Mapping string into integersDistinguishing slots in a function mapping Position across a listMapping over an association producing an unexpected resultConverting a list of associations into a single associationReplace single element of a nested list with multiple elementsFlattening a list desiring a specific formatString Join but not fullyManipulation of two lists of strings and integersDeleting unwanted elements in lists of wordsCoverting list of string into integers and reshaping the original list
Find number 8 with the least number of tries
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Mapping string into integers
Distinguishing slots in a function mapping Position across a listMapping over an association producing an unexpected resultConverting a list of associations into a single associationReplace single element of a nested list with multiple elementsFlattening a list desiring a specific formatString Join but not fullyManipulation of two lists of strings and integersDeleting unwanted elements in lists of wordsCoverting list of string into integers and reshaping the original list
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;
$begingroup$
Suppose I have the following list,
l = "b", "c", "d", "e", "b", "a", "b", "d", "e"
and further suppose I have the following association,
l1=<|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>
I wonder how can I replace the keys into my list such that I get,
2, 3, 4, 5, 2, 1, 2, 4, 5
list-manipulation map
edited 6 hours ago
Henrik Schumacher
69.1k5 gold badges101 silver badges193 bronze badges
asked 8 hours ago
WilliamWilliam
1,1736 silver badges9 bronze badges
$endgroup$
add a comment
|
$begingroup$
Suppose I have the following list,
l = "b", "c", "d", "e", "b", "a", "b", "d", "e"
and further suppose I have the following association,
l1=<|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>
I wonder how can I replace the keys into my list such that I get,
2, 3, 4, 5, 2, 1, 2, 4, 5
list-manipulation map
edited 6 hours ago
Henrik Schumacher
69.1k5 gold badges101 silver badges193 bronze badges
asked 8 hours ago
WilliamWilliam
1,1736 silver badges9 bronze badges
$endgroup$
add a comment
|
$begingroup$
Suppose I have the following list,
l = "b", "c", "d", "e", "b", "a", "b", "d", "e"
and further suppose I have the following association,
l1=<|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>
I wonder how can I replace the keys into my list such that I get,
2, 3, 4, 5, 2, 1, 2, 4, 5
list-manipulation map
edited 6 hours ago
Henrik Schumacher
69.1k5 gold badges101 silver badges193 bronze badges
asked 8 hours ago
WilliamWilliam
1,1736 silver badges9 bronze badges
$endgroup$
Suppose I have the following list,
l = "b", "c", "d", "e", "b", "a", "b", "d", "e"
and further suppose I have the following association,
l1=<|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>
I wonder how can I replace the keys into my list such that I get,
2, 3, 4, 5, 2, 1, 2, 4, 5
list-manipulation map
list-manipulation map
edited 6 hours ago
Henrik Schumacher
69.1k5 gold badges101 silver badges193 bronze badges
asked 8 hours ago
WilliamWilliam
1,1736 silver badges9 bronze badges
edited 6 hours ago
Henrik Schumacher
69.1k5 gold badges101 silver badges193 bronze badges
asked 8 hours ago
WilliamWilliam
1,1736 silver badges9 bronze badges
edited 6 hours ago
Henrik Schumacher
69.1k5 gold badges101 silver badges193 bronze badges
edited 6 hours ago
Henrik Schumacher
69.1k5 gold badges101 silver badges193 bronze badges
edited 6 hours ago
Henrik Schumacher
69.1k5 gold badges101 silver badges193 bronze badges
69.1k5 gold badges101 silver badges193 bronze badges
asked 8 hours ago
WilliamWilliam
1,1736 silver badges9 bronze badges
asked 8 hours ago
WilliamWilliam
1,1736 silver badges9 bronze badges
asked 8 hours ago
WilliamWilliam
1,1736 silver badges9 bronze badges
1,1736 silver badges9 bronze badges
add a comment
|
add a comment
|
4 Answers
4
active
oldest
votes
$begingroup$
The following gives you a "reversed" version of your association, with keys and values flipped:
l1 = <|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>;
lookup = First /@ PositionIndex@l1
(* <|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4, "e" -> 5|> *)
You can then use ReplaceAll (/.) to do the replacement:
l = "b", "c", "d", "e", "b", "a", "b", "d", "e";
l /. lookup
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
Alternatives
Other possible solutions for creating lookup:
lookup = <|Reverse /@ Normal@l1|>
lookup = Reverse /@ Normal@l1 (* doesn't need to be an association *)
answered 8 hours ago
Lukas LangLukas Lang
10.6k1 gold badge14 silver badges41 bronze badges
$endgroup$
add a comment
|
$begingroup$
For the specific numbering in OP, you can also use LetterNumber:
LetterNumber[l]
2, 3, 4, 5, 2, 1, 2, 4, 5
answered 6 hours ago
kglrkglr
220k10 gold badges250 silver badges506 bronze badges
$endgroup$
add a comment
|
$begingroup$
l /. Reverse /@ Normal[l1]
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
or
l /. AssociationMap[Reverse, l1]
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
answered 8 hours ago
Suba ThomasSuba Thomas
4,26411 silver badges20 bronze badges
$endgroup$
add a comment
|
$begingroup$
For this particular mapping, you could also use ToCharacterCode:
ToCharacterCode[StringJoin /@ l] - First@ToCharacterCode@"a" + 1
2, 3, 4, 5, 2, 1, 2, 4, 5
answered 6 hours ago
Carl WollCarl Woll
92k3 gold badges121 silver badges233 bronze badges
$endgroup$
add a comment
|
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The following gives you a "reversed" version of your association, with keys and values flipped:
l1 = <|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>;
lookup = First /@ PositionIndex@l1
(* <|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4, "e" -> 5|> *)
You can then use ReplaceAll (/.) to do the replacement:
l = "b", "c", "d", "e", "b", "a", "b", "d", "e";
l /. lookup
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
Alternatives
Other possible solutions for creating lookup:
lookup = <|Reverse /@ Normal@l1|>
lookup = Reverse /@ Normal@l1 (* doesn't need to be an association *)
answered 8 hours ago
Lukas LangLukas Lang
10.6k1 gold badge14 silver badges41 bronze badges
$endgroup$
add a comment
|
$begingroup$
The following gives you a "reversed" version of your association, with keys and values flipped:
l1 = <|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>;
lookup = First /@ PositionIndex@l1
(* <|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4, "e" -> 5|> *)
You can then use ReplaceAll (/.) to do the replacement:
l = "b", "c", "d", "e", "b", "a", "b", "d", "e";
l /. lookup
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
Alternatives
Other possible solutions for creating lookup:
lookup = <|Reverse /@ Normal@l1|>
lookup = Reverse /@ Normal@l1 (* doesn't need to be an association *)
answered 8 hours ago
Lukas LangLukas Lang
10.6k1 gold badge14 silver badges41 bronze badges
$endgroup$
add a comment
|
$begingroup$
The following gives you a "reversed" version of your association, with keys and values flipped:
l1 = <|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>;
lookup = First /@ PositionIndex@l1
(* <|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4, "e" -> 5|> *)
You can then use ReplaceAll (/.) to do the replacement:
l = "b", "c", "d", "e", "b", "a", "b", "d", "e";
l /. lookup
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
Alternatives
Other possible solutions for creating lookup:
lookup = <|Reverse /@ Normal@l1|>
lookup = Reverse /@ Normal@l1 (* doesn't need to be an association *)
answered 8 hours ago
Lukas LangLukas Lang
10.6k1 gold badge14 silver badges41 bronze badges
$endgroup$
The following gives you a "reversed" version of your association, with keys and values flipped:
l1 = <|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>;
lookup = First /@ PositionIndex@l1
(* <|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4, "e" -> 5|> *)
You can then use ReplaceAll (/.) to do the replacement:
l = "b", "c", "d", "e", "b", "a", "b", "d", "e";
l /. lookup
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
Alternatives
Other possible solutions for creating lookup:
lookup = <|Reverse /@ Normal@l1|>
lookup = Reverse /@ Normal@l1 (* doesn't need to be an association *)
answered 8 hours ago
Lukas LangLukas Lang
10.6k1 gold badge14 silver badges41 bronze badges
edited 8 hours ago
answered 8 hours ago
Lukas LangLukas Lang
10.6k1 gold badge14 silver badges41 bronze badges
answered 8 hours ago
Lukas LangLukas Lang
10.6k1 gold badge14 silver badges41 bronze badges
answered 8 hours ago
Lukas LangLukas Lang
10.6k1 gold badge14 silver badges41 bronze badges
10.6k1 gold badge14 silver badges41 bronze badges
add a comment
|
add a comment
|
$begingroup$
For the specific numbering in OP, you can also use LetterNumber:
LetterNumber[l]
2, 3, 4, 5, 2, 1, 2, 4, 5
answered 6 hours ago
kglrkglr
220k10 gold badges250 silver badges506 bronze badges
$endgroup$
add a comment
|
$begingroup$
For the specific numbering in OP, you can also use LetterNumber:
LetterNumber[l]
2, 3, 4, 5, 2, 1, 2, 4, 5
answered 6 hours ago
kglrkglr
220k10 gold badges250 silver badges506 bronze badges
$endgroup$
add a comment
|
$begingroup$
For the specific numbering in OP, you can also use LetterNumber:
LetterNumber[l]
2, 3, 4, 5, 2, 1, 2, 4, 5
answered 6 hours ago
kglrkglr
220k10 gold badges250 silver badges506 bronze badges
$endgroup$
For the specific numbering in OP, you can also use LetterNumber:
LetterNumber[l]
2, 3, 4, 5, 2, 1, 2, 4, 5
answered 6 hours ago
kglrkglr
220k10 gold badges250 silver badges506 bronze badges
edited 4 hours ago
answered 6 hours ago
kglrkglr
220k10 gold badges250 silver badges506 bronze badges
answered 6 hours ago
kglrkglr
220k10 gold badges250 silver badges506 bronze badges
answered 6 hours ago
kglrkglr
220k10 gold badges250 silver badges506 bronze badges
220k10 gold badges250 silver badges506 bronze badges
add a comment
|
add a comment
|
$begingroup$
l /. Reverse /@ Normal[l1]
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
or
l /. AssociationMap[Reverse, l1]
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
answered 8 hours ago
Suba ThomasSuba Thomas
4,26411 silver badges20 bronze badges
$endgroup$
add a comment
|
$begingroup$
l /. Reverse /@ Normal[l1]
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
or
l /. AssociationMap[Reverse, l1]
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
answered 8 hours ago
Suba ThomasSuba Thomas
4,26411 silver badges20 bronze badges
$endgroup$
add a comment
|
$begingroup$
l /. Reverse /@ Normal[l1]
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
or
l /. AssociationMap[Reverse, l1]
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
answered 8 hours ago
Suba ThomasSuba Thomas
4,26411 silver badges20 bronze badges
$endgroup$
l /. Reverse /@ Normal[l1]
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
or
l /. AssociationMap[Reverse, l1]
(* 2, 3, 4, 5, 2, 1, 2, 4, 5 *)
answered 8 hours ago
Suba ThomasSuba Thomas
4,26411 silver badges20 bronze badges
edited 8 hours ago
answered 8 hours ago
Suba ThomasSuba Thomas
4,26411 silver badges20 bronze badges
answered 8 hours ago
Suba ThomasSuba Thomas
4,26411 silver badges20 bronze badges
answered 8 hours ago
Suba ThomasSuba Thomas
4,26411 silver badges20 bronze badges
4,26411 silver badges20 bronze badges
add a comment
|
add a comment
|
$begingroup$
For this particular mapping, you could also use ToCharacterCode:
ToCharacterCode[StringJoin /@ l] - First@ToCharacterCode@"a" + 1
2, 3, 4, 5, 2, 1, 2, 4, 5
answered 6 hours ago
Carl WollCarl Woll
92k3 gold badges121 silver badges233 bronze badges
$endgroup$
add a comment
|
$begingroup$
For this particular mapping, you could also use ToCharacterCode:
ToCharacterCode[StringJoin /@ l] - First@ToCharacterCode@"a" + 1
2, 3, 4, 5, 2, 1, 2, 4, 5
answered 6 hours ago
Carl WollCarl Woll
92k3 gold badges121 silver badges233 bronze badges
$endgroup$
add a comment
|
$begingroup$
For this particular mapping, you could also use ToCharacterCode:
ToCharacterCode[StringJoin /@ l] - First@ToCharacterCode@"a" + 1
2, 3, 4, 5, 2, 1, 2, 4, 5
answered 6 hours ago
Carl WollCarl Woll
92k3 gold badges121 silver badges233 bronze badges
$endgroup$
For this particular mapping, you could also use ToCharacterCode:
ToCharacterCode[StringJoin /@ l] - First@ToCharacterCode@"a" + 1
2, 3, 4, 5, 2, 1, 2, 4, 5
answered 6 hours ago
Carl WollCarl Woll
92k3 gold badges121 silver badges233 bronze badges
answered 6 hours ago
Carl WollCarl Woll
92k3 gold badges121 silver badges233 bronze badges
answered 6 hours ago
Carl WollCarl Woll
92k3 gold badges121 silver badges233 bronze badges
answered 6 hours ago
Carl WollCarl Woll
92k3 gold badges121 silver badges233 bronze badges
92k3 gold badges121 silver badges233 bronze badges
add a comment
|
add a comment
|
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