s1=set('hello'),s2=set('world'), s1&s2的结果c是(),
时间: 2024-01-25 09:01:56 浏览: 22
intersection(s2)
The output of this code would be {'o', 'l'}, because these are the letters that are common to both 'hello' and 'world'.
Here's how it works:
- `set('hello')` creates a set containing the unique characters in the string 'hello', which are 'h', 'e', 'l', and 'o'.
- `set('world')` creates a set containing the unique characters in the string 'world', which are 'w', 'o', 'r', 'l', and 'd'.
- `s1.intersection(s2)` returns a new set containing the elements that are in both `s1` and `s2`. In this case, that means the letters 'o' and 'l', since those are the only letters that appear in both 'hello' and 'world'.
相关问题
s1=set('hello'),s2=set('world'), 程序s1&s2的运行结果是()
和s2的交集:
```python
s1 = set('hello')
s2 = set('world')
intersection = s1.intersection(s2)
print(intersection)
```
输出结果:
```
{'o', 'l'}
```
说明:程序使用`set()`函数将字符串转换为集合,然后使用`intersection()`方法获取两个集合的交集,最后输出结果。
有两个字符串s1和s2,删除s1中和s2重复的字符,形成新的s1,最后输出s1拼接上s2的结果,用Python实现
可以使用Python中的集合(set)数据结构和字符串拼接操作来实现。
代码如下:
```python
s1 = "hello world"
s2 = "python"
# 将s2转换为集合类型,方便后面判断s1中是否存在相同的字符
s2_set = set(s2)
# 遍历s1中的每个字符,如果不在s2_set中,则加入新的字符串中
new_s1 = ""
for c in s1:
if c not in s2_set:
new_s1 += c
# 拼接新的s1和s2
result = new_s1 + s2
print(result)
```
输出结果为:
```
he wrldpython
```