''
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@ -204,16 +204,16 @@ class Counter(dict):
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>>> sorted(c.elements())
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['A', 'A', 'B', 'B', 'C', 'C']
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# Knuth's example of prime factors of 1836: 2**2 * 3**3 * 17**1
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>>> import operator
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>>> prime_factors = Counter(dict([(2,2), (3,3), (17,1)]))
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>>> sorted(prime_factors.elements()) # list individual factors
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[2, 2, 3, 3, 3, 17]
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>>> reduce(operator.mul, prime_factors.elements(), 1) # multiply them
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# Knuth's example for prime factors of 1836: 2**2 * 3**3 * 17**1
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>>> prime_factors = Counter({2: 2, 3: 3, 17: 1})
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>>> product = 1
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>>> for factor in prime_factors.elements(): # loop over factors
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... product *= factor # and multiply them
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>>> product
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1836
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Note, if an element's count has been set to zero or a negative number,
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elements() will ignore it.
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Note, if an element's count has been set to zero or is a negative
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number, elements() will ignore it.
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'''
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# Emulate Bag.do from Smalltalk and Multiset.begin from C++.
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@ -233,15 +233,13 @@ class Counter(dict):
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def update(self, iterable=None):
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'''Like dict.update() but add counts instead of replacing them.
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Source can be an iterable, a dictionary, or another Counter.instance().
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Source can be an iterable, a dictionary, or another Counter instance.
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>>> c = Counter('which')
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>>> d = Counter('witch')
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>>> c.update(d) # add counts from d to those in c
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>>> c['h'] # count of 'h' is now three
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3
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>>> c.update('watch')
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>>> c['h']
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>>> c.update('witch') # add elements from another iterable
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>>> d = Counter('watch')
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>>> c.update(d) # add elements from another counter
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>>> c['h'] # four 'h' in which, witch, and watch
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4
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'''
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