Replacing each letter wich another letter
I am wondering how to replace each letter of a string with another letter. For example A -> D
.
I have tried using .replace()
(repeating it in each line for every letter) but when I replace the letter A with the letter D and then try to replace the letter D with the letter H it just replaces the letter A twice so instead of AD -> DH
I end up with AD -> HH
.
Im sure there is a better and simpler way of doing this.
Thank you in advance.
javascript node.js
add a comment |
I am wondering how to replace each letter of a string with another letter. For example A -> D
.
I have tried using .replace()
(repeating it in each line for every letter) but when I replace the letter A with the letter D and then try to replace the letter D with the letter H it just replaces the letter A twice so instead of AD -> DH
I end up with AD -> HH
.
Im sure there is a better and simpler way of doing this.
Thank you in advance.
javascript node.js
add a comment |
I am wondering how to replace each letter of a string with another letter. For example A -> D
.
I have tried using .replace()
(repeating it in each line for every letter) but when I replace the letter A with the letter D and then try to replace the letter D with the letter H it just replaces the letter A twice so instead of AD -> DH
I end up with AD -> HH
.
Im sure there is a better and simpler way of doing this.
Thank you in advance.
javascript node.js
I am wondering how to replace each letter of a string with another letter. For example A -> D
.
I have tried using .replace()
(repeating it in each line for every letter) but when I replace the letter A with the letter D and then try to replace the letter D with the letter H it just replaces the letter A twice so instead of AD -> DH
I end up with AD -> HH
.
Im sure there is a better and simpler way of doing this.
Thank you in advance.
javascript node.js
javascript node.js
asked Nov 20 '18 at 18:51
Artur DudekArtur Dudek
1
1
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add a comment |
1 Answer
1
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oldest
votes
Build up a dictionary:
const dict = { A: "D", D: "A", /*...*/ };
Then split the string into an array, map that to a new array by applying the replacement and join the array back to a string:
const output = input.split("").map(char => dict[char]).join("");
2
You may also use thereplace
method directly like so:input.replace(/./g, c => dict[c] || c)
.
– Elias Toivanen
Nov 20 '18 at 19:09
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Build up a dictionary:
const dict = { A: "D", D: "A", /*...*/ };
Then split the string into an array, map that to a new array by applying the replacement and join the array back to a string:
const output = input.split("").map(char => dict[char]).join("");
2
You may also use thereplace
method directly like so:input.replace(/./g, c => dict[c] || c)
.
– Elias Toivanen
Nov 20 '18 at 19:09
add a comment |
Build up a dictionary:
const dict = { A: "D", D: "A", /*...*/ };
Then split the string into an array, map that to a new array by applying the replacement and join the array back to a string:
const output = input.split("").map(char => dict[char]).join("");
2
You may also use thereplace
method directly like so:input.replace(/./g, c => dict[c] || c)
.
– Elias Toivanen
Nov 20 '18 at 19:09
add a comment |
Build up a dictionary:
const dict = { A: "D", D: "A", /*...*/ };
Then split the string into an array, map that to a new array by applying the replacement and join the array back to a string:
const output = input.split("").map(char => dict[char]).join("");
Build up a dictionary:
const dict = { A: "D", D: "A", /*...*/ };
Then split the string into an array, map that to a new array by applying the replacement and join the array back to a string:
const output = input.split("").map(char => dict[char]).join("");
answered Nov 20 '18 at 18:54
Jonas WilmsJonas Wilms
56k42851
56k42851
2
You may also use thereplace
method directly like so:input.replace(/./g, c => dict[c] || c)
.
– Elias Toivanen
Nov 20 '18 at 19:09
add a comment |
2
You may also use thereplace
method directly like so:input.replace(/./g, c => dict[c] || c)
.
– Elias Toivanen
Nov 20 '18 at 19:09
2
2
You may also use the
replace
method directly like so: input.replace(/./g, c => dict[c] || c)
.– Elias Toivanen
Nov 20 '18 at 19:09
You may also use the
replace
method directly like so: input.replace(/./g, c => dict[c] || c)
.– Elias Toivanen
Nov 20 '18 at 19:09
add a comment |
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