pyrtma.manager.Counter
- class Counter(iterable=None, /, **kwds)[source]
Bases:
dict
Dict subclass for counting hashable items. Sometimes called a bag or multiset. 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)]
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
Methods
Return a shallow copy.
Iterator over elements repeating each as many times as its count.
Create a new dictionary with keys from iterable and values set to value.
Return the value for key if key is in the dictionary, else default.
List the n most common elements and their counts from the most common to the least.
If key is not found, d is returned if given, otherwise KeyError is raised
Remove and return a (key, value) pair as a 2-tuple.
Insert key with a value of default if key is not in the dictionary.
Like dict.update() but subtracts counts instead of replacing them.
Like dict.update() but add counts instead of replacing them.
- __add__(other)[source]
Add counts from two counters.
>>> Counter('abbb') + Counter('bcc') Counter({'b': 4, 'c': 2, 'a': 1})
- clear() None. Remove all items from D.
- elements()[source]
Iterator 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.
- classmethod fromkeys(iterable, v=None)[source]
Create a new dictionary with keys from iterable and values set to value.
- get(key, default=None, /)
Return the value for key if key is in the dictionary, else default.
- items() a set-like object providing a view on D's items
- keys() a set-like object providing a view on D's keys
- most_common(n=None)[source]
List 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)]
- pop(k[, d]) v, remove specified key and return the corresponding value.
If key is not found, d is returned if given, otherwise KeyError is raised
- popitem()
Remove and return a (key, value) pair as a 2-tuple.
Pairs are returned in LIFO (last-in, first-out) order. Raises KeyError if the dict is empty.
- setdefault(key, default=None, /)
Insert key with a value of default if key is not in the dictionary.
Return the value for key if key is in the dictionary, else default.
- subtract(iterable=None, /, **kwds)[source]
Like dict.update() but subtracts counts instead of replacing them. Counts can be reduced below zero. Both the inputs and outputs are allowed to contain zero and negative counts.
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
- update(iterable=None, /, **kwds)[source]
Like dict.update() but add counts instead of replacing them.
Source can be an iterable, a dictionary, or another Counter instance.
>>> c = Counter('which') >>> 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
- values() an object providing a view on D's values