Removing elements of a table

Question

I want removing elements of a table. For example, I have the next table:

data = {{0, 3}, {1, 3}, {2, 5}, {4, 2}, {5, 1}, {6, 8}, {7, 4}};

and I want removing all elements where the 1st element of sublist is between 6 and 2 for I obtain this new table:

data = {{0, 3}, {1, 3}, {6, 8}, {7, 4}};

The next code don´t work:

DeleteCases[data, Flatten[Select[data, 6 > #[[1]] > 2 &]]]

If only I want removing one element (for example {0,3}), the code work:

DeleteCases[data, Flatten[Select[data, 1 > #[[1]] >= 0 &]]]

How can I remove elements where the 1st element of sublist is between two values?.

There has got to be a dupe for this, but in the meantime: why not stick to Select? E.g. Select[data, Not[6 > #[[1]] > 2] &]. — Marius Ladegård Meyer, Aug 10, 2016 at 9:46
DeleteCases[data,x/;(x[[1]]>xmin&&x[[1]]<xmax)] where xmin and xmax are your thresholds for first element — Rom38, Aug 10, 2016 at 9:47
The second argument of DeleteCases must be a pattern. I recommend reading through some of these chapters: reference.wolfram.com/language/tutorial/PatternsOverview.html — Szabolcs, Aug 10, 2016 at 9:54

ubpdqn · Accepted Answer · 2016-08-10 10:42:40Z

4

You can pick the desired elements, e.g.

Pick[data, 2 <= #[[1]] < 6 & /@ data, False]
Pick[data, Not[2 <= #[[1]] < 6] & /@ data]

or deleting:

DeleteCases[data, {_?(2 <= # < 6 &), _}]

answered Aug 10, 2016 at 10:42

ubpdqn

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$\begingroup$ When I import table of a file with two columns (x,y) I have data = {{{0, 3}, {1, 3}, {2, 5}, {4, 2}, {5, 1}, {6, 8}, {7, 4}}}. In these case only work the @MarkusDichtl code. I don´t understand why. $\endgroup$
– Manu
Aug 10, 2016 at 11:50
$\begingroup$ @Manu I am sorry I do not understand your comment. In my timezone I am off to sleep. I will aim to look after I get up but I am certain other uses will resolve :) $\endgroup$
– ubpdqn
Aug 10, 2016 at 11:53
$\begingroup$ @Manu. In that case, replace data in ubpdqn's code with First@data, and it should work. $\endgroup$
– march
Aug 10, 2016 at 16:03

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István Zachar · Accepted Answer · 2016-08-10 11:37:08Z

4

New approach since v10.2:

data /. {x_ /; 6 > x >= 2, y_} -> Nothing

and Nothing just vanishes, resulting in

{{0, 3}, {1, 3}, {6, 8}, {7, 4}}

István Zachar

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answered Aug 10, 2016 at 11:16

Markus Dichtl

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2

$\begingroup$ In the version 10 work too. $\endgroup$
– Manu
Aug 10, 2016 at 11:22
$\begingroup$ Nothing was introduced in 10.2, it's a special symbol for the standard evaluation process. $\endgroup$
– BoLe
Aug 10, 2016 at 11:35

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2 Answers 2