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# Ordered Rely of Characters in Python

## The problem#

Rely the variety of occurrences of every character and return it as a listing of tuples so as of look. For empty output return an empty record.

Instance:

``````ordered_count("abracadabra") == [('a', 5), ('b', 2), ('r', 2), ('c', 1), ('d', 1)]
``````

## The answer in Python code#

Possibility 1:

``````from collections import Counter
def ordered_count(enter):
c = Counter(enter)
return sorted(record(c.gadgets()), key=lambda x: enter.index(x[0]))
``````

Possibility 2:

``````def ordered_count(_input):
l = []
for i in _input:
if i not in l:
l.append(i)
return [(i, _input.count(i)) for i in l]
``````

Possibility 3:

``````def ordered_count(inp):
return [(i, inp.count(i)) for i in sorted(set(inp), key=inp.index)]
``````

## Take a look at instances to validate our resolution#

``````take a look at.describe("Primary Exams")

checks = (
('abracadabra', [('a', 5), ('b', 2), ('r', 2), ('c', 1), ('d', 1)])
)

for t in checks:
inp, exp = t
take a look at.assert_equals(ordered_count(inp), exp)

take a look at.describe("Random Exams")
def random_tests():
from string import (
ascii_letters,
punctuation,
digits
)

from collections import (
OrderedDict,
Counter
)

from random import (
randint,
alternative
)

class _OrderedCounter(Counter, OrderedDict):
cross

def reference(seq):
return record(_OrderedCounter(seq).gadgets())

CHARS = "".be a part of(("     ", ascii_letters, punctuation, digits))

for _ in vary(100):
test_case = "".be a part of(alternative(CHARS) for _ in vary(randint(1, 1000)))
take a look at.assert_equals(ordered_count(test_case), reference(test_case))

random_tests()
``````
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