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gabiteodoru
337 days ago
Or, even better, also from the cookbook: {(2#x)#1,x#0} But this really borders on obfuscation :P
1 comments
leprechaun1066
334 days ago
That does require someone to know that the take operator continues to treat the y list as circular when x is a list.
I think this form might be a bit easier: {(x,x)#(x*x)#1,x#0}
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gabiteodoru
334 days ago
But then you're not flexing your q muscles. Btw this is from the cookbook, these obfuscated patterns are actually recommended!
https://code.kx.com/phrases/matrix/#identity-matrix-of-order...
link
I think this form might be a bit easier: {(x,x)#(x*x)#1,x#0}