Мультимножество

Python3k

########################################################################

### Counter

########################################################################

def _count_elements(mapping, iterable):

'Tally elements from the iterable.'

mapping_get = mapping.get

for elem in iterable:

mapping[elem] = mapping_get(elem, 0) + 1

try: # Load C helper function if available

from _collections import _count_elements

except ImportError:

pass

class Counter(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

>>> c['a'] # count of letter 'a'

>>> 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'

>>> del c['b'] # remove all 'b'

>>> c['b'] # now there are zero 'b'

>>> d = Counter('simsalabim') # make another counter

>>> c.update(d) # add in the second counter

>>> c['a'] # now there are nine 'a'

>>> 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)]

'''

# References:

# http://en.wikipedia.org/wiki/Multiset

# http://www.gnu.org/software/smalltalk/manual-base/html_node/Bag.html

# http://www.demo2s.com/Tutorial/Cpp/0380__set-multiset/Catalog0380__set-multiset.htm

# http://code.activestate.com/recipes/259174/

# Knuth, TAOCP Vol. II section 4.6.3

def __init__(self, iterable=None, /, **kwds):

'''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

'''

super().__init__()

self.update(iterable, **kwds)

def __missing__(self, key):

'The count of elements not in the Counter is zero.'

# Needed so that self[missing_item] does not raise KeyError

return 0

def total(self):

'Sum of the counts'

return sum(self.values())

def most_common(self, n=None):

'''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)]

'''

# Emulate Bag.sortedByCount from Smalltalk

if n is None:

return sorted(self.items(), key=_itemgetter(1), reverse=True)

# Lazy import to speedup Python startup time

import heapq

return heapq.nlargest(n, self.items(), key=_itemgetter(1))

def elements(self):

'''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

>>> import math

>>> prime_factors = Counter({2: 2, 3: 3, 17: 1})

>>> math.prod(prime_factors.elements())

1836

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++.

return _chain.from_iterable(_starmap(_repeat, self.items()))

# Override dict methods where necessary

@classmethod

def fromkeys(cls, iterable, v=None):

# There is no equivalent method for counters because the semantics

# would be ambiguous in cases such as Counter.fromkeys('aaabbc', v=2).

# Initializing counters to zero values isn't necessary because zero

# is already the default value for counter lookups. Initializing

# to one is easily accomplished with Counter(set(iterable)). For

# more exotic cases, create a dictionary first using a dictionary

# comprehension or dict.fromkeys().

raise NotImplementedError(

'Counter.fromkeys() is undefined. Use Counter(iterable) instead.')

def update(self, iterable=None, /, **kwds):

'''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

'''

# The regular dict.update() operation makes no sense here because the

# replace behavior results in the some of original untouched counts

# being mixed-in with all of the other counts for a mismash that

# doesn't have a straight-forward interpretation in most counting

# contexts. Instead, we implement straight-addition. Both the inputs

# and outputs are allowed to contain zero and negative counts.

if iterable is not None:

if isinstance(iterable, _collections_abc.Mapping):

if self:

self_get = self.get

for elem, count in iterable.items():

self[elem] = count + self_get(elem, 0)

else:

# fast path when counter is empty

super().update(iterable)

else:

_count_elements(self, iterable)

if kwds:

self.update(kwds)

def subtract(self, iterable=None, /, **kwds):

'''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

>>> c['w'] # 1 in which, minus 1 in witch, minus 1 in watch

-1

'''

if iterable is not None:

self_get = self.get

if isinstance(iterable, _collections_abc.Mapping):

for elem, count in iterable.items():

self[elem] = self_get(elem, 0) - count

else:

for elem in iterable:

self[elem] = self_get(elem, 0) - 1

if kwds:

self.subtract(kwds)

def copy(self):

'Return a shallow copy.'

return self.__class__(self)

def __reduce__(self):

return self.__class__, (dict(self),)

def __delitem__(self, elem):

'Like dict.__delitem__() but does not raise KeyError for missing values.'

if elem in self:

super().__delitem__(elem)

def __repr__(self):

if not self:

return f'{self.__class__.__name__}()'

try:

# dict() preserves the ordering returned by most_common()

d = dict(self.most_common())

except TypeError:

# handle case where values are not orderable

d = dict(self)

return f'{self.__class__.__name__}({d!r})'

# Multiset-style mathematical operations discussed in:

# Knuth TAOCP Volume II section 4.6.3 exercise 19

# and at http://en.wikipedia.org/wiki/Multiset

# Outputs guaranteed to only include positive counts.

# To strip negative and zero counts, add-in an empty counter:

# c += Counter()

# Results are ordered according to when an element is first

# encountered in the left operand and then by the order

# encountered in the right operand.

# When the multiplicities are all zero or one, multiset operations

# are guaranteed to be equivalent to the corresponding operations

# for regular sets.

# Given counter multisets such as:

# cp = Counter(a=1, b=0, c=1)

# cq = Counter(c=1, d=0, e=1)

# The corresponding regular sets would be:

# sp = {'a', 'c'}

# sq = {'c', 'e'}

# All of the following relations would hold:

# set(cp + cq) == sp | sq

# set(cp - cq) == sp - sq

# set(cp | cq) == sp | sq

# set(cp & cq) == sp & sq

# (cp == cq) == (sp == sq)

# (cp != cq) == (sp != sq)

# (cp <= cq) == (sp <= sq)

# (cp < cq) == (sp < sq)

# (cp >= cq) == (sp >= sq)

# (cp > cq) == (sp > sq)

def __eq__(self, other):

'True if all counts agree. Missing counts are treated as zero.'

if not isinstance(other, Counter):

return NotImplemented

return all(self[e] == other[e] for c in (self, other) for e in c)

def __ne__(self, other):

'True if any counts disagree. Missing counts are treated as zero.'

if not isinstance(other, Counter):

return NotImplemented

return not self == other

def __le__(self, other):

'True if all counts in self are a subset of those in other.'

if not isinstance(other, Counter):

return NotImplemented

return all(self[e] <= other[e] for c in (self, other) for e in c)

def __lt__(self, other):

'True if all counts in self are a proper subset of those in other.'

if not isinstance(other, Counter):

return NotImplemented

return self <= other and self != other

def __ge__(self, other):

'True if all counts in self are a superset of those in other.'

if not isinstance(other, Counter):

return NotImplemented

return all(self[e] >= other[e] for c in (self, other) for e in c)

def __gt__(self, other):

'True if all counts in self are a proper superset of those in other.'

if not isinstance(other, Counter):

return NotImplemented

return self >= other and self != other

def __add__(self, other):

'''Add counts from two counters.

>>> Counter('abbb') + Counter('bcc')

Counter({'b': 4, 'c': 2, 'a': 1})

'''

if not isinstance(other, Counter):

return NotImplemented

result = Counter()

for elem, count in self.items():

newcount = count + other[elem]

if newcount > 0:

result[elem] = newcount

for elem, count in other.items():

if elem not in self and count > 0:

result[elem] = count

return result

def __sub__(self, other):

''' Subtract count, but keep only results with positive counts.

>>> Counter('abbbc') - Counter('bccd')

Counter({'b': 2, 'a': 1})

'''

if not isinstance(other, Counter):

return NotImplemented

result = Counter()

for elem, count in self.items():

newcount = count - other[elem]

if newcount > 0:

result[elem] = newcount

for elem, count in other.items():

if elem not in self and count < 0:

result[elem] = 0 - count

return result

def __or__(self, other):

'''Union is the maximum of value in either of the input counters.

>>> Counter('abbb') | Counter('bcc')

Counter({'b': 3, 'c': 2, 'a': 1})

'''

if not isinstance(other, Counter):

return NotImplemented

result = Counter()

for elem, count in self.items():

other_count = other[elem]

newcount = other_count if count < other_count else count

if newcount > 0:

result[elem] = newcount

for elem, count in other.items():

if elem not in self and count > 0:

result[elem] = count

return result

def __and__(self, other):

''' Intersection is the minimum of corresponding counts.

>>> Counter('abbb') & Counter('bcc')

Counter({'b': 1})

'''

if not isinstance(other, Counter):

return NotImplemented

result = Counter()

for elem, count in self.items():

other_count = other[elem]

newcount = count if count < other_count else other_count

if newcount > 0:

result[elem] = newcount

return result

def __pos__(self):

'Adds an empty counter, effectively stripping negative and zero counts'

result = Counter()

for elem, count in self.items():

if count > 0:

result[elem] = count

return result

def __neg__(self):

'''Subtracts from an empty counter. Strips positive and zero counts,

and flips the sign on negative counts.

'''

result = Counter()

for elem, count in self.items():

if count < 0:

result[elem] = 0 - count

return result

def _keep_positive(self):

'''Internal method to strip elements with a negative or zero count'''

nonpositive = [elem for elem, count in self.items() if not count > 0]

for elem in nonpositive:

del self[elem]

return self

def __iadd__(self, other):

'''Inplace add from another counter, keeping only positive counts.

>>> c = Counter('abbb')

>>> c += Counter('bcc')

>>> c

Counter({'b': 4, 'c': 2, 'a': 1})

'''

for elem, count in other.items():

self[elem] += count

return self._keep_positive()

def __isub__(self, other):

'''Inplace subtract counter, but keep only results with positive counts.

>>> c = Counter('abbbc')

>>> c -= Counter('bccd')

>>> c

Counter({'b': 2, 'a': 1})

'''

for elem, count in other.items():

self[elem] -= count

return self._keep_positive()

def __ior__(self, other):

'''Inplace union is the maximum of value from either counter.

>>> c = Counter('abbb')

>>> c |= Counter('bcc')

>>> c

Counter({'b': 3, 'c': 2, 'a': 1})

'''

for elem, other_count in other.items():

count = self[elem]

if other_count > count:

self[elem] = other_count

return self._keep_positive()

def __iand__(self, other):

'''Inplace intersection is the minimum of corresponding counts.

>>> c = Counter('abbb')

>>> c &= Counter('bcc')

>>> c

Counter({'b': 1})

'''

for elem, count in self.items():

other_count = other[elem]

if other_count < count:

self[elem] = other_count

return self._keep_positive()

Мультимножество

Report Page