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Elements are stored as dictionary keys and their counts are stored as dictionary values. >>> c = Counter('abcdeabcdabcaba') # count elements from a string >>> c.most_common(3) # three most common elements [('a', 5), ('b', 4), ('c', 3)] >>> sorted(c) # list all unique elements ['a', 'b', 'c', 'd', 'e'] >>> ''.join(sorted(c.elements())) # list elements with repetitions 'aaaaabbbbcccdde' >>> sum(c.values()) # total of all counts 15 >>> c['a'] # count of letter 'a' 5 >>> for elem in 'shazam': # update counts from an iterable ... c[elem] += 1 # by adding 1 to each element's count >>> c['a'] # now there are seven 'a' 7 >>> del c['b'] # remove all 'b' >>> c['b'] # now there are zero 'b' 0 >>> d = Counter('simsalabim') # make another counter >>> c.update(d) # add in the second counter >>> c['a'] # now there are nine 'a' 9 >>> c.clear() # empty the counter >>> c Counter() Note: If a count is set to zero or reduced to zero, it will remain in the counter until the entry is deleted or the counter is cleared: >>> c = Counter('aaabbc') >>> c['b'] -= 2 # reduce the count of 'b' by two >>> c.most_common() # 'b' is still in, but its count is zero [('a', 3), ('c', 1), ('b', 0)] Nc s t|j|fi|dS)a Create a new, empty Counter object. And if given, count elements from an input iterable. Or, initialize the count from another mapping of elements to their counts. >>> c = Counter() # a new, empty counter >>> c = Counter('gallahad') # a new counter from an iterable >>> c = Counter({'a': 4, 'b': 2}) # a new counter from a mapping >>> c = Counter(a=4, b=2) # a new counter from keyword args N)superr5rp)rrhr3rdrrr55s zCounter.__init__cCsdS)z1The count of elements not in the Counter is zero.r rr$rrr __missing__CszCounter.__missing__cCs t|S)zSum of the counts)sumrUrrrrtotalHs z Counter.totalcCs>|durt|tdddSddl}|j||tddS)zList the n most common elements and their counts from the most common to the least. If n is None, then list all element counts. >>> Counter('abracadabra').most_common(3) [('a', 5), ('b', 2), ('r', 2)] NrHT)r%reverser r%)sortedrTrheapqnlargest)rrNrrrr most_commonLs zCounter.most_commoncCsttt|S)aIterator over elements repeating each as many times as its count. >>> c = Counter('ABCABC') >>> sorted(c.elements()) ['A', 'A', 'B', 'B', 'C', 'C'] # 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 is a negative number, elements() will ignore it. )_chain from_iterable_starmap_repeatrTrrrrelements\szCounter.elementscCstd)Nz@Counter.fromkeys() is undefined. 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Source can be an iterable, a dictionary, or another Counter instance. >>> c = Counter('which') >>> c.subtract('witch') # subtract elements from another iterable >>> c.subtract(Counter('watch')) # subtract elements from another counter >>> c['h'] # 2 in which, minus 1 in witch, minus 1 in watch 0 >>> c['w'] # 1 in which, minus 1 in witch, minus 1 in watch -1 Nr rH)rrjrzrrTsubtractrrrrrs zCounter.subtractcCrc)zReturn a shallow copy.rdrrrrr_rez Counter.copycCs|jt|ffSr)r[rCrrrrrbzCounter.__reduce__cs||vr t|dSdS)zGLike dict.__delitem__() but does not raise KeyError for missing values.N)rr=)rrrdrrr=szCounter.__delitem__c*ttstStfddfDS)z=True if all counts agree. Missing counts are treated as zero.c3s*|]}|D] }||kVqqdSrrrcer2rrrr(z!Counter.__eq__..rjrrsrlrorrrrk zCounter.__eq__cCst|tstS||k S)z@True if any counts disagree. 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