Timeline for Symbolically replace part of expression if that expression meets a condition
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 24, 2016 at 0:39 | vote | accept | alessandro | ||
| Sep 22, 2016 at 12:40 | answer | added | user31159 | timeline score: 2 | |
| Sep 20, 2016 at 2:39 | comment | added | alessandro | Thank you, the : was just what I needed. Currently my solution to this problem is Replace[expr, subexpr : pre_. f_[I1_, I2_, I3_, I4_] post_. :> ReplaceAll[subexpr, {i -> I1, I1 -> i, j -> I2, I2 -> j, k -> I3, I3 -> k, l -> I4, I4 -> l}], {1}] | |
| Sep 19, 2016 at 21:08 | history | edited | alessandro | CC BY-SA 3.0 | Question, as asked, given the example was logically impossible to solve. |
| Sep 19, 2016 at 11:42 | comment | added | user31159 | Something like: Replace[expr, subexpr : ___ f2[___] :> Replace[subexpr, {l -> j, j -> l}, {2}], {1}]? | |
| Sep 19, 2016 at 11:34 | history | asked | alessandro | CC BY-SA 3.0 |