问题
I'm a bit surprised to find the results of the following code, where I simply want to remove all 3s from a sequence of ints:
var sequence = new [] { 1, 1, 2, 3 };
var result = sequence.SkipWhile(i => i == 3); // Oh noes! Returns { 1, 1, 2, 3 }
Why isn't 3 skipped?
My next thought was, OK, the Except operator will do the trick:
var sequence = new [] { 1, 1, 2, 3 };
var result = sequence.Except(i => i == 3); // Oh noes! Returns { 1, 2 }
In summary,
- Except removes the 3, but also removes non-distinct elements. Grr.
- SkipWhile doesn't skip the last element, even if it matches the condition. Grr.
Can someone explain why SkipWhile doesn't skip the last element? And can anyone suggest what LINQ operator I can use to remove the '3' from the sequence above?
回答1:
It's not broken. SkipWhile will only skip items in the beginning of the IEnumerable<T>. Once that condition isn't met it will happily take the rest of the elements. Other elements that later match it down the road won't be skipped.
int[] sequence = { 3, 3, 1, 1, 2, 3 };
var result = sequence.SkipWhile(i => i == 3);
// Result: 1, 1, 2, 3
回答2:
var result = sequence.Where(i => i != 3);
回答3:
The SkipWhile and TakeWhile operators skip or return elements from a sequence while a predicate function passes (returns True). The first element that doesn’t pass the predicate function ends the process of evaluation.
//Bypasses elements in a sequence as long as a specified condition is true and returns the remaining elements.
回答4:
One solution you may find useful is using List "FindAll" function.
List <int> aggregator = new List<int> { 1, 2, 3, 3, 3, 4 };
List<int> result = aggregator.FindAll(b => b != 3);
回答5:
Ahmad already answered your question, but here's another option:
var result = from i in sequence where i != 3 select i;
来源：https://stackoverflow.com/questions/2527047/linq-is-skipwhile-broken