Conditional replacement syntax

Question

Why does the following expression not do what I expect

Array[a, 6, 0] /. a[__?MemberQ[{1, 2}, #] &] -> 0

I expect

{a{0],0,0,a[3],a{4],a[5]}

but I get

{a{0],a[1],a[2],a[3],a{4],a[5]}

What do I need to change to get what I want?

Array[a, 6, 0] /. a[__?(MemberQ[{1, 2}, #] &)] -> 0 – ilian Jun 08 '15 at 13:44 — ilian, Jun 08 '15 at 13:44

score 3 · Answer 1 · edited Jun 08 '15 at 14:00

3

You should replace __?MemberQ[{1, 2}, #] & with __?(MemberQ[{1, 2}, #] &). Because the & has the low priority. So, the pattern is interpreted as part of the function, and not the function as part of the pattern. This can be easily seen using FullForm:

In[24]:= __?MemberQ[{1, 2}, #] & // FullForm

Out[24]//FullForm= 
 Function[PatternTest[BlankSequence[], MemberQ][List[1, 2], Slot[1]]]

vs

In[25]:= __?(MemberQ[{1, 2}, #] &) // FullForm

Out[25]//FullForm= 
 PatternTest[BlankSequence[], Function[MemberQ[List[1, 2], Slot[1]]]]

Using the modified function, you get

Array[a, 6, 0] /. a[__?(MemberQ[{1, 2}, #] &)] -> 0

{a[0], 0, 0, a[3], a[4], a[5]}

edited Jun 08 '15 at 14:00

rcollyer

33,976
7
92
191

answered Jun 08 '15 at 13:51

xyz

605
4
38
117

1

@rcollyer, Very nice:-) Thanks a lot – xyz Jun 08 '15 at 14:02
3

This should be part of the common pitfalls for new users, but I can't seem to find it. Also, I know it is a duplicate, but finding it is difficult, at best. So, I made this as canonical of answer as I could. – rcollyer Jun 08 '15 at 14:06

Conditional replacement syntax

1 Answers1