I have the following two lists
l1 = {{{2011, 3, 13}, 1}, {{2011, 3, 14}, 1}, {{2011, 3, 15}, 1}, {{2011, 3, 16}, 2}, {{2011, 3, 17}, 3}}; l2 = {{{2011, 3, 13}, 40}, {{2011, 3, 16}, 50}, {{2011, 3, 17}, 60}};
and I need to extract items from l2 whose date (the first element of each l2 element) matches dates in l1 (so as to produce two lists of exactly the same length)
I don't see why something like:
Select[l1, MemberQ[Transp开发者_StackOverflow中文版ose[l2][[1]], #[[1]]]]
should produce an empty list. Am I missing something trivial?
You forgot the ampersand. It should be
Select[l1, MemberQ[Transpose[l2][[1]], #[[1]]]&]
Sjoerd shows you how to make your method work, but it is not optimal. The problem is that Transpose[l2][[1]]
is evaluated again for every element in l1
. David give a method which does that step only once. You could also use:
Cases[l1, {Alternatives @@ l2[[All, 1]], _}]
A faster method for when your lists get bigger:
DeleteCases[GatherBy[Join[l1, l2], First], {_}][[All, 1]]
(* Out= {{{2011, 3, 13}, 40}, {{2011, 3, 16}, 50}, {{2011, 3, 17}, 60}} *)
If your list might contain duplicates, you can use the
l1 = GatherBy[l1, First][[All, 1]]
to remove the duplicates first.
Is this possibly what you have in mind?
dates=Transpose[l2][[1]];
Cases[l1, {x_, _} /; MemberQ[dates, x]]
Here is my non-sorting intersection code:
NonSortingIntersection[l1_, l2_, test_: SameQ] :=
Module[{res =
Last@Reap[
Scan[Extract[l1,
Position[l1, x_ /; test[x, #], {1}, 1, Heads -> False],
Sow] &, l2]]}, If[res === {}, res, First[res]]]
Here is the usage:
In[22]:= NonSortingIntersection[l1, l2,
Function[{x, y}, First[x] == First[y]]]
Out[22]= {{{2011, 3, 13}, 1}, {{2011, 3, 16}, 2}, {{2011, 3, 17}, 3}}
Notice that unlike other solutions the output is guaranteed to have length no longer that that of l2
. For instance:
In[24]:= Cases[Join[l1, l1], {x_, _} /; MemberQ[Evaluate[Transpose[l2][[1]]], x]]
Out[24]= {{{2011, 3, 13}, 1}, {{2011, 3, 16}, 2}, {{2011, 3, 17},
3}, {{2011, 3, 13}, 1}, {{2011, 3, 16}, 2}, {{2011, 3, 17}, 3}}
In[25]:= NonSortingIntersection[Join[l1, l1], l2,
Function[{x, y}, First[x] == First[y]]]
Out[25]= {{{2011, 3, 13}, 1}, {{2011, 3, 16}, 2}, {{2011, 3, 17}, 3}}
This may or may not be desirable, but this is up to radhat who knows his problem better.
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