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Datewhale学习笔记05

发布时间 2023-11-27 14:48:46作者: 瑟兰迪尔·绿叶

Datewhale学习笔记5

$\textcolor{blue}{Datewhale学习笔记}$$\textcolor{red}{chap5}$

聪明办法学 Python 2nd Edition

Chapter 5 循环 Loop

for 循环和循环范围

for 循环的特点

基于提供的范围，重复执行特定次数的操作

In [1]

def sumFromMToN(m, n):
    total = 0
    # 注意： range(x, y) 是左闭右开区间，包含 x，不包含 y
    for x in range(m, n+1):
        total += x
    return total

In [2]

sumFromMToN(5, 10)
45

In [3]

sumFromMToN(5, 10) == 5+6+7+8+9+10
True

range() 是个什么东西？

其实在这里，我们也可以不用循环来完成同样的任务

In [4]

def sumFromMToN(m, n):
    return sum(range(m, n+1))

In [5]

sumFromMToN(5, 10)
45

如果我们省略第一个参数会发生什么？

In [6]

def sumToN(n):
    total = 0
    # range 起始范围默认为 0
    for x in range(n+1):
        total += x
    return total

In [7]

sumToN(5)
15

那如果我们添加第三个参数呢？

In [8]

def sumEveryKthFromMToN(m, n, k):
    total = 0
    # 第三个参数为 “步长” step
    for x in range(m, n+1, k):
        total += x
    return total

In [9]

sumEveryKthFromMToN(5, 20, 7) == (5 + 12 + 19)
True

只对从 m 到 n 的 奇数求和

In [10]

# 我们也可以通过修改循环内部的代码来改变步长

def sumOfOddsFromMToN(m, n):
    total = 0
    for x in range(m, n+1):
        if x % 2 == 1:
            total += x
    return total

In [11]

sumOfOddsFromMToN(4, 10) == sumOfOddsFromMToN(5,9) == (5+7+9)
True

现在我们反着来试一下！

In [12]

# 我们将生成一个反向数字序列
# (仅供演示使用，代码实践中不建议这么做)

def sumOfOddsFromMToN(m, n):
    total = 0
    for x in range(n, m-1, -1):
        if x % 2 == 1:
            total += x
    return total

In [13]

sumOfOddsFromMToN(4, 10) == sumOfOddsFromMToN(5,9) == (5+7+9)
True

还有更多方法来解决这个问题

for 循环嵌套

In [14]

# 下面的代码将输出二维坐标

def printCoordinates(xMax, yMax):
    for x in range(1, xMax+1):
        for y in range(1, yMax+1):
            print(f"( {x} , {y} )  ", end="")
        print()

In [15]

printCoordinates(5, 5)
( 1 , 1 )  ( 1 , 2 )  ( 1 , 3 )  ( 1 , 4 )  ( 1 , 5 )  
( 2 , 1 )  ( 2 , 2 )  ( 2 , 3 )  ( 2 , 4 )  ( 2 , 5 )  
( 3 , 1 )  ( 3 , 2 )  ( 3 , 3 )  ( 3 , 4 )  ( 3 , 5 )  
( 4 , 1 )  ( 4 , 2 )  ( 4 , 3 )  ( 4 , 4 )  ( 4 , 5 )  
( 5 , 1 )  ( 5 , 2 )  ( 5 , 3 )  ( 5 , 4 )  ( 5 , 5 )

In [16]

from IPython.display import IFrame
IFrame('https://pythontutor.com/visualize.html#code=%23%20%E4%B8%8B%E9%9D%A2%E7%9A%84%E4%BB%A3%E7%A0%81%E5%B0%86%E8%BE%93%E5%87%BA%E4%BA%8C%E7%BB%B4%E5%9D%90%E6%A0%87%0A%0Adef%20printCoordinates%28xMax,%20yMax%29%3A%0A%20%20%20%20for%20x%20in%20range%281,%20xMax%2B1%29%3A%0A%20%20%20%20%20%20%20%20for%20y%20in%20range%281,%20yMax%2B1%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20print%28f%22%28%20%7Bx%7D%20,%20%7By%7D%20%29%20%20%22,%20end%3D%22%22%29%0A%20%20%20%20%20%20%20%20print%28%29%0A%20%20%20%20%20%20%20%20%0AprintCoordinates%283,%203%29&cumulative=false&curInstr=0&heapPrimitives=nevernest&mode=display&origin=opt-frontend.js&py=3&rawInputLstJSON=%5B%5D&textReferences=false', width=1300, height=800)
<IPython.lib.display.IFrame at 0x233782b5850>

如果换成 * 呢？

In [17]

def Stars(n, m):
    # 输出一个 n*m 的星型矩阵图
    for row in range(n):
        for col in range(m):
            print("*", end="")
        print()

In [18]

Stars(5, 5)
*****
*****
*****
*****
*****

换一种写法

In [19]

# be careful! 这些代码与之前的有什么不同？

def printMysteryStarShape(n):
    for row in range(n):
        print(row, end=" ")
        for col in range(row):
            print("*", end=" ")
        print()

In [20]

printMysteryStarShape(5)
0 
1 * 
2 * * 
3 * * * 
4 * * * *

In [21]

from IPython.display import IFrame
IFrame('https://pythontutor.com/visualize.html#code=%23%20be%20careful!%20%E8%BF%99%E4%BA%9B%E4%BB%A3%E7%A0%81%E4%B8%8E%E4%B9%8B%E5%89%8D%E7%9A%84%E6%9C%89%E4%BB%80%E4%B9%88%E4%B8%8D%E5%90%8C%EF%BC%9F%0A%0Adef%20printMysteryStarShape%28n%29%3A%0A%20%20%20%20for%20row%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20print%28row,%20end%3D%22%20%22%29%0A%20%20%20%20%20%20%20%20for%20col%20in%20range%28row%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20print%28%22*%22,%20end%3D%22%20%22%29%0A%20%20%20%20%20%20%20%20print%28%29%0A%20%20%20%20%20%20%20%20%0AprintMysteryStarShape%285%29&cumulative=false&curInstr=0&heapPrimitives=nevernest&mode=display&origin=opt-frontend.js&py=3&rawInputLstJSON=%5B%5D&textReferences=false', width=1300, height=800)
<IPython.lib.display.IFrame at 0x233781b3a60>

while 循环

嘿！

当你不知道循环什么时候停下来的时候，为什么不试试 while

找出一个数最左边的那一位的数值（12345 的 1 ）

In [22]

# 我不知道它什么时候停下来

def leftmostDigit(n):
    n = abs(n)
    while n >= 10:
        n = n//10
    return n

In [23]

leftmostDigit(46535248)
4

举个例子：依次找出 n 个 4 或者 7 的整数倍非负整数

In [24]

def isMultipleOf4or7(x):
    return ((x % 4) == 0) or ((x % 7) == 0)

def nthMultipleOf4or7(n):
    found = 0
    guess = -1
    while found <= n:
        guess += 1
        if isMultipleOf4or7(guess):
            found += 1
    return guess

In [25]

print("4 或 7 的倍数: ", end="")
for n in range(15):
    print(nthMultipleOf4or7(n), end=" ")
4 或 7 的倍数: 0 4 7 8 12 14 16 20 21 24 28 32 35 36 40

Bad Style：在知道循环范围的情况下使用 while

In [26]

def sumToN(n):
    # 尽管它能正确运行，但是这是非常不推荐的做法！
    # 应该使用 for 循环而不是 while 循环
    total = 0
    counter = 1
    while counter <= n:
        total += counter
        counter += 1
    return total

In [27]

sumToN(5) == 1+2+3+4+5
True

break 与 continue 语句

In [28]

for n in range(200):
    if n % 3 == 0:
        continue # 跳过这次循环
    elif n == 8:
        break # 跳出当前整个循环
    else:
        pass # 啥也不做，占位符（不会被运行）
    print(n, end=" ")
1 2 4 5 7

假·死循环

与环境交互后，在特定条件下终止的循环

In [29]

# 不需要看懂这些代码，关注演示的过程

def readUntilDone():
    linesEntered = 0
    while True:
        response = input("输入一个字符串（输入 done 则退出）: ")
        if response == "done":
            break
        print("你输入了: ", response)
        linesEntered += 1
    print("Bye!")
    return linesEntered

In [30]

linesEntered = readUntilDone()
print("你输入了", linesEntered, "行 (不包括 'done').")
你输入了:  learn
你输入了:  python
你输入了:  the
你输入了:  smart
你输入了:  way
Bye!
你输入了 5 行 (不包括 'done').

isPrime

判断一个数是不是质数

In [31]

# 不是最快的写法，但最容易理解

def isPrime(n):
    if n < 2:
        return False
    for factor in range(2,n):
        if n % factor == 0:
            return False
    return True

In [32]

for n in range(100):
    if isPrime(n):
        print(n, end=" ")
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

faster IsPrime:

In [33]

# 快了一点

def fasterIsPrime(n):
    if n < 2:
        return False
    if n == 2:
        return True
    if n % 2 == 0:
        return False
    maxFactor = round(n**0.5)
    for factor in range(3, maxFactor+1, 2):
        if n % factor == 0:
            return False
    return True

In [34]

for n in range(100):
    if fasterIsPrime(n):
        print(n, end=" ")
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

真的快了？

In [35]

# 验证他它俩结果是一样的
for n in range(100):
    assert(isPrime(n) == fasterIsPrime(n))
print("两种解法的结果一致")
两种解法的结果一致

In [39]

import time

bigPrime = 102030407
print("Timing isPrime(",bigPrime,")", end=" ")

# isPrime
time0 = time.time()
print(", returns ", isPrime(bigPrime), end=" ")

time1 = time.time()
print(", time = ",(time1-time0)*1000,"ms\n")

# fasterIsPrime
print("Timing fasterIsPrime(",bigPrime,")", end=" ")
time0 = time.time()

print(", returns ", fasterIsPrime(bigPrime), end=" ")
time1 = time.time()

# result
print(", time = ",(time1-time0)*1000,"ms")
Timing isPrime( 102030407 ) , returns  True , time =  4708.568811416626 ms

Timing fasterIsPrime( 102030407 ) , returns  True , time =  0.4515647888183594 ms

nth Prime

依次找出第 n 位质数

In [37]

def nthPrime(n):
    found = 0
    guess = 0
    while found <= n:
        guess += 1
        if fasterIsPrime(guess):
            found += 1
    return guess

In [38]

for n in range(10):
    print(n, nthPrime(n))
print("Done!")
0 2
1 3
2 5
3 7
4 11
5 13
6 17
7 19
8 23
9 29
Done!

总结

  • For 循环用于指定范围的重复操作。
  • range() 可以生成一个数字范围。
  • 在不知道循环什么时间停止的时候，应该试试 While 循环。
  • 循环同样也是可以嵌套的。
  • 巧妙地使用 break 和 continue 语句。
  • 合理的剪枝，缩小搜索范围/循环范围，可以大幅提高程序运行效率。

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