【Python基础语法学习】 

 1 import math
 2 '''
 3 n="987654321"
 4 
 5 print(l[1:7]*2) #repeat
 6 m=(1,6,2.9,'a')
 7 print(m[2])
 8 
 9 d={} #先声明（？
10 d[98]=356
11 d['a']="nihao"
12 print(d.values())
13 
14 print(9/2) #4.5
15 print(3^2) #1
16 if 1 and 0:
17     print(123)
18 
19 if 'a' in l:
20     print(123)
21 a=1
22 while a<9:
23     print(a)
24     a+=1
25 else:
26     print(0)
27     a+=1
28 l=['x','y','z','a','b','c','d','o',]
29 
30 for le in l:
31     if le=='a':
32         continue
33     #print(le)
34 else:
35     pass
36     print(0)
37 
38 def sum(a,b):
39     #pass
40     print(a+b)
41     #return
42 print(sum(9,8),"hhh")
43 '''
44 fp=open("123.txt","a")
45 fp.write("ooooo")
46 fp.close

 也把课上留的作业贴上来

 1 chars={}
 2 print("请输入若干字符（英文或中文）：")
 3 string=input()
 4 
 5 for c in string:
 6     if c not in chars.keys():
 7         chars[c]=1
 8     else:
 9         chars[c]+=1
10 
11 print("字符总数为：",len(string))
12 print("字符种类数量为：",len(chars))
13 for c in chars.keys():
14     print(c,"出现了",chars[c],"次")
15 
16 a=1
17 while a<len(string):
18     cnt=0
19     for c in string:
20         if chars[c]==a:
21             cnt+=1
22     if cnt:
23         print("出现",a,"次的字符个数(非种类数量)为：",cnt)
24   #      print("出现",a,"次的字符个数(种类数量)为：",cnt//a)
25     a+=1

 

posted @ 2021-05-04 23:20  ranzai  阅读(145)  评论(0)    收藏  举报