Timeline for Transform a rule to a list of rules
Current License: CC BY-SA 3.0
10 events
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Apr 13, 2017 at 12:56 | history | edited | CommunityBot |
replaced http://mathematica.stackexchange.com/ with https://mathematica.stackexchange.com/
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Nov 14, 2014 at 21:17 | comment | added | higgy |
This answers my question. I forgot that x -> y could be expressed as Rule[x,y], and therefore be used with Thread.
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Nov 14, 2014 at 21:16 | vote | accept | higgy | ||
Nov 14, 2014 at 20:40 | comment | added | higgy | In chat. Any way to freeze this question until after we've chatted? | |
Nov 14, 2014 at 20:36 | comment | added | Szabolcs | @higgy No, it's not necessary. I think we're having a communication breakdown. Why don't you join chat? | |
Nov 14, 2014 at 20:35 | comment | added | Szabolcs |
@higgy You don't need to know the first part of the rule to apply Thread to it. If you have x = a -> {1,2,3} then you can just use Thread[x] , no need to know what's inside x , other than that it's a rule with a list on the RHS. Can you give an example input and a corresponding example output? I thought a -> {1, 2, 3} and {a -> 1, a -> 2, a -> 3} could be considered such. Thread takes you from one to the other.
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Nov 14, 2014 at 20:34 | comment | added | higgy | If extracting the left-hand side is a necessary step in all possible solutions to this question, then edit your answer I and will accept it. I'm also curious as to whether there are solutions that don't require extracting the left-hand side first, i.e., something more direct. | |
Nov 14, 2014 at 20:29 | comment | added | Szabolcs |
@higgy a -> b is just an alternate notation for Rule[a,b] , so you can extract a using First , Part , Extract , etc. See here: reference.wolfram.com/language/tutorial/PartsOfExpressions.html
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Nov 14, 2014 at 20:27 | comment | added | higgy | How would this be done without knowing that "a" is "a". In your answer, you refer to "a" in your code. I'll update the question to make it clearer. A related question would be: How do you extract the left-hand side of a rule efficiently? | |
Nov 14, 2014 at 20:25 | history | answered | Szabolcs | CC BY-SA 3.0 |