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We have a list containing IntegerDigits elements as 

list0={{0,0,1,0,0,0},
 {0,0,1,0,1,0},
 {0,0,1,1,0,0},
 {0,0,1,1,1,0},
 {0,1,1,0,0,0},
 {0,1,1,0,1,0},
 {0,1,1,1,1,0},
 {1,0,1,0,0,0},
  ...};

Two any arbitrary positions of each element can be definitely determined by 0 or 1. For instants, in the above list, we have 0 in the first position and 1 in the fourth position. They are fixed and are not changeable. We have to create a list whose elements are -1 or +1. If the Total between two mentioned positions isEven (if evenQ=True) it must be +1, otherwise it must be -1. Or we can say (-1)^(Total[elements between first and fourth positions]) I mean:

list1={{1}, (*0+0=0*)
       {-1}, (*0+1=1*)
       {-1}, (*1+0=1*)
       {+1},  (*1+1=2*)
       {+1},
       {-1},
       {+1},
       {+1},
       ....}

How can I do this aim?

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6
  • $\begingroup$ Why do you write (*0+0=0*) for the first one? It has 1 in the 3rd position, and 0 in the 6th position. $\endgroup$
    – Coolwater
    Aug 17, 2017 at 16:23
  • $\begingroup$ Should the position be taken backwards because IntegerDigits puts the coefficient of the smallest power of the base at the end of the output list $\endgroup$
    – Coolwater
    Aug 17, 2017 at 16:25
  • 2
    $\begingroup$ Something like (-1)^Total[list0[[All, {4, 5}]], {2}]? $\endgroup$
    – Carl Woll
    Aug 17, 2017 at 16:26
  • $\begingroup$ For your second comment you are right. So sorry. But for the first comment: there are two 0s between 1 and 0. $\endgroup$
    – Inzo Babaria
    Aug 17, 2017 at 16:27
  • $\begingroup$ So sorry I corrected them. $\endgroup$
    – Inzo Babaria
    Aug 17, 2017 at 16:29

2 Answers 2

Reset to default
2
$\begingroup$ 
list0 =
 {{0, 0, 1, 0, 0, 0}, {0, 0, 1, 0, 1, 0}, {0, 0, 1, 1, 0, 0},
  {0, 0, 1, 1, 1, 0}, {0, 1, 1, 0, 0, 0}, {0, 1, 1, 0, 1, 0},
  {0,1, 1, 1, 1, 0}, {1, 0, 1, 0, 0, 0}};

If[#, 1, -1]& @* EvenQ @* Total /@ list0[[All, 4 ;; 5]]

{1, -1, -1, 1, 1, -1, 1, 1}

Or, similar to Carl Woll's comment,

(-1)^(Plus @@@ list0[[All, 4 ;; 5]])
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2
  • $\begingroup$ Thank you so much but because the positions are arbitrary the Total maybe exceeds from 2 or 0 $\endgroup$
    – Inzo Babaria
    Aug 17, 2017 at 16:31
  • $\begingroup$ See updated answer $\endgroup$
    – eldo
    Aug 17, 2017 at 16:41
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I suspect you started out with the integers themselves:

ints = FromDigits[#, 2] & /@ list0

{8, 10, 12, 14, 24, 26, 30, 40}

If that's the case, then you don't need IntegerDigits:

(-1)^Sign[Mod[BitAnd[2^1 + 2^2, #], 2^1 + 2^2]] & /@ ints 
(-1)^Sign[Mod[BitAnd[6, #], 6]] & /@ ints

{1, -1, -1, 1, 1, -1, 1, 1}

where 2^0 corresponds to the last element of IntegerDigits and 2^1 corresponds to the second last, etc.

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7
  • $\begingroup$ What is dd? It was not defined! $\endgroup$
    – Inzo Babaria
    Aug 17, 2017 at 16:34
  • $\begingroup$ @InzoBabaria I removed them $\endgroup$
    – Coolwater
    Aug 17, 2017 at 16:35
  • $\begingroup$ I cannot understand why 6?! $\endgroup$
    – Inzo Babaria
    Aug 17, 2017 at 16:39
  • $\begingroup$ I cannot generalize to an arbitrary case $\endgroup$
    – Inzo Babaria
    Aug 17, 2017 at 16:39
  • 2
    $\begingroup$ You might be interested in BitGet[]. $\endgroup$
    – J. M.'s slightly less busy ♦
    Aug 17, 2017 at 16:49

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