AM Posted May 7, 2017 Share Posted May 7, 2017 perhaps there's a OM-solution... in this case it's to hard to find... (search-engine?) otherwise... (defun integer-to-binary-lengths* (alist) (loop for i in alist when (> (abs i) 1) append (append (list 1) (loop repeat (- (abs i) 1) collect 0)) else collect 1)) (integer-to-binary-lengths* '(2 2 2 1 1 4 4 4 4)) (integer-to-binary-lengths* '(6 4 8 5 2 1 10 2)) Quote Link to comment Share on other sites More sharing options...
opmo Posted May 7, 2017 Share Posted May 7, 2017 (binary-to-decimal '(1 1 0)) => 6 (decimal-to-binary '(9 3 6 1 10 11)) => ((1 0 0 1) (1 1) (1 1 0) (1) (1 0 1 0) (1 0 1 1)) Quote Link to comment Share on other sites More sharing options...
AM Posted May 7, 2017 Author Share Posted May 7, 2017 it's not decimal-to-binary!!! another idea... for example... 6 => 1 0 0 0 0 0 = a "1" and 5 times a "0") Quote Link to comment Share on other sites More sharing options...
Stephane Boussuge Posted May 8, 2017 Share Posted May 8, 2017 I've made this long time ago, bit different but may be also of interest, this function use a specs of number of "1" and "0" in input: ;;; GEN-BINARY_INTEGER (defun gen-binary-integer (one-list zero-list &key (flatten t)) (do-verbose ("gen-binary-integer") (if flatten (flatten (loop for i in one-list for o in (gen-trim (length one-list) zero-list) collect (append (gen-repeat i '(1)) (gen-repeat o '(0)) ))) (loop for i in one-list for o in (gen-trim (length one-list) zero-list) collect (append (gen-repeat i '(1)) (gen-repeat o '(0)) ))))) #| USAGE (gen-binary-integer '(3 4 3 2 4 2 5) '(1 1 3)) => (1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 1 1 1 1 0 1 1 0 0 0 1 1 1 1 1 0) (gen-binary-integer '(3 4 3 2 4 2 5) '(1 1 3) :flatten nil) => ((1 1 1 0) (1 1 1 1 0) (1 1 1 0 0 0) (1 1 0) (1 1 1 1 0) (1 1 0 0 0) (1 1 1 1 1 0)) (setf intg '(3 2 4 5 3)) (setf zero-k '(1 2 1 3)) (gen-binary-integer intg zero-k) => '(1 1 1 0 1 1 0 0 1 1 1 1 0 1 1 1 1 1 0 0 0 1 1 1 0) |# S. AM 1 Quote Link to comment Share on other sites More sharing options...
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