Timeline for Replace x, but not x[a] [duplicate]
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 4, 2015 at 14:43 | history | closed | Mr.Wizard | Duplicate of Pattern to match L, but not L[_] | |
| Jun 3, 2015 at 21:04 | answer | added | Mr.Wizard | timeline score: 2 | |
| Nov 13, 2014 at 7:19 | vote | accept | Martin J.H. | ||
| Nov 13, 2014 at 2:15 | answer | added | Basheer Algohi | timeline score: 1 | |
| Nov 13, 2014 at 0:13 | history | tweeted | twitter.com/#!/StackMma/status/532687573458509824 | ||
| Nov 12, 2014 at 21:44 | answer | added | Szabolcs | timeline score: 11 | |
| Nov 12, 2014 at 21:29 | comment | added | Martin J.H. |
@ChipHurst: Yes, that of course works. I feel sufficiently foolish now. (I think I had tried {x[a] -> x, x -> x[a]}, which was nonsense.)
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| Nov 12, 2014 at 21:23 | comment | added | Greg Hurst |
How about x[a] + x /. {x[a] -> x[a], x -> x[a]}
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| Nov 12, 2014 at 21:13 | comment | added | kglr |
... also x[a] + x /. {x -> y, z : _[_] :> z}?
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| Nov 12, 2014 at 21:08 | history | edited | Martin J.H. | CC BY-SA 3.0 |
realized my problem was more specific
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| Nov 12, 2014 at 21:07 | comment | added | kglr |
Replace[x[a] + x, x -> y, {1}]?
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| Nov 12, 2014 at 20:59 | history | asked | Martin J.H. | CC BY-SA 3.0 |