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I have a very long nested list. Here is a short example:
list = {{1.2, {1, 1, 1, 1}, {2, 2, 2}}, {0.9, {3, 3, 3, 3}, {4, 4, 4}}, {1.4, {4, 4, 4, 4}, {5, 5, 5}}}
Now I want to cancel all entries if the first entry of each part of the list (list[[i]]) is > 1. So that the new list would only consist of the second part of list (list[[2]])
newlist = {0.9, {3, 3, 3, 3}, {4, 4, 4}}
I have already tried my luck with Position and Delete, but with no success.
Thanks a lot for your help!
Rather than think of it as deleting certain unwanted parts of the list, you can equivalently think of it as selecting the parts you wish to keep. Accordingly, Select can be straightforwardly applied
Select[list, #[[1]] < 1 &]
{{0.9, {3, 3, 3, 3}, {4, 4, 4}}}
A. G. points out that this can be made more transparent by replacing the perhaps somewhat cryptic Part command [[1]] with
Select[list, First[#] < 1 &]
You can Flatten[%, 1] if you wish to get the levels exactly as described in the question.
#[[1]] can also be written First[#] or First@# or even (# // First) !!!
$\endgroup$
A solution using Pick:
Pick[list, UnitStep[list[[All, 1]] - 1], 0]
(I subtract one from the first element and then grab all positions with a negative such element)
Cases[list, {x_?NumericQ, a__} /; x <= 1 -> {x, a}]$\endgroup$Pick, e.g. my own answers to (900) and (30155) $\endgroup$Cases[list, {x_, a__} /; x <= 1]. Also don't forget to use:>with named patterns! $\endgroup$list, it should therefore be{{0.9, {3, 3, 3, 3}, {4, 4, 4}}}(note the extra{}) -- Unless you know that only one item will be selected. $\endgroup$