Python学习入门(1)---MIT导论

学习所用的视频和文档

• MIT 6.00.1x
• Python教程
• Think Python
• python 官方文档

工具

Nodepad++

使用之前需要设置格式
设置->首选项->语言->勾选替换为空格

python安装包

官网，直接搜索安装文档
centos7安装python

语法

输入输出

1.input–输入,output–输出

x=3
x=x*x #square value of x
print(x)
y=float(input("Enter a number"))
print(y*y)

2.输出结果在一行

print "Hi",
print "there"

3.range用法

1)

range(stop)

range(5)

[0,1,2,3,4]

2)

range(start, stop)

range(2,5)

[2,3,4]

3)

range(start, stop, stepSize)

range(2,10,2)

[2,4,6,8]

range返回类型

(1)

type(range(5))

运行结果

class 'range'

(2)

type(list(range(4)))

运行结果

class 'List'

4.Sum

The function sum is a built-in function that takes one list as an argument and returns the sum of that list. The items in the list must be numeric types.

条件判断

代码1

x=int(input("Enter an integer:"))
if x%2 == 0:
    print("")
    print('Even')
else:
    print(" ")
    print("ODD")
print("Done with conditional")

注意:缩进

代码2

x=6
if x%2 ==0:
    if x%3 ==0:
       print("Divisible by 2 and 3")
    else:
       print("Divisible by 2 and  not by 3")
elif x%3 ==0:
    print("Division by 3 but not by 2")

注:elif=else if

简单算法

迭代

x=3
ans=0
itersLeft=x
while(itersLeft !=0):
    ans=ans+x
    itersLeft=itersLeft-1;
print(str(x)+'*'+str(x)+'='+str(ans))

for LOOP

for...in ...

Guess and Check- - -Finding a cube root of an integer

using “while”

x=int(input("Enter an integer:"))
ans=0
while ans**3<x:
    ans=ans+1
if ans**3!=x :
   print(str(x)+'is not a perfect cube')
else:
   print("Cube root of"+str(x)+" is "+str(ans))

using “for”

x=int(input("Enter an integer:"))
for ans in range(0,abs(x)+1):
    if ans**3==abs(x):
        break
if(ans**3!=abs(x)):
    print(str(x)+" is not a perfect cube")
else:
    if x<0:
        ans=-ans
    print("Cube root of "+str(x)+" is "+str(ans))

浮点数的表示

二进制表示

将一个整数变为二进制数

# lecture 3.4, slide 3

num = 302

if num < 0:
    isNeg = True
    num = abs(num)
else:
    isNeg = False
result = ''
if num == 0:
    result = '0'
while num > 0:
    result = str(num%2) + result
    num = num/2
if isNeg:
    result = '-' + result

print(result)

将一个浮点数变为二进制数

求0.375的二进制表示
步骤：
1)0.375*2^3=3
2)3的二进制数为11
3)用2^3=8来除以3
4)获得小数点和2的幂次数之间的关系
代码如下 ：

# lecture 3.4, slide 4
#获取输入的浮点数
x = float(raw_input('Enter a decimal number between 0 and 1: '))

p = 0
#判定某个数是否为整数,即x*2^p是否为整数，循环结束直至为整数
while ((2**p)*x)%1 != 0:
    print('Remainder = ' + str((2**p)*x - int((2**p)*x)))
    p += 1

num = int(x*(2**p))

#将num这个整数用二进制数表示出来
result = ''
if num == 0:
    result = '0'
while num > 0:
    result = str(num%2) + result
    num = num/2

#将小数点加进最后的结果中
for i in range(p - len(result)):
    result = '0' + result
result = result[0:-p] + '.' + result[-p:]

print('The binary representation of the decimal ' + str(x) + ' is ' + str(result))

注意点

1.如果某个浮点数与任意的2的幂次方相乘，都不能为整数，那么PYTHON中表示出的该浮点数为一个无限接近的数，并不是该浮点数本身。
2.在测试两个浮点数是否相同是，用下面代码

 abs(x-­y)<0.0001,  

而非

  x==y  

1. 3.

print(0.1)

显示结果为 0.1
原因:Python的设计者进行了automaMcally round

Guess Number的方法

Approximation Methods

应用于：GUASS NUMBER中
CODE:

# lecture 3.5, slide 2

x = 25
#essilon是精确度
epsilon = 0.01
#step是步长
step = epsilon**2
numGuesses = 0
ans = 0.0
#两个循环条件必须同时满足 1.数字^2与25相差大于精确度  2.数字不大于步长
while (abs(ans**2 - x)) >= epsilon and ans <= x:
    ans += step
    numGuesses += 1
print('numGuesses = ' + str(numGuesses))
if abs(ans**2-x) >= epsilon:
    print('Failed on square root of ' + str(x))
else:
    print(str(ans) + ' is close to the square root of ' + str(x))

Bisection Search

应用于：GUASS NUMBER中
CODE:

# lecture 3.6, slide 2
# bisection search for square root

x = 12345
epsilon = 0.01
numGuesses = 0
low = 0.0
high = x
ans = (high + low)/2.0
while abs(ans**2 - x) >= epsilon:
    print('low = ' + str(low) + ' high = ' + str(high) + ' ans = ' + str(ans))
    numGuesses += 1
    if ans**2 < x:
        low = ans
    else:
        high = ans
    ans = (high + low)/2.0
print('numGuesses = ' + str(numGuesses))
print(str(ans) + ' is close to square root of ' + str(x))

Newton-Raphson Root Finding

应用于：GUASS NUMBER中
- 原理

• CODE:

# Lecture 3.7, slide 3

# Newton-Raphson for square root

epsilon = 0.01
y = 24.0
guess = y/2.0

while abs(guess*guess - y) >= epsilon:
    guess = guess - (((guess**2) - y)/(2*guess))
    print(guess)
print('Square root of ' + str(y) + ' is about ' + str(guess))