Selecting elements from lists based on string pattern
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I have a list:
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6
I would like to select list items that have first elements that end in the string "bc" to give:
res = 1,2,5,6
This is a variant on this question which has to do with strings only. Thank you as always for suggestions.
list-manipulation string-manipulation
asked Feb 2 at 4:46
Suite401Suite401
998312
$endgroup$
add a comment |
$begingroup$
I have a list:
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6
I would like to select list items that have first elements that end in the string "bc" to give:
res = 1,2,5,6
This is a variant on this question which has to do with strings only. Thank you as always for suggestions.
list-manipulation string-manipulation
asked Feb 2 at 4:46
Suite401Suite401
998312
$endgroup$
add a comment |
$begingroup$
I have a list:
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6
I would like to select list items that have first elements that end in the string "bc" to give:
res = 1,2,5,6
This is a variant on this question which has to do with strings only. Thank you as always for suggestions.
list-manipulation string-manipulation
asked Feb 2 at 4:46
Suite401Suite401
998312
$endgroup$
I have a list:
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6
I would like to select list items that have first elements that end in the string "bc" to give:
res = 1,2,5,6
This is a variant on this question which has to do with strings only. Thank you as always for suggestions.
list-manipulation string-manipulation
list-manipulation string-manipulation
asked Feb 2 at 4:46
Suite401Suite401
998312
asked Feb 2 at 4:46
Suite401Suite401
998312
asked Feb 2 at 4:46
Suite401Suite401
998312
asked Feb 2 at 4:46
Suite401Suite401
998312
asked Feb 2 at 4:46
Suite401Suite401
998312
998312
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Pick[lis[[All, 2 ;;]], StringEndsQ[lis[[All, 1]], "bc"]]
1, 2, 5, 6
Also
f[_String?(StringEndsQ["bc"]), x__] := x
f[_] := Sequence
f /@ lis
1, 2, 5, 6
answered Feb 2 at 7:26
kglrkglr
185k10202421
$endgroup$
add a comment |
$begingroup$
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6;
Using Cases
Cases[lis, _?(StringTake[#, -2] == "bc" &), x_, y_ :> x, y,
Infinity]
(* 1, 2, 5, 6 *)
or
Rest /@ Cases[lis, _?(StringTake[#, -2] == "bc" &), __, Infinity]
(* 1, 2, 5, 6 *)
Or using Select
Rest /@ Select[lis, StringTake[#[[1]], -2] == "bc" &]
(* 1, 2, 5, 6 *)
Or using DeleteCases
Rest /@ DeleteCases[lis, _?(StringTake[#, -2] != "bc" &), __,
Infinity]
(* 1, 2, 5, 6 *)
answered Feb 2 at 5:54
Bob HanlonBob Hanlon
60.6k33597
$endgroup$
add a comment |
$begingroup$
This uses string matching on the first element of each sublist and then discards the first element from the matches.
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6;
Rest/@Select[lis,StringMatchQ[First[#],RegularExpression["\w*bc"]]&]
(* 1, 2, 5, 6 *)
You may need to adjust the pattern given to RegularExpression if your strings do not all begin with word characters.
answered Feb 2 at 5:33
BillBill
5,75069
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Pick[lis[[All, 2 ;;]], StringEndsQ[lis[[All, 1]], "bc"]]
1, 2, 5, 6
Also
f[_String?(StringEndsQ["bc"]), x__] := x
f[_] := Sequence
f /@ lis
1, 2, 5, 6
answered Feb 2 at 7:26
kglrkglr
185k10202421
$endgroup$
add a comment |
$begingroup$
Pick[lis[[All, 2 ;;]], StringEndsQ[lis[[All, 1]], "bc"]]
1, 2, 5, 6
Also
f[_String?(StringEndsQ["bc"]), x__] := x
f[_] := Sequence
f /@ lis
1, 2, 5, 6
answered Feb 2 at 7:26
kglrkglr
185k10202421
$endgroup$
add a comment |
$begingroup$
Pick[lis[[All, 2 ;;]], StringEndsQ[lis[[All, 1]], "bc"]]
1, 2, 5, 6
Also
f[_String?(StringEndsQ["bc"]), x__] := x
f[_] := Sequence
f /@ lis
1, 2, 5, 6
answered Feb 2 at 7:26
kglrkglr
185k10202421
$endgroup$
Pick[lis[[All, 2 ;;]], StringEndsQ[lis[[All, 1]], "bc"]]
1, 2, 5, 6
Also
f[_String?(StringEndsQ["bc"]), x__] := x
f[_] := Sequence
f /@ lis
1, 2, 5, 6
answered Feb 2 at 7:26
kglrkglr
185k10202421
answered Feb 2 at 7:26
kglrkglr
185k10202421
answered Feb 2 at 7:26
kglrkglr
185k10202421
answered Feb 2 at 7:26
kglrkglr
185k10202421
185k10202421
add a comment |
add a comment |
$begingroup$
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6;
Using Cases
Cases[lis, _?(StringTake[#, -2] == "bc" &), x_, y_ :> x, y,
Infinity]
(* 1, 2, 5, 6 *)
or
Rest /@ Cases[lis, _?(StringTake[#, -2] == "bc" &), __, Infinity]
(* 1, 2, 5, 6 *)
Or using Select
Rest /@ Select[lis, StringTake[#[[1]], -2] == "bc" &]
(* 1, 2, 5, 6 *)
Or using DeleteCases
Rest /@ DeleteCases[lis, _?(StringTake[#, -2] != "bc" &), __,
Infinity]
(* 1, 2, 5, 6 *)
answered Feb 2 at 5:54
Bob HanlonBob Hanlon
60.6k33597
$endgroup$
add a comment |
$begingroup$
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6;
Using Cases
Cases[lis, _?(StringTake[#, -2] == "bc" &), x_, y_ :> x, y,
Infinity]
(* 1, 2, 5, 6 *)
or
Rest /@ Cases[lis, _?(StringTake[#, -2] == "bc" &), __, Infinity]
(* 1, 2, 5, 6 *)
Or using Select
Rest /@ Select[lis, StringTake[#[[1]], -2] == "bc" &]
(* 1, 2, 5, 6 *)
Or using DeleteCases
Rest /@ DeleteCases[lis, _?(StringTake[#, -2] != "bc" &), __,
Infinity]
(* 1, 2, 5, 6 *)
answered Feb 2 at 5:54
Bob HanlonBob Hanlon
60.6k33597
$endgroup$
add a comment |
$begingroup$
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6;
Using Cases
Cases[lis, _?(StringTake[#, -2] == "bc" &), x_, y_ :> x, y,
Infinity]
(* 1, 2, 5, 6 *)
or
Rest /@ Cases[lis, _?(StringTake[#, -2] == "bc" &), __, Infinity]
(* 1, 2, 5, 6 *)
Or using Select
Rest /@ Select[lis, StringTake[#[[1]], -2] == "bc" &]
(* 1, 2, 5, 6 *)
Or using DeleteCases
Rest /@ DeleteCases[lis, _?(StringTake[#, -2] != "bc" &), __,
Infinity]
(* 1, 2, 5, 6 *)
answered Feb 2 at 5:54
Bob HanlonBob Hanlon
60.6k33597
$endgroup$
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6;
Using Cases
Cases[lis, _?(StringTake[#, -2] == "bc" &), x_, y_ :> x, y,
Infinity]
(* 1, 2, 5, 6 *)
or
Rest /@ Cases[lis, _?(StringTake[#, -2] == "bc" &), __, Infinity]
(* 1, 2, 5, 6 *)
Or using Select
Rest /@ Select[lis, StringTake[#[[1]], -2] == "bc" &]
(* 1, 2, 5, 6 *)
Or using DeleteCases
Rest /@ DeleteCases[lis, _?(StringTake[#, -2] != "bc" &), __,
Infinity]
(* 1, 2, 5, 6 *)
answered Feb 2 at 5:54
Bob HanlonBob Hanlon
60.6k33597
answered Feb 2 at 5:54
Bob HanlonBob Hanlon
60.6k33597
answered Feb 2 at 5:54
Bob HanlonBob Hanlon
60.6k33597
answered Feb 2 at 5:54
Bob HanlonBob Hanlon
60.6k33597
60.6k33597
add a comment |
add a comment |
$begingroup$
This uses string matching on the first element of each sublist and then discards the first element from the matches.
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6;
Rest/@Select[lis,StringMatchQ[First[#],RegularExpression["\w*bc"]]&]
(* 1, 2, 5, 6 *)
You may need to adjust the pattern given to RegularExpression if your strings do not all begin with word characters.
answered Feb 2 at 5:33
BillBill
5,75069
$endgroup$
add a comment |
$begingroup$
This uses string matching on the first element of each sublist and then discards the first element from the matches.
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6;
Rest/@Select[lis,StringMatchQ[First[#],RegularExpression["\w*bc"]]&]
(* 1, 2, 5, 6 *)
You may need to adjust the pattern given to RegularExpression if your strings do not all begin with word characters.
answered Feb 2 at 5:33
BillBill
5,75069
$endgroup$
add a comment |
$begingroup$
This uses string matching on the first element of each sublist and then discards the first element from the matches.
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6;
Rest/@Select[lis,StringMatchQ[First[#],RegularExpression["\w*bc"]]&]
(* 1, 2, 5, 6 *)
You may need to adjust the pattern given to RegularExpression if your strings do not all begin with word characters.
answered Feb 2 at 5:33
BillBill
5,75069
$endgroup$
This uses string matching on the first element of each sublist and then discards the first element from the matches.
lis = "abc", 1, 2, "cde", 3, 4, "fgbc", 5, 6;
Rest/@Select[lis,StringMatchQ[First[#],RegularExpression["\w*bc"]]&]
(* 1, 2, 5, 6 *)
You may need to adjust the pattern given to RegularExpression if your strings do not all begin with word characters.
answered Feb 2 at 5:33
BillBill
5,75069
answered Feb 2 at 5:33
BillBill
5,75069
answered Feb 2 at 5:33
BillBill
5,75069
answered Feb 2 at 5:33
BillBill
5,75069
5,75069
add a comment |
add a comment |
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