Declination of word (amount)

# -*- coding: utf-8 -*-
"""
Числительное склонение слова, в зависимости от количества
"""

def f(count, name):
    if count in range(11,15):
        return str(count) + ' ' + name + 'ов'
    else:
        D = [{'':[1]},{'а':[2,3,4]},{'ов':[0,5,6,7,8,9]}]
        def func(count):
            count = str(count)
            for _dict in D:
                for value in _dict.items():
                    if int(count[-1]) in value[1]:
                        return value[0]
        _end = func(count)
        return str(count) + ' ' + name + _end


if __name__ == '__main__':

    assert f(0, 'товар') == '0 товаров'
    assert f(1, 'товар') == '1 товар'
    assert f(6, 'товар') == '6 товаров'
    assert f(11, 'товар') == '11 товаров'
    assert f(13, 'товар') == '13 товаров'
    assert f(41, 'товар') == '41 товар'
    assert f(55, 'товар') == '55 товаров'
    assert f(4, 'товар') == '4 товара'
    assert f(44, 'товар') == '44 товара'

    print('All Tests ok')