高阶字典

# setfault()处理字典缺失的键
d = {'hydrogen':1, 'helium':2}

try:
    a = d['carbon']

except KeyError:
    print('key not exist')

b = d.setdefault('carbon', 12)
print('after setfault: \n', d)

# setfault(）不会改变已存在的键值对
c =d.setdefault('helium', 1234)
print('will it change?\n', d, '\n of course not!')

# 用defaultdict()处理字典缺失的键

from collections import defaultdict

# defaultdict() 的参数为一个函数
d = defaultdict(int)
print(d['C4TMAN'])


def hehe():
    return '大概是个傻子吧'


d = defaultdict(hehe)
print(d['C4TMAN'])

# 使用int定义计数器
# 若指代defaultdict的变量为非字典，将返回一个空子典

dict_counter = defaultdict(int)
L = ['sb', 'sb', 'sb', 'hehe', 'hehe']

for ele in L:
    dict_counter[ele] += 1

for key, value in dict_counter.items():
    print('there are ', value, key + 's')

# Counter（） 接受一个列表，并返回一个以元素及其数量为键值对的字典( 其实是一个Counter的对象 )
from collections import Counter

for key, value in Counter(L).items():
    print('there are ', value, key + 's')

# most_common方法以降序返回元素
L_counter = Counter(L)
print(L_counter.most_common())

# most_common 接受一个整数做参数，并返回那个参数前的所有元素
print(L_counter.most_common(1))

L2 = ['hehe', 'hehe', 'hehe', 'C4TMAN', 'C4TMAN', 'WJY']
L2_counter = Counter(L2)

# Counter之间的运算
print(L_counter + L2_counter)  # Counter({'hehe': 5, 'sb': 3, 'C4TMAN': 2, 'WJY': 1})
print(L2_counter - L_counter)  # Counter({'C4TMAN': 2, 'hehe': 1, 'WJY': 1})
print(L2_counter & L_counter)  # Counter({'hehe': 2})
print(L2_counter | L_counter)  # Counter({'hehe': 3, 'sb': 3, 'C4TMAN': 2, 'WJY': 1})
                               # 对于共有项，计数不会相加，而是取最大值