How to find elements in a list satisfying different criteria

Question

I am not good at Mathematica. I have the following list already ordered according to the second part:

list1 = {{3, 2}, {5, 2}, {7, 2}, {7, 4}, {8, 4}, {12, 8}, {94, 8}, {4, 10}, {42, 10}};

I want to select the first elements where the second part gets bigger. The result would be:

list2 = {{3, 2}, {7, 4}, {12, 8}, {4, 10}}

Αλέξανδρος Ζεγγ · Accepted Answer · 2020-01-27 15:28:19Z

3

One approach goes like this:

First /@ GatherBy[list1, Last]

It turns out that GatherBy can be replaced by SplitBy.

answered Dec 18 '18 at 14:12

8,1512 gold badges15 silver badges36 bronze badges

$\begingroup$ Wow! Thank you very much! $\endgroup$ – user57467 Dec 18 '18 at 14:16
$\begingroup$ @user57467 Glad to be helpful. $\endgroup$ – Αλέξανδρος Ζεγγ Dec 18 '18 at 14:24

add a comment |

kglr · Accepted Answer · 2018-12-19 11:52:03Z

3

You can also use SequenceReplace:

SequenceReplace[list1, {p : ({_, a_} ..)} :> First[{p}]]

{{3, 2}, {7, 4}, {12, 8}, {4, 10}}

answered Dec 19 '18 at 10:50

kglr

283k10 gold badges325 silver badges624 bronze badges

$\begingroup$ This function is still not included in my version. I will study it in the future. Thank you very much! $\endgroup$ – user57467 Dec 19 '18 at 16:10

add a comment |

OkkesDulgerci · Accepted Answer · 2020-01-27 15:37:04Z

2

This seems to work.

DeleteDuplicatesBy[list1, Last]

{{3, 2}, {7, 4}, {12, 8}, {4, 10}}

answered Jan 27 at 15:37

OkkesDulgerci

9,2291 gold badge14 silver badges30 bronze badges

add a comment |

3 Answers 3