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Current File : /usr/lib/python3.14/collections/__pycache__/__init__.cpython-314.pyc
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* ChainMap     dict-like class for creating a single view of multiple mappings
* Counter      dict subclass for counting hashable objects
* OrderedDict  dict subclass that remembers the order entries were added
* defaultdict  dict subclass that calls a factory function to supply missing values
* UserDict     wrapper around dictionary objects for easier dict subclassing
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* UserString   wrapper around string objects for easier string subclassing

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>>> p[0] + p[1]                     # indexable like a plain tuple
33
>>> x, y = p                        # unpack like a regular tuple
>>> x, y
(11, 22)
>>> p.x + p.y                       # fields also accessible by name
33
>>> d = p._asdict()                 # convert to a dictionary
>>> d['x']
11
>>> Point(**d)                      # convert from a dictionary
Point(x=11, y=22)
>>> p._replace(x=100)               # _replace() is like str.replace() but targets named fields
Point(x=100, y=22)

�,� �_z*Type names and field names must be stringsz6Type names and field names must be valid identifiers: z0Type names and field names cannot be a keyword: z-Field names cannot start with an underscore: z"Encountered duplicate field name: z(Got more default values than field names�, r�c3�*"�TF	qR2x�K	R#5i)z=%rNr)�.0�names& r�	<genexpr>�namedtuple.<locals>.<genexpr>�s���D����s�|��s�r��
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ltRtRtRtRtRtRtRtRtRtRtRtRtRtRtRtRt Rt!Rt"R t#Vt$V;t%#)"ri'aDict 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)]

c�J<�\SV`4VP!V3/VBR#)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)�superrJr�)rr�rDr�s"",�rrJ�Counter.__init__Zs!���	�������H�%��%r)c��^#)z1The count of elements not in the Counter is zero.rr-s&&r�__missing__�Counter.__missing__hs��r)c�4�\VP44#)zSum of the counts)�sumr�rs&r�total�
Counter.totalms���4�;�;�=�!�!r)c���Vf&\VP4\^4RR7#\f^RIs\P	WP4\^4R7#)z�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)]

NT)r.�reverse)r.)�sortedrr��heapq�nlargest�rrws&&r�most_common�Counter.most_commonqsI��
�9��$�*�*�,�K��N�D�I�I��=���~�~�a����;�q�>�~�B�Br)c�f�\P!\\VP	444#)a�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.

)�_chain�
from_iterable�_starmap�_repeatrrs&r�elements�Counter.elements�s"��&�#�#�H�W�d�j�j�l�$C�D�Dr)c��\R4h)z@Counter.fromkeys() is undefined.  Use Counter(iterable) instead.)�NotImplementedError)rBr��vs&&&rr��Counter.fromkeys�s��"�N�P�	Pr)c�:<�Ve{\V\P4'dPV'd8VPpVP	4FwrEWS!V^4,W&K	M\
SV`V4M\W4V'dVP
V4R#R#)a�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

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>>> 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

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>>> Counter('abbb') + Counter('bcc')
Counter({'b': 4, 'c': 2, 'a': 1})

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

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

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Counter({'b': 1})

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

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

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Counter({'b': 1})

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