使用元组来保存下表中员工的薪水,并尝试将元组中第三个数据修改为6200,在控制台输出这个代码怎么写
直接运行
#表中有三个员工的薪水分别为5000、5500、6000元,可以使用元组来保存
salaries = (5000, 5500, 6000)
salaries_list = list(salaries)
salaries_list[2] = 6200
salaries = tuple(salaries_list)
print(salaries)
#include<opencv2/opencv.hpp>
#include<time.h>
#include<iostream>
using namespace std;
using namespace cv;
int main() {
system("color 2F");
Mat image(600, 600, CV_8UC3);
RNG& rng = theRNG();
while (1) {
int count = rng.uniform(3, 103); //随机生成点的数量
vector<Point> points;
//随机生成点的坐标
for (int i = 0; i < count; i++) {
Point point;
point.x = rng.uniform(image.cols / 4, image.cols * 3 / 4);
point.y = rng.uniform(image.rows / 4, image.rows * 3 / 4);
points.push_back(point);
}
//对给定的 2D 点集,寻找最小面积的包围圆形
Point2f center;
float radius = 0;
minEnclosingCircle(Mat(points), center, radius);
//绘制出随机颜色的点
image = Scalar::all(0);
for (int i = 0; i < count; i++) {
circle(image, points[i], 3, Scalar(rng.uniform(0, 255),
rng.uniform(0, 255), rng.uniform(0, 255)), FILLED, LINE_AA);
}
//绘制出最小面积的包围圆
circle(image, center, cvRound(radius), Scalar(rng.uniform(0, 255),
rng.uniform(0, 255), rng.uniform(0, 255)), 2, LINE_AA);
imshow("圆形包围示例", image);
char key = (char)waitKey();
if (key == 27 || key == 'q' || key == 'Q') {
break;
}
}
return 0;
}
运行图:
解决方案:
使用元组保存员工薪水表格,每个元组包含员工的姓名、职位和薪水,可以使用以下代码:
staff_salary = (('Tom', 'Manager', 8000), ('Jack', 'Engineer', 6000), ('Tony', 'Assistant', 4000))
要修改元组的第三个数据,可以将元组转换成列表,修改完后再转换回元组,需要使用以下代码:
staff_salary_list = list(staff_salary)
staff_salary_list[2] = ('Tony', 'Assistant', 6200)
staff_salary = tuple(staff_salary_list)
最后将修改后的元组输出到控制台,可以使用以下代码:
print(staff_salary)
完整代码如下:
staff_salary = (('Tom', 'Manager', 8000), ('Jack', 'Engineer', 6000), ('Tony', 'Assistant', 4000))
staff_salary_list = list(staff_salary)
staff_salary_list[2] = ('Tony', 'Assistant', 6200)
staff_salary = tuple(staff_salary_list)
print(staff_salary)
输出结果为:
(('Tom', 'Manager', 8000), ('Jack', 'Engineer', 6000), ('Tony', 'Assistant', 6200))