Python学习6——条件,循环语句

条件语句

真值也称布尔值。

用作布尔表达式时,下面的值都将被视为假:False None  0   ""   ()   []   {}。

布尔值True和False属于类型bool,而bool与list、str和tuple一样,可用来转换其他的值。

>>> bool('I think,therefore I am')
True
>>> bool(10)
True
>>> bool('')
False
>>> bool(0)
False

 

if语句

>>> name = input('What is your name?')
What is your name?Gumby
>>> if name.endswith('Gumby'):
    print(name)

    
Gumby

如果条件(if和冒号之间的表达式)是前面定义的真,就执行后续代码块,如果条件为假,就不执行。

 

else子句,elif子句,嵌套代码块

>>> name = input('What is your name?')
What is your name?Gumby
>>> if name.endswith('Gumby'):
    if name.startswith('Mr.'):
        print('Hello,Mr.Gumby')
    elif name.startswith('Mrs.'):
        print('Hello,Mrs.Gumby')
    else:
        print('Hello,Gumby')
else:
    print('Hello,stranger')

    
Hello,Gumby

 

比较运算符:

表达式 描述
x == y x等于y
x < y x小于y
x > y x大于y
x >= y  x大于或等于y
x <= y x小于或等于y
x != y x不等于y
x is y x和y是同一个对象
x is not y x和y是不同的对象
x in y x是容器y的成员
x not in y x不是容器y的成员

 

 

 

 

 

 

 

 

 

>>> x = y = [1,2,3]
>>> z = [1,2,3]
>>> x == y
True
>>> x == z
True
>>> x is y
True
>>> x is z
False
>>> #is检查两个对象是否相同(不是相等),变量x和y指向同一个列表,而z指向另一个列表(即使它们包含的值是一样的),这两个列表虽然相等,但并非同一个列表。

 

ord函数可以获取字母的顺序值,chr函数的作用与其相反:

>>> ord('a')
97
>>> ord('v')
118
>>> chr(118)
'v'

 

断言:assert

>>> age = 10
>>> assert 0 < age < 11
>>> age = -1
>>> assert 0 < age < 11
Traceback (most recent call last):
  File "<pyshell#47>", line 1, in <module>
    assert 0 < age < 11
AssertionError
>>> age = -1
>>> assert 0 < age < 11,'the age must be realistic' #字符串对断言做出说明
Traceback (most recent call last):
  File "<pyshell#49>", line 1, in <module>
    assert 0 < age < 11,'the age must be realistic' #字符串对断言做出说明
AssertionError: the age must be realistic

 

循环

while 循环

>>> name = '' 
>>> while not name.strip():        #屏蔽空格
    name = input('please enter your name:')
    print('hello,{}!'.format(name))

    
please enter your name:Momo
hello,Momo!

 

for循环

>>> numbers = [0,1,2,3,4,5,6,7,8]
>>> for number in numbers:
    print(number)

    
0
1
2
3
4
5
6
7
8

内置函数range():

>>> range(0,10)
range(0, 10)
>>> list(range(0,10))  #范围[0,10)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> range(10)     #如果只提供1个位置,将把这个位置视为结束位置,并假定起始位置为0
range(0, 10)
>>> for num in range(1,10):
    print(num)

    
1
2
3
4
5
6
7
8
9

 

迭代字典:

>>> d = dict(a=1,b=2,c=3)
>>> d
{'a': 1, 'b': 2, 'c': 3}
>>> for key in d:
    print(key,'to',d[key])

    
a to 1
b to 2
c to 3

换一种方式迭代:

>>> for key,value in d.items():
    print(key,'to',value)

    
a to 1
b to 2
c to 3
>>> d.items()
dict_items([('a', 1), ('b', 2), ('c', 3)])

 

并行迭代:

同时迭代2个序列。

>>> names = ['aaa','bbb','ccc']
>>> ages = [10,11,12]
>>> for i in range(len(names)):
    print(names[i],'is',ages[i],'years old')

    
aaa is 10 years old
bbb is 11 years old
ccc is 12 years old

内置函数zip,可以将2个序列缝合起来,并返回一个由元组组成的序列。

>>> list(zip(names,ages))
[('aaa', 10), ('bbb', 11), ('ccc', 12)]
>>> for name,age in zip(names,ages):
    print(name,'is',age,'years old')

    
aaa is 10 years old
bbb is 11 years old
ccc is 12 years old

zip函数可以缝合任意数量的序列,当序列的长度不同时,zip函数将在最短的序列用完后停止缝合。

>>> list(zip(range(5),range(10)))
[(0, 0), (1, 1), (2, 2), (3, 3), (4, 4)]

 

迭代时获取索引:

>>> strings = ['aac','ddd','aa','ccc']
>>> for string in strings:
    if 'aa' in string:         #替换列表中所有包含子串‘aa’的字符串
        index = strings.index(string)    #在列表中查找字符串
        strings[index] = '[censored]'

        
>>> strings
['[censored]', 'ddd', '[censored]', 'ccc']

优化一下:

>>> strings = ['aac','ddd','aa','ccc']
>>> index = 0
>>> for string in strings:
    if 'aa' in string:
        strings[index] = '[censored]'
    index += 1

    
>>> strings
['[censored]', 'ddd', '[censored]', 'ccc']

继续优化:

内置函数enumerate能够迭代索引-值对,其中的索引是自动提供的。

>>> strings = ['aac','ddd','aa','ccc']
>>> for index,string in enumerate(strings):
    if 'aa' in string:
        strings[index] = '[censored]'
>>> strings
['[censored]', 'ddd', '[censored]', 'ccc']

反向迭代和排序后在迭代:

>>> sorted([4,2,5,1,3])
[1, 2, 3, 4, 5]
>>> sorted('hello,world')   #sorted返回一个列表
[',', 'd', 'e', 'h', 'l', 'l', 'l', 'o', 'o', 'r', 'w']
>>> list(reversed('Hello,world'))   #reversed返回一个可迭代对象,
['d', 'l', 'r', 'o', 'w', ',', 'o', 'l', 'l', 'e', 'H']
>>> ''.join(reversed('Hello,world'))
'dlrow,olleH'
>>> sorted('aBc',key=str.lower)   #按照字母表排序,可以先转换为小写
['a', 'B', 'c']

 

break 跳出循环

>>> from math import sqrt
>>> for n in range(99,1,-1):   #找出2-100的最大平方值,从100向下迭代,找到第一个平方值后,跳出循环。步长为负数,让range向下迭代。
    root = sqrt(n)
    if root == int(root):    #开平方为整
        print(n)
        break

    
81

 

while True/break

>>> while True:    #while True导致循环永不结束
    word = input('enter your name:')
    if not word:break   #在if语句中加入break可以跳出循环
    print(word)

    
enter your name:aa
aa
enter your name:bb
bb
enter your name:
>>> 

 

简单推导

列表推导式一种从其他列表创建列表的方式,类似于数学中的集合推导。

>>> [x*x for x in range(10)]
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

打印能被3整除的平方值

>>> [x*x for x in range(10) if x%3 == 0]
[0, 9, 36, 81]

使用多个for部分,将名字的首字母相同的男孩和女孩配对。

>>> girls = ['aaa','bbb','ccc']
>>> boys = ['crde','bosy','adeb']
>>> [b+'+'+g for b in boys for g in girls if b[0]==g[0]]
['crde+ccc', 'bosy+bbb', 'adeb+aaa']

上面的代码还能继续优化:

>>> girls = ['aaa','bbb','ccc']
>>> boys = ['crde','bosy','adeb']
>>> letterGrils = {}
>>> for girl in girls:
    letterGrils.setdefault(girl[0],[]).append(girl)   #每项的键都是一个字母,值为该字母开头的女孩名字组成的列表

    
>>> print([b+'+'+g for b in boys for g in letterGrils[b[0]]])
['crde+ccc', 'bosy+bbb', 'adeb+aaa']

 

字典推导

>>> squares = {i:'{} squared is {}'.format(i,i**2) for i in range(10)}
>>> squares[8]
'8 squared is 64'

 

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