Python Day-List 理解-练习

百变鹏仔 5天前 #Python

文章标签 Python

列表理解

当您想要基于现有列表的值创建新列表时，列表理解提供了更短的语法。（参考-https://www.w3schools.com/python/python_lists_comprehension.asp)

示例：1
方法：1

fruits = ["apple", "banana", "cherry", "kiwi", "mango"]newlist = []for x in fruits:  if "a" in x:    newlist.append(x)print(newlist)

方法：2（综合）

fruits = ["apple", "banana", "cherry", "kiwi", "mango"]newlist = [x for x in fruits if "a" in x]print(newlist)

输出：

['apple', 'banana', 'mango']

示例：2

l = [10,20,30,40]newlist = []#using normal loopfor num in l:    newlist.append(num**2)print(newlist)#using loop in comprehensive waynewlist = [num**2 for num in l]print(newlist)

输出：

[100, 400, 900, 1600][100, 400, 900, 1600]

练习：
1.从2个列表中查找相似的数字以及从相同的2个列表中查找不同的数字。
l1 = [10,20,30,40]
l2 = [30,40,50,60]
得到这个输出：
a) 30,40

#30,40l1 = [10,20,30,40]l2 = [30,40,50,60]#normal methodfor num in l1:    for no in l2:        if num== no:            print(num,end=' ')#comprehensiveprint([num for num in l1 for no in l2 if num==no])

输出：

[30, 40]

b) 10,20,50,60

l1 = [10,20,30,40]l2 = [30,40,50,60]#comprehensiveoutput = [num for num in l1 if num not in l2]output = output + [num for num in l2 if num not in l1]print(output)#normal methodfor num in l1:    if num not in l2:        print(num,end=' ')for num in l2:    if num not in l1:        print(num,end=' ')

输出：

[10, 20, 50, 60]10 20 50 60

2。以综合方法查找给定输出的程序
l1 = [1,2,3]
l2 = [5,6,7]
输出：[(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7), (3, 5), (3, 6) , (3, 7)]

l1 = [1,2,3]l2 = [5,6,7]l = [(i,j) for i in l1 for j in l2 if i!=j]print(l)

输出：

[(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7), (3, 5), (3, 6), (3, 7)]

3。查找给定输出的程序：
s = "a1b2c3"
输出：abc123

方法：1

s = "a1b2c3"alpha_list = []num_list = []for letter in s:    if letter.isalpha():        alpha_list.append(letter)    else:        num_list.append(letter)print("".join(alpha_list+num_list))

方法：2

s = "a1b2c3"letter=''.join([i for i in s if i.isalpha()])no=''.join([i for i in s if i.isdigit()])print(letter+no)

输出：

abc123

4。查找给定输出的程序：

s = "a4k3b2"
输出：aeknbd

s = "a4k3b2"i = 0 while i<len(s):    first = s[i]    second = int(s[i+1])    print(first, chr(ord(first)+second),sep='',end='')    i+=2

输出：

aeknbd

说明：

立即学习“Python免费学习笔记（深入）”；

-->使用ord(first)获取first的ascii值，然后将其添加到其中以查找新字符。
-->ord() 用于查找 ascii 值。
-->hr() 转换 ascii 值-->character.

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