Is there a better way of calling LINQ Any + NOT All?

Question

I need to check if a sequence has any items satisfying some condition but at the same time NOT all items satisfying the same condition.

For example, for a sequence of 10 items I want to have TRUE if the sequence has at least one that satisfy the condition but not all:

• 10 items satisfying, 0 items not, result is FALSE
• 0 items satisfying, 10 items not, result is FALSE
• 1 item satisfying, 9 items not, result is TRUE
• 9 items satisfying, 1 item not, result is TRUE

I know I could to this:

mySequence.Any (item => item.SomeStatus == SomeConst) && !mySequence.All (item => item.SomeStatus == SomeConst) 

But this is not optimal.

Is there a better way?

Answer 1

You'll like this.

var anyButNotAll = mySequence     .Select(item => item.SomeStatus == SomeConst)     .Distinct()     .Take(2)     .Count() == 2; 

The Take(2) stops it iterating over any more elements than it has to.

Answer 2

If your concern is iterating through all of the elements in a large collection, you're okay - Any and All will short-circuit as soon as possible.

The statement

mySequence.Any (item => item.SomeStatus == SomeConst) 

will return true as soon as one element that meets the condition is found, and

!mySequence.All (item => item.SomeStatus == SomeConst) 

will return true as soon as one element does not.

Since both conditions are mutually exclusive, one of the statements is guaranteed to return after the first element, and the other is guaranteed to return as soon as the first element is found.

---

As pointed out by others, this solution requires starting to iterate through the collection twice. If obtaining the collection is expensive (such as in database access) or iterating over the collection does not produce the same result every time, this is not a suitable solution.