有两个列表:
List<string> files;
List<Filter> filters;
Run Code Online (Sandbox Code Playgroud)
我希望结果如下:
List<KeyValuePair<string, Filter>> fileFilterMap;
Run Code Online (Sandbox Code Playgroud)
我尝试了几个东西(lambda表达式,linq)但失败了.我真的不想要
for(int i = 0; i< files.count; i++)
方法.
Guf*_*ffa 13
您可以使用:
List<KeyValuePair<string, Filter>> fileFilterMap =
Enumerable.Range(0, files.Count)
.Select(i => new KeyValuePair<string, Filter>(files[i], filters[i]))
.ToList();
Run Code Online (Sandbox Code Playgroud)
要么:
List<KeyValuePair<string, Filter>> fileFilterMap =
Enumerable.Zip(
files,
filters,
(file, filter) => new KeyValuePair<string, Filter>(file, filter)
)
.ToList();
Run Code Online (Sandbox Code Playgroud)
var fileFilterMap = Enumerable.Zip(files, filters, (file, filter) => new KeyValuePair(file, filter));
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
3508 次 |
| 最近记录: |