aList[1::2]=map(lambda x:x!=5,range(3))
时间: 2024-06-07 17:09:22 浏览: 15
这行代码的作用是将 aList 列表中从下标为1开始的每隔一个元素替换成 True 或 False,具体取决于 range(3) 中对应下标的元素是否等于 5。
更详细地说,range(3) 生成的是一个包含 0、1、2 的整数序列,lambda 函数将这个序列中等于 5 的元素映射成 False,其余元素映射成 True。然后使用 map() 函数将这个映射后的序列与 aList 列表的每隔一个元素(即下标为1、3、5...的元素)进行替换。最终替换后的列表赋值回 aList。
需要注意的是,这行代码可能会引发 IndexError 异常,具体取决于 aList 列表的长度是否足够长以覆盖从下标为1开始每隔一个元素的区间。
相关问题
for num in range(100,1000): r=map(lambda x:int(x)**3,str(num))
if sum(r)==num: print(num)
This code finds all the Armstrong numbers between 100 and 999. An Armstrong number is a number that is equal to the sum of its own digits raised to the power of the number of digits. For example, 153 is an Armstrong number because 1^3 + 5^3 + 3^3 = 153.
The code uses a for loop to iterate through all the numbers between 100 and 999. It then converts each number to a string and applies a lambda function to each digit to raise it to the power of 3. The resulting list of cubed digits is then summed up and checked against the original number. If they are equal, the number is printed out as an Armstrong number.
for num in range(100,1000): r=map(lambda x:int(x)**3,str(num)) if sum(r)==num: print(num)
As an AI language model, I cannot execute code, but I can explain what this code does.
This code is checking for all three-digit numbers that are equal to the sum of the cubes of their digits.
The code first iterates through all numbers from 100 to 999 using the `range()` function.
For each number, it converts it to a string and then maps the `int()` function to each digit in the string to convert it to an integer. It then cubes each integer using the `**` operator and returns the list of cubes.
The `sum()` function is then used to add up all the cubes, and if the sum equals the original number, it is printed to the console.
This code is an example of a brute-force approach to solving a problem, as it simply checks every possible three-digit number until it finds the ones that meet the criteria.