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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"}}
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lis = {{M, 1, 0, 0, 1, 1, 1, 0}, {A, 0, 0, 0}};
{First@#, StringJoin[ToString /@ Rest@#]} & /@ lis

(* {{M, "1001110"}, {A, "000"}} *)
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  • 3
    $\begingroup$ Similarly, {#1, StringJoin @@ ToString /@ {##2}} & @@@ lis. $\endgroup$ – march Apr 5 '16 at 21:38
  • $\begingroup$ @march. That's a good one too. $\endgroup$ – RunnyKine Apr 5 '16 at 22:02
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Cases[{{M, 1, 0, 0, 1, 1, 1, 0}, {A, 0, 0, 0}}, {x_, y__} :> {x, 
   StringJoin[ToString /@ {y}]}]
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  • $\begingroup$ Why RuleDelayed required? (It is, I know; but why?) For example, if the second pattern mapped to just {y}, it would not be required. $\endgroup$ – Alan Apr 5 '16 at 21:01
  • $\begingroup$ @Alan I assume to keep y from being shadowed by another variable named y $\endgroup$ – Daniel Apr 5 '16 at 22:12
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Transpose[{a[[;; , 1]], StringJoin @@@ IntegerString @ a[[;; , 2 ;;]]}]
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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

enter image description here

Hope this helps.

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