您可以使用
Enumerable.Zip
组合两个的函数
List
s:
var ans = mealtraiDeserializeObject.MealID.Zip(mealtraiDeserializeObject.Day, (m, d) => new {
mealtraiDeserializeObject.TraineeID,
MealID = m,
Day = d
}).ToList();
您还可以使用(很少提及)的双参数版本
Enumerable.Select
,但我不认为效率的微小提高值得(IMO)可读性的降低:
var ans = mealtraiDeserializeObject.MealID.Select((m, i) => new {
mealtraiDeserializeObject.TraineeID,
MealID = m,
Day = mealtraiDeserializeObject.Day[i]
}).ToList();