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elements
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'e']
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with repetitions
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    >>> c['a']                          # count of
letter 'a'
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from an iterable
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element's count
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counter
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counter
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nine 'a'
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    >>> c.clear()                       # empty the counter
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    Note:  If a count is set to zero or reduced to zero, it will remain
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counter
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        >>> d = Counter('watch')
        >>> c.update(d)                 # add elements from
another counter
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'h' in which, witch, and watch
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dict.update() but subtracts counts instead of replacing them.
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        >>> c = Counter('which')
        >>> c.subtract('witch')             # subtract
elements from another iterable
        >>> c.subtract(Counter('watch'))    # subtract
elements from another counter
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which, minus 1 in witch, minus 1 in watch
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counts from two counters.

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Subtract count, but keep only results with positive counts.

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Intersection is the minimum of corresponding counts.

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Counter('bcc')
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cC@s|t�S(sEAdds an empty counter, effectively stripping
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