python 表达式和语句及for、while循环练习实例

(编辑:jimmy 日期: 2024/11/14 浏览:2)

Python中表达式和语句及for、while循环练习

1)表达式

常用的表达式操作符:
x + y, x - y
x * y, x / y, x // y, x % y

逻辑运算:
x or y, x and y, not x

成员关系运算:
x in y, x not in y

对象实例测试:
x is y, x not is y

比较运算:
x < y, x > y, x <= y, x >= y, x == y, x != y

位运算:
x | y, x & y, x ^ y, x << y, x  y

一元运算:
-x, +x, ~x:

幂运算:
x ** y

索引和分片:
x[i], x[i:j], x[i:j:stride]

调用:
x(...)

取属性:
  x.attribute

元组:(...)
序列:[...]
字典:{...}

三元选择表达式:x if y else z

匿名函数:lambda args: expression

生成器函数发送协议:yield x

 运算优先级:
(...), [...], {...}
s[i], s[i:j]
s.attribute
s(...)
+x, -x, ~x
x ** y
*, /, //, %
+, -
<<,  
&
^
|
<, <=, >, >=, ==, !=
is, not is
in, not in
not
and
or
lambda 

2)语句:

赋值语句

  调用
  print: 打印对象
  if/elif/else: 条件判断
  for/else: 序列迭代
  while/else: 普通循环
  pass: 占位符
  break: 
  continue
  def
  return
  yield
  global: 命名空间
  raise: 触发异常
  import: 
  from: 模块属性访问
  class: 类
  try/except/finally: 捕捉异常
  del: 删除引用
  assert: 调试检查
  with/as: 环境管理器
  
赋值语句:

  隐式赋值:import, from, def, class, for, 函数参数

  元组和列表分解赋值:当赋值符号(=)的左侧为元组或列表时,Python会按照位置把右边的对象和左边的目标自左而右逐一进行配对儿;个数不同时会触发异常,此时可以切片的方式进行;

  多重目标赋值

  增强赋值: +=, -=, *=, /=, //=, %=, 

3)for循环练习

练习1:逐一分开显示指定字典d1中的所有元素,类似如下
k1 v1
k2 v2
...
  
  > d1 = { 'x':1,'y':2,'z':3,'m':4 }
  > for (k,v) in d1.items():
  print k,v 
  y 2
  x 1
  z 3
  m 4
  
  练习2:逐一显示列表中l1=["Sun","Mon","Tue","Wed","Thu","Fri","Sat"]中的索引为奇数的元素;
  
  > l1 = ["Sun","Mon","Tue","Wed","Thu","Fri","Sat"]
  > for i in range(1,len(l1),2):
  print l1[i]
  
  Mon
  Wed
  Fri
  
  练习3:将属于列表l1=["Sun","Mon","Tue","Wed","Thu","Fri","Sat"],但不属于列表l2=["Sun","Mon","Tue","Thu","Sat"]的所有元素定义为一个新列表l3; 
  
  > l1 = ["Sun","Mon","Tue","Wed","Thu","Fri","Sat"]
  > l2 = ["Sun","Mon","Tue","Thu","Sat"]
  > l3 = [ ]
  > for i in l1:
  if i not in l2:
l3.append(i)
  > l3
  ['Wed', 'Fri']
  
   练习4:已知列表namelist=['stu1','stu2','stu3','stu4','stu5','stu6','stu7'],删除列表removelist=['stu3', 'stu7', 'stu9'];请将属于removelist列表中的每个元素从namelist中移除(属于removelist,但不属于namelist的忽略即可);
   
  > namelist= ['stu1','stu2','stu3','stu4','stu5','stu6','stu7']
  > removelist = ['stu3', 'stu7', 'stu9']  
  > for i in namelist:
  if i in removelist :
namelist.remove(i)
  > namelist
  ['stu1', 'stu2', 'stu4', 'stu5', 'stu6']

4)while循环练习

练习1:逐一显示指定列表中的所有元素;

  > l1 = [1,2,3,4,5]
  > i = 0
  > while i < len(l1)
  print l1[i]
  i += 1
  
  1
  2
  3
  4
  5

  > l1 = [1,2,3,4,5]
  > while l1:
  print l1.pop(0)
  
  1
  2
  3
  4
  5
  
练习2:求100以内所有偶数之和;
  
  > i = 0
  > sum = 0 
  > while i < 101:
  sum += i
  i += 2
print sum
  
  2550
  
  > for i in range(0,101,2):
  sum+=i  
 print sum
  
  2550
  
    练习3:逐一显示指定字典的所有键;并于显示结束后说明总键数;
    
  > d1 = {'x':1, 'y':23, 'z': 78}
  > i1 = d1.keys()
  > while i1:
  print i1.pop(0)
else:
  print len(d1)
  x
  y
  z
  3

    练习4:创建一个包含了100以内所有奇数的列表;
    
  > d1 = [ ]
  > i = 1
  > while i < 101:
  d1.append(i)
  i+=2
  > print d1
  [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99]  
  
  > d1 = [ ] 
  > for i in range(1,101,2)
  d1.append(i)
  > print d1
  [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99]
  
练习5:列表l1=[0,1,2,3,4,5,6], 列表l2=["Sun","Mon","Tue","Wed","Thu","Fri","Sat"],以第一个列表中的元素为键,以第二个列表中的元素为值生成字典d1;
   
  > l1 = [0,1,2,3,4,5,6] 
  > l2 = ["Sun","Mon","Tue","Wed","Thu","Fri","Sat"]
  > d1 = {}
  > count = 0
  > if len(l1) == len(l2):
  while count < len(l1):
d1[l1[count]] = l2[count] 
count += 1

以上这篇python 表达式和语句及for、while循环练习实例就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持。

一句话新闻

高通与谷歌联手!首款骁龙PC优化Chrome浏览器发布
高通和谷歌日前宣布,推出首次面向搭载骁龙的Windows PC的优化版Chrome浏览器。
在对骁龙X Elite参考设计的初步测试中,全新的Chrome浏览器在Speedometer 2.1基准测试中实现了显著的性能提升。
预计在2024年年中之前,搭载骁龙X Elite计算平台的PC将面世。该浏览器的提前问世,有助于骁龙PC问世就获得满血表现。
谷歌高级副总裁Hiroshi Lockheimer表示,此次与高通的合作将有助于确保Chrome用户在当前ARM兼容的PC上获得最佳的浏览体验。