How do I delete items at the same position from every sub-list within a list?

Question

We have data in a list called megadata where megadata[[1]] for instance are the temperatures of each data point, megadata[[2]] are the pressures, etc. We are trying to selectively remove data points to see how it changes our fits but I have been unable to get Delete or Drop or anything else to work. I need something to let me remove the same position from every nested list within the list. So for instance if I wanted to delete positions 2 and 3 I need something to change megadata = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}} into `megadata = {{1, 4}, {5, 8}, {9, 12}.

Edit: I'm really new to Mathematica (learning ahead of the semester as part of undergraduate research) so I'm having trouble understanding some of the comments made. I used the Drop[#, {2, 3}] & /@ megadata idea and it worked quite nicely for removing adjacent points but my nested lists have a length of 12 and when I would try using it as Drop[#, {4, 7}] & /@ megadata, it seemed to remove all the points between 4 and 7 (so it would remove 4,5,6, and 7). Is there anyway to modify this format slightly to make it not remove that whole range of points and only remove 4 and 7?

Edit 2: I actually think I found the answer. Didn't see the link to the duplicate answer above. Sorry real new to all of this. Thanks for all the help though!

This is closely related to Manipulating a sublist based on position although there are even closer posts, you should find them. — Artes, Jul 15 '14 at 23:00
If you want to remove points i order to test a fit then I'd suggest using Cases with some sort of condition applied to a pattern. But we'd need an example of your real data. — Mike Honeychurch, Jul 15 '14 at 23:03
@artes Your +1 and accepted answer in the cited link is somehow but not "closely" related :) — eldo, Jul 15 '14 at 23:21
@eldo Thanks for an upvote, however a remark is needed, there is some kind of relativity in the concept of distance, so I don't insist that it is universal. When it comes to this problem I have to ask why you don't use a simpler and nicer approach: list /. {a_, b_, c_, d_} :> {a, d}? — Artes, Jul 15 '14 at 23:37

score 2 · Answer 1 · answered Jul 15 '14 at 23:33

Both Delete and Drop should certainly work if used with Map.

data = Partition[Range[4 4], 4]

{{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}}

Drop[#, {2, 3}] & /@ data
Delete[#, {{2}, {3}}] & /@ data

Both return

{{1, 4}, {5, 8}, {9, 12}, {13, 16}}

But as eldo points out, Map can be avoided.

data[[All, {1, 4}]]

also works. Notice in this last case, you specify the columns you want to keep, rather than the ones you want to drop.

eldo · Answer 2 · 2014-07-15T23:12:27.307

1

data={{1,2,3,4},{5,6,7,8},{9,10,11,12}}

Part[#, {1, -1}] & /@ data

{{1, 4}, {5, 8}, {9, 12}}

Cases[data, {a_, __, b_} :> {a, b}]

and many more possibilities

edited Jul 15 '14 at 23:12

answered Jul 15 '14 at 23:03

eldo

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you don't need to use Map. see my comment above – Mike Honeychurch Jul 15 '14 at 23:04
@Mike Thanks for the tip :) – eldo Jul 15 '14 at 23:14
data[[All,{1,-1}]] but my other comment was more like DeleteCases[data, {a_/;some condition, __, b_/;maybe condition here too}] etc because he wants to remove points from a fit – Mike Honeychurch Jul 15 '14 at 23:18

How do I delete items at the same position from every sub-list within a list?

2 Answers2