# How to delete more than one (specific) elements in a list

lst = {a, b, c}
DeleteCases[lst, a]


Delete one specific element a.

How do I delete elements a and c?

DeleteCases[lst, a || c]
(* does not work *)


How to I delete more than one (specific) elements efficiently?

• One | too much :) a|c. – Kuba Jan 20 '15 at 21:30
• Yup, as Kuba and soandos pointed out, pattern-matching uses |, whereas Boolean expressions use ||. You can read more at guide/Patterns, which contains a list of most of the pattern-matching syntax that is commonly used. It may seem weird that they use different syntax, but in Mathematica, expressions are considered different objects than patterns. Unlike expressions, which are used all the time, patterns are not commonly encountered by beginning users, and are strange to learn about at first (I'm still getting used to them), but they can sometimes be useful. – DumpsterDoofus Jan 20 '15 at 21:35
• @DumpsterDoofus I have mislead by the useless help files many times. Wolfram should really consider using some of the examples here, which are far more useful than its own examples. – Chen Stats Yu Jan 20 '15 at 21:39
• More generally, given a list A and b, you could get the elements of A that aren't in b using either Complement[A, b] or DeleteCases[A, Alternatives @@ b]. – DumpsterDoofus Jan 20 '15 at 21:40
• There is a large chapter on patterns in the documentaion. One of its sections deals with patterns involving alternatives. – Sjoerd C. de Vries Jan 20 '15 at 22:06

You can use Alternatives:
DeleteCases[lst, a|c]

• That's just what I thought. But why || does not represent OR in this case? – Chen Stats Yu Jan 20 '15 at 21:32
• @ChenStatsYu It does. The problem is that a||b is not a pattern object. Its not even a function, and so DeleteCases doesn't know what to do with it – soandos Jan 20 '15 at 21:34
• @soandos, a||c is a perfectly legitimate pattern. DeleteCases treats it just like any other pattern, e.g. DeleteCases[{a || c, a && c}, a || c] – Simon Woods Jan 20 '15 at 21:38
• @SimonWoods I think I mean that in the context of a pattern object, the || and && are not evaluated – soandos Jan 20 '15 at 21:39