"collections.Counter(string).most_common" Code Answer's
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collections.Counter(string).most_common
Fantastic Frog
on May 04, 2020
# arithmetic operations
c1 = Counter(a=2, b=0, c=-1)
c2 = Counter(a=1, b=-1, c=2)
c = c1 + c2 # return items having +ve count only
print(c) # Counter({'a': 3, 'c': 1})
c = c1 - c2 # keeps only +ve count elements
print(c) # Counter({'a': 1, 'b': 1})
c = c1 & c2 # intersection min(c1[x], c2[x])
print(c) # Counter({'a': 1})
c = c1 | c2 # union max(c1[x], c2[x])
print(c) # Counter({'a': 2, 'c': 2})
Source: www.journaldev.com
collections.Counter(string).most_common
Fantastic Frog
on May 04, 2020
counter = Counter({'Dog': 2, 'Cat': -1, 'Horse': 0})
# most_common()
most_common_element = counter.most_common(1)
print(most_common_element) # [('Dog', 2)]
least_common_element = counter.most_common()[:-2:-1]
print(least_common_element) # [('Cat', -1)]
Source: www.journaldev.com
collections.Counter(string).most_common
Fantastic Frog
on May 04, 2020
# Counter works with non-numbers too
special_counter = Counter(name='Pankaj', age=20)
print(special_counter) # Counter({'name': 'Pankaj', 'age': 20})
Source: www.journaldev.com
collections.Counter(string).most_common
Fantastic Frog
on May 04, 2020
# Iterable as argument for Counter
counter = Counter('abc')
print(counter) # Counter({'a': 1, 'b': 1, 'c': 1})
# List as argument to Counter
words_list = ['Cat', 'Dog', 'Horse', 'Dog']
counter = Counter(words_list)
print(counter) # Counter({'Dog': 2, 'Cat': 1, 'Horse': 1})
# Dictionary as argument to Counter
word_count_dict = {'Dog': 2, 'Cat': 1, 'Horse': 1}
counter = Counter(word_count_dict)
print(counter) # Counter({'Dog': 2, 'Cat': 1, 'Horse': 1})
Source: www.journaldev.com
collections.Counter(string).most_common
Fantastic Frog
on May 04, 2020
counter = Counter({'a': 3, 'b': 3, 'c': 0})
# miscellaneous examples
print(sum(counter.values())) # 6
print(list(counter)) # ['a', 'b', 'c']
print(set(counter)) # {'a', 'b', 'c'}
print(dict(counter)) # {'a': 3, 'b': 3, 'c': 0}
print(counter.items()) # dict_items([('a', 3), ('b', 3), ('c', 0)])
# remove 0 or negative count elements
counter = Counter(a=2, b=3, c=-1, d=0)
counter = +counter
print(counter) # Counter({'b': 3, 'a': 2})
# clear all elements
counter.clear()
print(counter) # Counter()
Source: www.journaldev.com
collections.Counter(string).most_common
Fantastic Frog
on May 04, 2020
counter = Counter({'Dog': 2, 'Cat': -1, 'Horse': 0})
# elements()
elements = counter.elements() # doesn't return elements with count 0 or less
for value in elements:
print(value)
Source: www.journaldev.com
collections.Counter(string).most_common
Fantastic Frog
on May 04, 2020
counter = Counter('ababab')
print(counter) # Counter({'a': 3, 'b': 3})
c = Counter('abc')
print(c) # Counter({'a': 1, 'b': 1, 'c': 1})
# subtract
counter.subtract(c)
print(counter) # Counter({'a': 2, 'b': 2, 'c': -1})
# update
counter.update(c)
print(counter) # Counter({'a': 3, 'b': 3, 'c': 0})
Source: www.journaldev.com
collections.Counter(string).most_common
Fantastic Frog
on May 04, 2020
# Delete element from Counter
del counter['Unicorn']
print(counter) # Counter({'Dog': 2, 'Cat': 1, 'Horse': 0})
Source: www.journaldev.com
collections.Counter(string).most_common
Fantastic Frog
on May 04, 2020
counter = Counter({'Dog': 2, 'Cat': 1, 'Horse': 1})
# setting count
counter['Horse'] = 0
print(counter) # Counter({'Dog': 2, 'Cat': 1, 'Horse': 0})
# setting count for non-existing key, adds to Counter
counter['Unicorn'] = 1
print(counter) # Counter({'Dog': 2, 'Cat': 1, 'Unicorn': 1, 'Horse': 0})
Source: www.journaldev.com
collections.Counter(string).most_common
Fantastic Frog
on May 04, 2020
# getting count for non existing key, don't cause KeyError
print(counter['Unicorn']) # 0
Source: www.journaldev.com
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