输入数字or 字符串，统计重复次数---字典统计练习

'''
输入一个数字，打印每一位及其重复次数
'''
num = input("input a num:").strip().lstrip('0') # '00100'

# NO 1
counter = {}
for i in num:
    counter[i] = counter.get(i, 0) + 1
print(counter)
# NO 2
counter = {}
for i in num:
    if i not in counter:
        counter[i] = 0
    counter[i] += 1
print(counter)

# NO 3
counter = {}
for i in num:
    counter[i] = counter.setdefault(i, 0) + 1
print(counter)

# NO 4
counter = dict.fromkeys(range(10),0) 
for i in num:
    counter[int(i)] += 1
print(counter)

'''
{'1': 1, '2': 1, '3': 2, '4': 2, '5': 2}
{'1': 1, '2': 1, '3': 2, '4': 2, '5': 2}
{'1': 1, '2': 1, '3': 2, '4': 2, '5': 2}
{0: 0, 1: 1, 2: 1, 3: 2, 4: 2, 5: 2, 6: 0, 7: 0, 8: 0, 9: 0}
'''

'''
数字重复统计：
    随机产生100个整数
    数字的范围 【-1000,1000】
    升序输出及重复次数
'''

# NO 1
import random
from collections import defaultdict

nums = [random.randint(-1000,1000) for _ in range(100)]
counter = defaultdict(int)

for i in nums:
    counter[i] += 1
#使用内建方法sorted() 排序），会生成一个新的list
print(sorted(counter.items(),reverse=True)) # items() 是按照key 排序的，如果value排序，需要高阶函数配合

# NO 2 
# 使用了  list 的sort() 等效于内建方法sorted() 
# 通过测试，事实上，这个效率比内建函数要高
import random
nums = [random.randint(-1000,1000) for _ in range(100)]

counter = defaultdict(int)

for i in nums:
    counter[i] += 1    
keys = list(counter.keys()) # 取出keys
keys.sort() # 就地排序
newlist = [0] * len(keys)
for i,v in enumerate(keys):
    newlist[i] = v, counter[v]
print(newlist)

'''
字符串重复统计
    'abcdefghijklmnopqrstuvwxyz'
    随机挑选两个字母，共挑选100个
    降序输出所有不同的字符串，重复次数
'''

import string
import random

# 取26个字母；
a = string.ascii_lowercase  #  'abcdefghijklmnopqrstuvwxyz'
# # s = ''
# # for i in range(97,123): 
# #     s += chr(i)  # 'abcdefghijklmnopqrstuvwxyz'
# 取100个
words = [random.choice(a) +random.choice(a) for _ in range(100)]

counter = {}
for word in words:
    counter[word] = counter.get(word, 0) + 1

# print(counter)
print(sorted(counter.items(), reverse=True))


# dic = dict({1:2,2:2,5:1,3:9})
# dic[9] = 1
# dic[0] = 4
# print(dic)
# l = dic.items()
# print(l) #  dict_items([(1, 2), (2, 2), (5, 1), (3, 9), (9, 1), (0, 4)])
# print(sorted(l)) # [(0, 4), (1, 2), (2, 2), (3, 9), (5, 1), (9, 1)]
# print(l) # dict_items([(1, 2), (2, 2), (5, 1), (3, 9), (9, 1), (0, 4)])