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Consider the following list:

{{0.1, 5}, {0.3, 18}, {0.4, 6}, {0.7, 9}, {1.1, 5}}

I want to obtain the second element of a sub-list with respect to the first element of the sub-list.

for example: given 0.4 I would like to retrieve 6, given 0.1 I would like to retrieve 5.

The first element in the sub-lists are monotonically increasing and non repetitive.

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  • 2
    $\begingroup$ Have you tried anything? $\endgroup$
    – Kuba
    Commented May 16, 2018 at 13:55
  • 1
    $\begingroup$ @Kuba, now there are many things to try and to learn from. best $\endgroup$
    – jarhead
    Commented May 16, 2018 at 15:17

6 Answers 6

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Two possible examples (say for example you are interested in 0.1, but it works with any entry):

list = {{0.1, 5}, {0.3, 18}, {0.4, 6}, {0.7, 9}, {1.1, 5}};
Cases[list, {0.1, _}]
Select[list, #[[1]] == 0.1 &]

Edit: if you only want the second element, go for:

Cases[list, {0.1, _}][[1, 2]]
Select[list, #[[1]] == 0.1 &][[1, 2]]
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  • 5
    $\begingroup$ Consider also Cases[list, {0.1, b_} -> b] to return only the second element of each sublist. $\endgroup$
    – MarcoB
    Commented May 16, 2018 at 14:05
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Another approach is to convert your List to an Association:

list = {{0.1, 5}, {0.3, 18}, {0.4, 6}, {0.7, 9}, {1.1, 5}};
arse = AssociationThread @@ Transpose@list;
arse[0.4] (* 6 *)
arse[0.1] (* 5 *)

This assumes that each Key (each first element of a pair) are unique, but otherwise your question would be ill-posed.

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  • $\begingroup$ Please, don't call associations that way... ;) $\endgroup$ Commented May 16, 2018 at 15:21
  • $\begingroup$ @HenrikSchumacher Aah, I was going to ask for explanations (like it's a bad habit to call "association[something]`). Then I got it :D I'll edit to raise the ambiguity ;) $\endgroup$
    – anderstood
    Commented May 16, 2018 at 15:29
  • $\begingroup$ Not sure wether that is better... =D $\endgroup$ Commented May 16, 2018 at 15:31
  • $\begingroup$ It raises the ambiguity! $\endgroup$
    – anderstood
    Commented May 16, 2018 at 15:32
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Given that you only want the second element, and there are no repeats, you could use FirstCase

secondElement[firstElement_] := 
 FirstCase[{firstElement, other_} :> other]

secondElement[.1]@{{0.1, 5}, {0.3, 18}, {0.4, 6}, {0.7, 9}, {1.1, 5}}
(* 5 *)

secondElement[0.4]@{{0.1, 5}, {0.3, 18}, {0.4, 6}, {0.7, 9}, {1.1, 5}}
(* 6 *)
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Pick [#[[All, 2]], #[[All, 1]], 0.4] &@lst

{6}

Pick [#[[All, 2]], #[[All, 1]], 0.1] &@lst

{5}

Pick [#[[All, 2]], #[[All, 1]], 0.1 | 0.4] &@lst 

{5, 6}

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list = {{0.1, 5}, {0.3, 18}, {0.4, 6}, {0.7, 9}, {1.1, 5}};

Create an Association

look = <|Rule @@@ list|>

<|0.1 -> 5, 0.3 -> 18, 0.4 -> 6, 0.7 -> 9, 1.1 -> 5|>

Query one value

look[0.4]

6

Use Lookup to query more than one value

Lookup[look, {0.1, 0.4}]

{5, 6}

Handling of missing keys

Lookup[look, {0.1, 0.4, -1}]

{5, 6, Missing["KeyAbsent", -1]}

Lookup[look, {0.1, 0.4, -1}, Nothing]

{5, 6}

Use KeyTake to get keys and values (missing keys are ignored)

KeyTake[{0.1, 0.4, -1}] @ look

<|0.1 -> 5, 0.4 -> 6|>

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list = {{0.1, 5}, {0.3, 18}, {0.4, 6}, {0.7, 9}, {1.1, 5}};

Using ReplaceList:

ReplaceList[list, {___, s : {0.1, _}, ___} :> s]

{{0.1, 5}}

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