下面代码的执行结果是 ls1="hello" ls2="world" for i in ls2: if i not in ls1: print(i,end='') 单选题 (2 分) A. ol B. helo C. he D. wrd
时间: 2024-02-19 10:59:30 浏览: 35
这段代码的执行结果是:r。
代码中首先定义了两个字符串变量 ls1 和 ls2,分别赋值为 "hello" 和 "world"。然后使用 for 循环遍历 ls2 中的每个字符,将字符赋值给变量 i。在循环体中,使用 if 语句判断变量 i 是否存在于 ls1 中,如果不存在,则使用 print 函数输出 i 的值,并在输出后不换行。
因此,在循环的执行过程中,字符 'w'、'o'、'r' 和 'l' 都存在于字符串 ls2 中,但只有字符 'r' 不在字符串 ls1 中,因此只有字符 'r' 会被输出。最终的输出结果为 r。
因此,答案为 D. r。
相关问题
for i in range(16): sm+=int(ls1[i])*int(ls2[i])
As an AI language model, I can see that this code snippet is incomplete and does not have a clear objective. However, I can explain what this code is doing so far.
This is a for loop that will iterate 16 times. During each iteration, it will multiply the integer value at index i of ls1 with the integer value at index i of ls2 and assign the result to the variable 'sm'. However, since 'sm' is not used or stored anywhere, this loop is not doing anything useful.
It is possible that there is more code following this snippet that will make use of 'sm', but without that context, it is difficult to say.
列表ls1=[1,43],ls2=ls1,ls1[0]=22,计算两个列表运算结果
根据你提供的信息,我们可以得出以下代码:
```python
ls1 = [1, 43]
ls2 = ls1
ls1[0] = 22
```
此时,`ls1` 和 `ls2` 都指向同一个列表对象 `[22, 43]`。因此,对 `ls1` 进行的修改也会影响 `ls2`。
如果你想计算两个列表的运算结果,可以使用以下代码:
```python
ls1 = [1, 43]
ls2 = ls1.copy() # 复制列表
ls1[0] = 22
print(ls1) # 输出 [22, 43]
print(ls2) # 输出 [1, 43]
```
这样,`ls1` 和 `ls2` 就指向两个不同的列表对象,互不影响,可以分别进行操作。