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Suppose I have a nested list of the form:

test = {{1,2},{2,3},{1,6},{2,5},{2,2},{1,7},{1,4}}

How do I find the position of the list in which the elements are equal, which in this example is {5} -> position of {2,2}

Update 1

Suppose my nested list is:

{{0.99258878, 0.99258845}, {0.97357684, 0.97357712}, {0.95312105, 
  0.95386979}, {0.93345214, 0.93345301}, {0.91233411, 0.91233415}, {0.89050811, 
  0.89050798}, {0.86800801, 0.86800954}, {0.84486887, 0.84486964}, {0.82114124, 
  0.82114187}, {0.79689807, 0.79689753}}

I wish to find the position for which elements are equal with a 6 significant digit accuracy, like the position of the list {0.99258878, 0.99258845}

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5
  • $\begingroup$ Position[test, {a_, a_}] $\endgroup$
    – Syed
    Commented Mar 22, 2023 at 7:53
  • $\begingroup$ @Syed thank you, suppose I have a very big nested list in which some elements are approximately equal, say a list like {1.5,1.4999} or {1.88812,1.88913} how to include the positions of these as well? How to state the level of accuracy? $\endgroup$
    – codebpr
    Commented Mar 22, 2023 at 7:58
  • $\begingroup$ Please update the post and present a list with real numbers and the accuracy required. This is new information. $\endgroup$
    – Syed
    Commented Mar 22, 2023 at 8:01
  • $\begingroup$ @Syed I have updated my question! $\endgroup$
    – codebpr
    Commented Mar 22, 2023 at 8:20
  • 2
    $\begingroup$ A variation on the method given below by Syed: Round[lst,10^-6]//MapApply[Subtract]//Position[0] $\endgroup$
    – user1066
    Commented Mar 23, 2023 at 18:39

1 Answer 1

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Using Round with the second argument:

test = {{0.99258878, 0.99258845}, {0.97357684, 
   0.97357712}, {0.95312105, 0.95386979}, {0.93345214, 
   0.93345301}, {0.91233411, 0.91233415}, {0.89050811, 
   0.89050798}, {0.86800801, 0.86800954}, {0.84486887, 
   0.84486964}, {0.82114124, 0.82114187}, {0.79689807, 0.79689753}}

Position[test, _?(SameQ @@ (Round[#, 10^-6]) &)]

{{2}, {5}, {6}, {10}}

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