foreach (Feature fe in features)
{
if (fileNames.Any(file => file.Contains(fe.FeatureName)))
{
fe.MatchCount = fe.MatchCount == 0 ? 1 : fe.MatchCount;
}
}
答案 0 :(得分:9)
你是在循环变量的末尾变异对象,所以你不能在 pure LINQ中这样做(干净利落)。保持循环;理解它会更简单,但也许它可以减少位:
var qry = features.Where(fe => fe.MatchCount == 0 &&
fileNames.Any(file => file.Contains(fe.FeatureName));
foreach (Feature fe in qry) { fe.MatchCount == 1; }
答案 1 :(得分:3)
features.Where(f => fileNames.Any(file => file.Contains(f.FeatureName)))
.ToList()
.ForEach(x => x.MatchCount = x.MatchCount == 0 ? 1 : x.MatchCount);
答案 2 :(得分:1)
值得一提的是,如果您的列表非常大,将查询实现到列表,然后使用'ForEach'再次迭代它可能是一个非常昂贵的调用。我建议添加以下扩展方法,为IEnumerable提供'ForEach'方法:
public static void Map<T>(this IEnumerable<T> source, Action<T> func)
{
foreach (T i in source)
func(i);
}
我称之为地图,但如果您愿意,可以将其称为ForEach。这将Danny的答案变为:
features.Where(f => fileNames.Any(file => file.Contains(f.FeatureName)))
.Map(x => x.MatchCount = x.MatchCount == 0 ? 1 : x.MatchCount);
答案 3 :(得分:0)
Func<Feature, Feature> updateMatchCount = (fe) =>{
fe.MatchCount = fe.MatchCount == 0 ? 1 : fe.MatchCount;
return fe;
};
var updatedFeatures = from fe in features
where fileNames.Any(file => file.Contains(fe.FeatureName))
select updateMatchCount(fe);