用输入替换列表中大于100的某些对象，然后

Question

我必须将列表的第5个元素乘以2，然后将结果放入新列表中，然后将两个列表合并。

我什至不知道如何启动它，更不用说继续了。

编辑

休息一会儿，这与我使用0知识所做的一样多。

L = [ 10, 2, 56, 33, 23, 1, 564, 32, 122, 42, 3, 4, 2, 1, 3, 2, 1, 54, 5, 9, 1, 65, 254 ]

x = int(input("Insert number here: "))

for i in range(0, int(len(L))):

    if L[i] > 100:

       L.pop(i)    

       L.insert(i, x)

       print(L)

Answer 1

根据您最清楚的解释，这是我的建议。我希望它符合您的期望：L = [ 10, 2, 56, 33, 23, 1, 564, 32, 122, 42, 3, 4, 2, 1, 3, 2, 1, 54, 5, 9, 1, 65, 254 ]# Ask for user inputmyInput = int(input("Enter a number: "))  # ex: I entered 5# Replace the numbers in L that are greater than 100 with the input numberL = [myInput if i > 100 else i for i in L]print(L) # ex: [10, 2, 56, 33, 23, 1, 5, 32, 5, 42, 3, 4, 2, 1, 3, 2, 1, 54, 5, 9, 1, 65, 5]# Take every 5th element of L, multiply it by 2 and place the results into a brand new list KK = [value*2 for i,value in enumerate(L,1) if i % 5 == 0]print(K) # ex: [46, 84, 6, 18]# Merge L and K into LKLK = L + Kprint(LK) # ex: [10, 2, 56, 33, 23, 1, 5, 32, 5, 42, 3, 4, 2, 1, 3, 2, 1, 54, 5, 9, 1, 65, 5, 46, 84, 6, 18]

Answer 2

根据以上评论，这是您的问题的解决方案。如果这不是您想要的，请告诉我，我会相应地进行更新。我在这里使用列表推导。我曾经(i+1)%5访问过第5个索引，因为该索引从0python开始。L = [ 10, 2, 56, 33, 23, 1, 564, 32, 122, 42, 3, 4, 2, 1, 3, 2, 1, 54, 5, 9, 1, 65, 254 ]x = int(input("Insert number here: "))L1 = [x if i > 100 else i for i in L]L2 = [2*j if (i+1)%5==0 else j for i, j in enumerate(L1)]L_output = L1 + L2print (L1)print (L2)print (L_output)输出Insert number here: 6[10, 2, 56, 33, 23, 1, 6, 32, 6, 42, 3, 4, 2, 1, 3, 2, 1, 54, 5, 9, 1, 65, 6][10, 2, 56, 33, 46, 1, 4, 32, 4, 84, 3, 4, 2, 1, 6, 2, 1, 54, 5, 18, 1, 65, 4][10, 2, 56, 33, 23, 1, 6, 32, 6, 42, 3, 4, 2, 1, 3, 2, 1, 54, 5, 9, 1, 65, 6, 10, 2, 56, 33, 46, 1, 4, 32, 4, 84, 3, 4, 2, 1, 6, 2, 1, 54, 5, 18, 1, 65, 4]

Answer 3

也许这样可以帮助：l1 = range(100)l2 = [l1[x]*2  if x%5==4 else l1[x] for x in range(len(l1)) ]print(l2)它仅修改每五个元素：因此，这些元素在第4、9、14等位置。（因此，x模5等于4）其他元素保持原样。