1. arr1 = [1,2,3,4,5,6,78,8] 2. arr2 = [1,2,4,5,777,98]实现数组交集并集和差集
时间: 2023-08-19 21:26:46 浏览: 47
好的,针对你的问题,我可以提供以下实现:
1. 数组交集:
```python
arr1 = [1,2,3,4,5,6,78,8]
arr2 = [1,2,4,5,777,98]
intersection = list(set(arr1).intersection(set(arr2)))
print(intersection)
```
输出结果为:`[1, 2, 4, 5]`
2. 数组并集:
```python
arr1 = [1,2,3,4,5,6,78,8]
arr2 = [1,2,4,5,777,98]
union = list(set(arr1).union(set(arr2)))
print(union)
```
输出结果为:`[1, 2, 3, 4, 5, 6, 98, 777, 8, 78]`
3. 数组差集:
```python
arr1 = [1,2,3,4,5,6,78,8]
arr2 = [1,2,4,5,777,98]
diff_arr1 = list(set(arr1).difference(set(arr2)))
diff_arr2 = list(set(arr2).difference(set(arr1)))
print("arr1中独有的元素:", diff_arr1)
print("arr2中独有的元素:", diff_arr2)
```
输出结果为:
```
arr1中独有的元素: [8, 3, 6, 78]
arr2中独有的元素: [98, 777]
```
希望我的回答对你有所帮助!
相关问题
1. 输出打印数组 arr = {{1,2,3},{4,5,6},{7,8,9}}
以下是打印二维数组的方法:
```java
int[][] arr = {{1,2,3},{4,5,6},{7,8,9}};
for(int i=0;i<arr.length;i++){
for(int j=0;j<arr[i].length;j++){
System.out.print(arr[i][j] + " ");
}
System.out.println();
}
```
输出结果为:
```
1 2 3
4 5 6
7 8 9
```
recomMv = arr2[arr2.index.intersection(arr1.index)]
这段代码的目的是从arr2中选择那些在arr1中也存在的索引值,并将对应的元素存储在recomMv中。
- `arr1.index`表示arr1的索引值。
- `arr2.index`表示arr2的索引值。
- `arr2.index.intersection(arr1.index)`表示arr2和arr1索引值的交集,即在arr2和arr1中都存在的索引值。
- `arr2[arr2.index.intersection(arr1.index)]`通过索引值的交集,从arr2中选择相应的元素,存储在recomMv中。