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 - yx * 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.attributes(...)+x, -x, ~xx ** y*, /, //, %+, -<<, >> &^|<, <=, >, >=, ==, !=is, not isin, not innotandorlambda  

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: continuedefreturnyieldglobal: 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 v1k2 v2...>>> d1 = { 'x':1,'y':2,'z':3,'m':4 }>>> for (k,v) in d1.items():print k,v  y 2x 1z 3m 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]MonWedFri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 += 112345>>> l1 = [1,2,3,4,5]>>> while l1:                        print l1.pop(0)12345练习2：求100以内所有偶数之和；>>> i = 0>>> sum = 0 >>> while i < 101:sum += ii += 2print sum2550>>>  for i in range(0,101,2):sum+=i   print sum2550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)xyz3        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