一、if语句

语法：

if condition：

indented statement（s）

如果条件为真，就执行这些缩进的语句，否则跳过这些语句

例子：

answer = input("Do you want to see a spiral? y/n:")
if answer == 'y':print("Working...")import turtlet = turtle.Pen()t.width(2)for x in range(100):t.forward(x*2)t.left(89)
print("Okay, we're done!")

如果你输入”y“，就绘制出螺旋线，否则直接输出”Okay, we're done!“

二、布尔表达式和布尔值

布尔表达式，或者说条件表达式，结果为布尔值，要么True，要么False，首字母都要大写。

语法：expression1 conditional_operator expression2

2.1.比较操作符

运算符	描述	实例
==	等于 - 比较对象是否相等	(a == b) 返回 False。
!=	不等于 - 比较两个对象是否不相等	(a != b) 返回 True。
>	大于 - 返回x是否大于y	(a > b) 返回 False。
<	小于 - 返回x是否小于y。所有比较运算符返回1表示真，返回0表示假。这分别与特殊的变量True和False等价。注意，这些变量名的大写。	(a < b) 返回 True。
>=	大于等于 - 返回x是否大于等于y。	(a >= b) 返回 False。
<=	小于等于 - 返回x是否小于等于y。	(a <= b) 返回 True。

2.2.你还不够大

# OldEnough.py
driving_age = eval(input("What is the legal driving age where you live? "))
your_age = eval(input("How old are you? "))
if your_age >= driving_age:print("You're old enough to drive!")
if your_age < driving_age:print("Sorry, you can drive in", driving_age - your_age, "years.")

第一个条件表达式的结果知道后，第二个条件表达式的结果肯定也知道了，不需要再次计算，所以最后的if可以用else代替，显得更加简洁。

三、else语句

语法：

if condition：

indented statement（s）

else:

other indented statement（s）

如果条件表达式为True，那么执行缩进的语句，否则执行其他缩进的语句

修改上例：

# OldEnoughOrElse.py
driving_age = eval(input("What is the legal driving age where you live? "))
your_age = eval(input("How old are you? "))
if your_age >= driving_age:print("You're old enough to drive!")
else:print("Sorry, you can drive in", driving_age - your_age, "years.")

3.1.多边形或玫瑰花瓣

# PolygonOrRosette.py
import turtle
t = turtle.Pen()
# Ask the user for the number of sides or circles, default to 6
number = int(turtle.numinput("Number of sides or circles","How many sides or circles in your shape?", 6))
# Ask the user whether they want a polygon or rosette
shape = turtle.textinput("Which shape do you want?","Enter 'p' for polygon or 'r' for rosette:")
for x in range(number):if shape == 'r':        # User selected rosettet.circle(100)else:                   # Default to polygont.forward (150)t.left(360/number)

3.2.偶数还是奇数

# RosettesAndPolygons.py - a spiral of polygons AND rosettes!
import turtle
t = turtle.Pen()
# Ask the user for the number of sides, default to 4
sides = int(turtle.numinput("Number of sides","How many sides in your spiral?", 4))
# Our outer spiral loop for polygons and rosettes, from size 5 to 75
for m in range(5,75):   t.left(360/sides + 5)t.width(m//25+1)t.penup()        # Don't draw lines on spiralt.forward(m*4)   # Move to next cornert.pendown()      # Get ready to draw# Draw a little rosette at each EVEN corner of the spiralif (m % 2 == 0):for n in range(sides):t.circle(m/3)t.right(360/sides)# OR, draw a little polygon at each ODD corner of the spiralelse:for n in range(sides):t.forward(m)t.right(360/sides)

四、elif语句

# WhatsMyGrade.py
grade = eval(input("Enter your number grade (0-100): "))
if grade >= 90:print("You got an A! :) ")
elif grade >= 80:print("You got a B!")
elif grade >= 70:print("You got a C.")
elif grade >= 60:print("You got a D...")
else:print("You got an F. :( ")

五、复杂条件-------if、and、or和not

以下假设变量 a 为 10, b为 20:

运算符	逻辑表达式	描述	实例
and	x and y	布尔"与" - 如果 x 为 False，x and y 返回 x 的值，否则返回 y 的计算值。	(a and b) 返回 20。
or	x or y	布尔"或" - 如果 x 是 True，它返回 x 的值，否则它返回 y 的计算值。	(a or b) 返回 10。
not	not x	布尔"非" - 如果 x 为 True，返回 False 。如果 x 为 False，它返回 True。	not(a and b) 返回 False

# WhatToWear.py
rainy = input("How's the weather? Is it raining? (y/n)").lower()
cold = input("Is it cold outside? (y/n)").lower()
if (rainy == 'y' and cold == 'y'):      # Rainy and cold, yuck!print("You'd better wear a raincoat.")
elif (rainy == 'y' and cold != 'y'):    # Rainy, but warmprint("Carry an umbrella with you.")
elif (rainy != 'y' and cold == 'y'):    # Dry, but coldprint("Put on a jacket, it's cold out!")
elif (rainy != 'y' and cold != 'y'):    # Warm and sunny, yay!print("Wear whatever you want, it's beautiful outside!")

六、秘密消息

对称式加密，即加密和解密的密钥是相同的

message = input("Enter a message to encode or decode: ") # Get a message
message = message.upper()           # Make it all UPPERCASE :)
output = ""                         # Create an empty string to hold output
for letter in message:              # Loop through each letter of the messageif letter.isupper():            # If the letter is in the alphabet (A-Z),value = ord(letter) + 13    # shift the letter value up by 13,letter = chr(value)         # turn the value back into a letter,if not letter.isupper():    # and check to see if we shifted too farvalue -= 26             # If we did, wrap it back around Z->Aletter = chr(value)     # by subtracting 26 from the letter valueoutput += letter                # Add the letter to our output string
print("Output message: ", output)   # Output our coded/decoded message

作业

1.彩色玫瑰花瓣和螺旋线

修改# RosettesAndPolygons.py，达到如下效果：

# RosettesAndSpirals.py - a spiral of shapes AND rosettes!
import turtle
t=turtle.Pen()
t.penup()
t.speed(0)
turtle.bgcolor('black')
# Ask the user for the number of sides, default to 4, min 2, max 6
sides = int(turtle.numinput("Number of sides","How many sides in your spiral?", 4,2,6))
colors=['red', 'yellow', 'blue', 'green', 'purple', 'orange']
# Our outer spiral loop
for m in range(1,60):t.forward(m*4)t.width(m//25+1)position = t.position() # remember this corner of the spiralheading = t.heading()   # remember the direction we were heading# Our 'inner' spiral loop,# draws a little rosette at each corner of the big spiralif (m % 2 == 0):for n in range(sides):t.pendown()t.pencolor(colors[n%sides])t.circle(m/4)t.right(360/sides - 2)t.penup()# OR, draws a little spiral at each corner of the big spiralelse:for n in range(3,m):t.pendown()t.pencolor(colors[n%sides])t.forward(n)t.right(360/sides - 2)t.penup()t.setx(position[0])     # go back to the big spiral's x locationt.sety(position[1])     # go back to the big spiral's y locationt.setheading(heading)   # point in the big spirals direction/headingt.left(360/sides + 4)   # move to the next point on the big spiral

2.用户定义的密钥

修改EncoderDecoder.py,允许用户输入他们自己的密钥值，从1到25，例如：5，那么加密时密钥为5，解密时用相反的密钥21（26-5=21）。

message = input("Enter a message to encode or decode: ") # get a message
key = eval(input("Enter a key value from 1-25: ")) # get a key
message = message.upper()           # make it all UPPERCASE :)
output = ""                         # create an empty string to hold output
for letter in message:              # loop through each letter of the messageif letter.isupper():            # if the letter is in the alphabet (A-Z)value = ord(letter) + key   # shift the letter value up by keyletter = chr(value)         # turn the value back into a letterif not letter.isupper():    # check to see if we shifted too farvalue -= 26             # if we did, wrap it back around Z->Aletter = chr(value)     # by subtracting 26 from the letter valueoutput += letter                # add the letter to our output string
print ("Output message: ", output)  # output our coded/decoded message

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