Using Cases to get elements in a list

Question

I have the following list:

list1 = {"O                    21   B21      20   A20      19   D19      0", \
"90  O            -0.28487   1.23835", "91  O            -0.23761   \
1.29381", "92  O            -0.23093   1.19437", "93  O            \
-0.19657   1.53843", "94  O            -0.17235   1.46275", "O        \
            21   B21      20   A20      19   D19      0", "90  O      \
      -0.28601   1.23805", "91  O            -0.23863   1.29489", "92 \
 O            -0.23206   1.19334", "93  O            -0.19730   \
1.53926"}

I need only the elements that do not have the expression: B21, D19, etc.... Like that:

list1 = { \
"90  O            -0.28487   1.23835", "91  O            -0.23761   \
1.29381", "92  O            -0.23093   1.19437", "93  O            \
-0.19657   1.53843", "94  O            -0.17235   1.46275",  "90  O      \
      -0.28601   1.23805", "91  O            -0.23863   1.29489", "92 \
 O            -0.23206   1.19334", "93  O            -0.19730   \
1.53926"};

So I did:

Cases[%, Except[
  RegularExpression["[A-Z]" ~~ __Integer]], \[Infinity]].

But It did no work.

score 5 · Answer 1 · answered Oct 13 '14 at 17:08

Select[list1,  StringFreeQ[#, ___ ~~ LetterCharacter ~~ DigitCharacter .. ~~ ___] &]

(*
90  O            -0.28487   1.23835
91  O            -0.23761   1.29381
92  O            -0.23093   1.19437
93  O            -0.19657   1.53843
94  O            -0.17235   1.46275
90  O            -0.28601   1.23805
91  O            -0.23863   1.29489
92  O            -0.23206   1.19334
93  O            -0.19730   1.53926
*)

kglr · Accepted Answer · 2014-10-13T19:37:08.317

4

Pick[list1, StringFreeQ[#, RegularExpression["[A-Z]\\d"]] & /@ list1] 
Cases[list1, x_?(StringFreeQ[#, RegularExpression["[A-Z]\\d"]] &)]
Select[list1, StringFreeQ[#, RegularExpression["[A-Z]\\d"]] &]
DeleteCases[list1, x_?(!StringFreeQ[#, RegularExpression["[A-Z]\\d"]] &)]

and all of the above with RegularExpression["[A-Z]\\d"] replaced with

Alternatives @@ CharacterRange["A", "Z"] ~~ DigitCharacter

all give

Pick[list1, StringFreeQ[#, RegularExpression["[A-Z]\\d"]] & /@ list1] //Column

where

list1 //Column

edited Oct 13 '14 at 19:37

answered Oct 13 '14 at 17:13

kglr

394,356
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I did not know th pick comand, is it new? – locometro Oct 13 '14 at 18:31
@user7096, Pick has been around since version 5; it is often more convenient/powerful than its older siblings Select and Cases. – kglr Oct 13 '14 at 18:52

score 2 · Answer 3 · answered Oct 13 '14 at 17:28

2

One can also use StringReplace

DeleteCases[StringReplace[list1, x___ ~~ LetterCharacter ~~ NumberString ~~ y___ :> ""], ""]

answered Oct 13 '14 at 17:28

RunnyKine

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Basheer Algohi · Answer 4 · 2014-10-13T18:39:20.853

0

I am not sure about this but it looks like (for such specific list) that you can do it like this:

Select[list1, Xnor[StringMatchQ[#, "O" ~~ __]] &]

edited Oct 13 '14 at 18:39

answered Oct 13 '14 at 18:32

Basheer Algohi

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Using Cases to get elements in a list

4 Answers4