有没有办法在VSCode中映射每个文件的特定启动配置?

如果我想跑步怎么办 streamlit 对于 app.py 和 python 的其余部分 *.py 文件。如何跳过选择步骤并仅使用的快速组合 Ctrl+F5 (或 F5 )让VSCode根据文件名来决定?

我在中添加了多个启动配置 .vscode/launch.json :

{
    // Use IntelliSense to learn about possible attributes.
    // Hover to view descriptions of existing attributes.
    // For more information, visit: https://go.microsoft.com/fwlink/?linkid=830387
    "version": "0.2.0",
    "configurations": [
        {
            "name": "Python: app.py",
            "type": "python",
            "request": "launch",
            "module": "streamlit",
            "console": "integratedTerminal",
            "justMyCode": true,
            "args": [
                "run",
                "app.py"
            ],
            "serverReadyAction": {
                "pattern": "You can now view your Streamlit app in your browser",
                "uriFormat": "http://localhost:8501",
                "action": "openExternally"
              }
        },
        {
            "name": "Python: Current File",
            "type": "python",
            "request": "launch",
            "program": "${file}",
            "console": "integratedTerminal",
            "justMyCode": true
        }
    ]
}

所以现在我可以选择其中的任何一个来运行当前在VSCode中打开的文件,但我想节省一些时间。