Day 5 🧭行动：学习列表（list）容器 🤓体会：列表是有序的，可重复的，可修改的 🧑‍💻代码： #合并两个列表并去重 # num_list1 = [19, 23, 54, 64, 875, 20, 109, 232, 123, 54] # num_list2 = [55, 80, 72, 35, 60, 123, 54, 29, 91] # for num in num_list1: # num_list2.append(num) # print("合并后的列表：",num_list2) # new_list = [] # for num in num_list2: # if num not in new_list: # new_list.append(num) # print("去重后的列表：",new_list) #上一个代码的简易版 # num_list1 = [19, 23, 54, 64, 875, 20, 109, 232, 123, 54] # num_list2 = [55, 80, 72, 35, 60, 123, 54, 29, 91] # num_list = [*num_list1,*num_list2] # print("合并后的列表：",num_list) # new_list = [] # for num in num_list2: # if num not in new_list: # new_list.append(num) # print("去重后的列表：",new_list) #减少遍历次数 # num_list1 = [19, 23, 54, 64, 875, 20, 109, 232, 123, 54] # num_list2 = [55, 80, 72, 35, 60, 123, 54, 29, 91] # # num_list = num_list1 + num_list2 # print("合并后的列表：",num_list) # new_list = [] # for num in num_list2: # if num not in new_list: # new_list.append(num) # print("去重后的列表：",new_list) #1到20的二次方 # num_list = [i**2 for i in range(1,21)] # print(num_list) #求指定列表中的偶数的二次方 # num_list2 = [55, 80, 72, 35, 60, 123, 54, 29, 91] # new_list = [i**2 for i in num_list2 if i%2==0] # print(new_list) #合并三个列表并去重 # list1 = ['M', 'A', 'C', 'E', 'F', 'G', 'H', 'L', 'N', 'I', 'J', 'K', 'O'] # list2 = ['X', 'Z', 'T', 'Y', 'D', 'E', 'F', 'G'] # list3 = ['W', 'A', 'S', 'D'] # old_list = [*list1, *list2, *list3] # print(old_list) # new_list =[] # for i in old_list: # if i not in new_list: # new_list.append(i) # new_list.sort() # print(new_list) #求列表中3或5倍数的平方 # list1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30] # list2 = [i**2 for i in list1 if i%3 == 0 or i%5 == 0] # print(list2) #取出列表中的正数 # list1 = [11, 2, 31, 4, -5, 15, 17, 28, 49, 10, -11, 16, 54, -14, 36, -16, 87, -19] # list = [i for i in list1 if i > 0] # print(list)