What is the idiomatic way of counting the number of elements matching predicate?

问题

Is there a better way to count the number of elements for which the predicate function is true, other than this:

PredCount[lst_, pred_] := Length@Select[lst, pred];

I'm asking because it seems inefficient to construct a subset of lst with Select[], and because Count[] only works with patterns. In my use case, the function PredCount is called many times with a large lst.

回答1:

You can often do this by turning your predicate into a pattern with a condition. For example:

Count[list, x_/;x>5]

would count the number of elements in list which are greater than 5.

回答2:

I would use PatternTest

PredCount = Count[#, _?#2] &;

PredCount[Range@30, PrimeQ]

(*out*) 10

This pattern is simple enough that you might use Count directly.

来源：https://stackoverflow.com/questions/10344339/what-is-the-idiomatic-way-of-counting-the-number-of-elements-matching-predicate