From 783d73fc2f5ca7ca0a2c84380f3ddd6bce01de93 Mon Sep 17 00:00:00 2001 From: Raymond Hettinger Date: Tue, 13 Jan 2009 04:13:53 +0000 Subject: [PATCH] Fixup and simplify docstrings and doctests. --- Lib/collections.py | 30 ++++++++++++++---------------- 1 file changed, 14 insertions(+), 16 deletions(-) diff --git a/Lib/collections.py b/Lib/collections.py index 349f884b46..1d32d67df0 100644 --- a/Lib/collections.py +++ b/Lib/collections.py @@ -173,7 +173,7 @@ class Counter(dict): of elements to their counts. >>> c = Counter() # a new, empty counter - >>> c = Counter('hocus pocus') # a new counter from an iterable + >>> c = Counter('gallahad') # a new counter from an iterable >>> c = Counter({'a': 4, 'b': 2}) # a new counter from a mapping ''' @@ -204,16 +204,16 @@ class Counter(dict): >>> sorted(c.elements()) ['A', 'A', 'B', 'B', 'C', 'C'] - # Knuth's example of prime factors of 1836: 2**2 * 3**3 * 17**1 - >>> import operator - >>> prime_factors = Counter(dict([(2,2), (3,3), (17,1)])) - >>> sorted(prime_factors.elements()) # list individual factors - [2, 2, 3, 3, 3, 17] - >>> reduce(operator.mul, prime_factors.elements(), 1) # multiply them + # Knuth's example for prime factors of 1836: 2**2 * 3**3 * 17**1 + >>> prime_factors = Counter({2: 2, 3: 3, 17: 1}) + >>> product = 1 + >>> for factor in prime_factors.elements(): # loop over factors + ... product *= factor # and multiply them + >>> product 1836 - Note, if an element's count has been set to zero or a negative number, - elements() will ignore it. + Note, if an element's count has been set to zero or is a negative + number, elements() will ignore it. ''' # Emulate Bag.do from Smalltalk and Multiset.begin from C++. @@ -233,15 +233,13 @@ class Counter(dict): def update(self, iterable=None): '''Like dict.update() but add counts instead of replacing them. - Source can be an iterable, a dictionary, or another Counter.instance(). + Source can be an iterable, a dictionary, or another Counter instance. >>> c = Counter('which') - >>> d = Counter('witch') - >>> c.update(d) # add counts from d to those in c - >>> c['h'] # count of 'h' is now three - 3 - >>> c.update('watch') - >>> c['h'] + >>> c.update('witch') # add elements from another iterable + >>> d = Counter('watch') + >>> c.update(d) # add elements from another counter + >>> c['h'] # four 'h' in which, witch, and watch 4 ''' -- 2.50.1