how easily can I convert elements of this list to string. If I have output like this:
{{M, 1, 0, 0, 1, 1, 1, 0}, {A, 0, 0, 0}}
And I want something like this:
{{M, 1001110}, {A, 000}}
Or
{{M, "1001110"}, {A, "000"}}
$\begingroup$
$\endgroup$
Add a comment
|
$\begingroup$
$\endgroup$
2
lis = {{M, 1, 0, 0, 1, 1, 1, 0}, {A, 0, 0, 0}};
{First@#, StringJoin[ToString /@ Rest@#]} & /@ lis
(* {{M, "1001110"}, {A, "000"}} *)
-
3$\begingroup$ Similarly,
{#1, StringJoin @@ ToString /@ {##2}} & @@@ lis. $\endgroup$– marchApr 5 '16 at 21:38 -
$\begingroup$
$\endgroup$
2
Cases[{{M, 1, 0, 0, 1, 1, 1, 0}, {A, 0, 0, 0}}, {x_, y__} :> {x,
StringJoin[ToString /@ {y}]}]
answered Apr 5 '16 at 19:11
-
$\begingroup$ Why RuleDelayed required? (It is, I know; but why?) For example, if the second pattern mapped to just
{y}, it would not be required. $\endgroup$– AlanApr 5 '16 at 21:01 -
$\begingroup$ @Alan I assume to keep
yfrom being shadowed by another variable namedy$\endgroup$– DanielApr 5 '16 at 22:12
$\begingroup$
$\endgroup$
Transpose[{a[[;; , 1]], StringJoin @@@ IntegerString @ a[[;; , 2 ;;]]}]
$\begingroup$
$\endgroup$
If on the off chance you are seeking binary output for numbers then:
dat = {{M, 1, 0, 0, 1, 1, 1, 0}, {A, 0, 0, 0}};
{First@#, BaseForm[FromDigits[Rest@#, 2], 2]} & /@ dat
Hope this helps.
