输入数字or 字符串，统计重复次数---字典统计练习 

 1 '''
 2 输入一个数字，打印每一位及其重复次数
 3 '''
 4 num = input("input a num:").strip().lstrip('0') # '00100'
 5 
 6 # NO 1
 7 counter = {}
 8 for i in num:
 9     counter[i] = counter.get(i, 0) + 1
10 print(counter)
11 # NO 2
12 counter = {}
13 for i in num:
14     if i not in counter:
15         counter[i] = 0
16     counter[i] += 1
17 print(counter)
18 
19 # NO 3
20 counter = {}
21 for i in num:
22     counter[i] = counter.setdefault(i, 0) + 1
23 print(counter)
24 
25 # NO 4
26 counter = dict.fromkeys(range(10),0) 
27 for i in num:
28     counter[int(i)] += 1
29 print(counter)
30 
31 '''
32 {'1': 1, '2': 1, '3': 2, '4': 2, '5': 2}
33 {'1': 1, '2': 1, '3': 2, '4': 2, '5': 2}
34 {'1': 1, '2': 1, '3': 2, '4': 2, '5': 2}
35 {0: 0, 1: 1, 2: 1, 3: 2, 4: 2, 5: 2, 6: 0, 7: 0, 8: 0, 9: 0}
36 '''
 1 '''
 2 数字重复统计：
 3     随机产生100个整数
 4     数字的范围 【-1000,1000】
 5     升序输出及重复次数
 6 '''
 7 
 8 # NO 1
 9 import random
10 from collections import defaultdict
11 
12 nums = [random.randint(-1000,1000) for _ in range(100)]
13 counter = defaultdict(int)
14 
15 for i in nums:
16     counter[i] += 1
17 #使用内建方法sorted() 排序），会生成一个新的list
18 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
19 
20 # NO 2 
21 # 使用了  list 的sort() 等效于内建方法sorted() 
22 # 通过测试，事实上，这个效率比内建函数要高
23 import random
24 nums = [random.randint(-1000,1000) for _ in range(100)]
25 
26 counter = defaultdict(int)
27 
28 for i in nums:
29     counter[i] += 1    
30 keys = list(counter.keys()) # 取出keys
31 keys.sort() # 就地排序
32 newlist = [0] * len(keys)
33 for i,v in enumerate(keys):
34     newlist[i] = v, counter[v]
35 print(newlist)
 1 '''
 2 字符串重复统计
 3     'abcdefghijklmnopqrstuvwxyz'
 4     随机挑选两个字母，共挑选100个
 5     降序输出所有不同的字符串，重复次数
 6 '''
 7 
 8 import string
 9 import random
10 
11 # 取26个字母；
12 a = string.ascii_lowercase  #  'abcdefghijklmnopqrstuvwxyz'
13 # # s = ''
14 # # for i in range(97,123): 
15 # #     s += chr(i)  # 'abcdefghijklmnopqrstuvwxyz'
16 # 取100个
17 words = [random.choice(a) +random.choice(a) for _ in range(100)]
18 
19 counter = {}
20 for word in words:
21     counter[word] = counter.get(word, 0) + 1
22 
23 # print(counter)
24 print(sorted(counter.items(), reverse=True))
25 
26 
27 # dic = dict({1:2,2:2,5:1,3:9})
28 # dic[9] = 1
29 # dic[0] = 4
30 # print(dic)
31 # l = dic.items()
32 # print(l) #  dict_items([(1, 2), (2, 2), (5, 1), (3, 9), (9, 1), (0, 4)])
33 # print(sorted(l)) # [(0, 4), (1, 2), (2, 2), (3, 9), (5, 1), (9, 1)]
34 # print(l) # dict_items([(1, 2), (2, 2), (5, 1), (3, 9), (9, 1), (0, 4)])

 

posted @ 2018-08-15 11:46  JerryZao  阅读(561)  评论(0)  编辑  收藏  举报