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
>>> 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.
'''
# Emulate Bag.do from Smalltalk and Multiset.begin from C++.
return _chain.from_iterable(_starmap(_repeat, self.items()))
# Override dict methods where necessary
python类items()的实例源码
def __repr__(self):
if not self:
return '%s()' % self.__class__.__name__
try:
items = ', '.join(map('%r: %r'.__mod__, self.most_common()))
return '%s({%s})' % (self.__class__.__name__, items)
except TypeError:
# handle case where values are not orderable
return '{0}({1!r})'.format(self.__class__.__name__, dict(self))
# 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()
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 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
>>> 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.
'''
# Emulate Bag.do from Smalltalk and Multiset.begin from C++.
return _chain.from_iterable(_starmap(_repeat, self.items()))
# Override dict methods where necessary
def __repr__(self):
if not self:
return '%s()' % self.__class__.__name__
try:
items = ', '.join(map('%r: %r'.__mod__, self.most_common()))
return '%s({%s})' % (self.__class__.__name__, items)
except TypeError:
# handle case where values are not orderable
return '{0}({1!r})'.format(self.__class__.__name__, dict(self))
# 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()
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 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
>>> 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.
'''
# Emulate Bag.do from Smalltalk and Multiset.begin from C++.
return _chain.from_iterable(_starmap(_repeat, self.items()))
# Override dict methods where necessary
def __repr__(self):
if not self:
return '%s()' % self.__class__.__name__
try:
items = ', '.join(map('%r: %r'.__mod__, self.most_common()))
return '%s({%s})' % (self.__class__.__name__, items)
except TypeError:
# handle case where values are not orderable
return '{0}({1!r})'.format(self.__class__.__name__, dict(self))
# 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()
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 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
>>> 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.
'''
# Emulate Bag.do from Smalltalk and Multiset.begin from C++.
return _chain.from_iterable(_starmap(_repeat, self.items()))
# Override dict methods where necessary
def __repr__(self):
if not self:
return '%s()' % self.__class__.__name__
try:
items = ', '.join(map('%r: %r'.__mod__, self.most_common()))
return '%s({%s})' % (self.__class__.__name__, items)
except TypeError:
# handle case where values are not orderable
return '{0}({1!r})'.format(self.__class__.__name__, dict(self))
# 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()
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 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
>>> 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.
'''
# Emulate Bag.do from Smalltalk and Multiset.begin from C++.
return _chain.from_iterable(_starmap(_repeat, self.items()))
# Override dict methods where necessary
def __repr__(self):
if not self:
return '%s()' % self.__class__.__name__
try:
items = ', '.join(map('%r: %r'.__mod__, self.most_common()))
return '%s({%s})' % (self.__class__.__name__, items)
except TypeError:
# handle case where values are not orderable
return '{0}({1!r})'.format(self.__class__.__name__, dict(self))
# 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()
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 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
>>> 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.
'''
# Emulate Bag.do from Smalltalk and Multiset.begin from C++.
return _chain.from_iterable(_starmap(_repeat, self.items()))
# Override dict methods where necessary
def __repr__(self):
if not self:
return '%s()' % self.__class__.__name__
try:
items = ', '.join(map('%r: %r'.__mod__, self.most_common()))
return '%s({%s})' % (self.__class__.__name__, items)
except TypeError:
# handle case where values are not orderable
return '{0}({1!r})'.format(self.__class__.__name__, dict(self))
# 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()
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
