replace - Find and replace one or more substrings - MATLAB (original) (raw)

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Find and replace one or more substrings

Syntax

Description

newStr = replace([str](#bu5rv7y%5Fsep%5Fbu4l86d-str),[old](#bu5rv7y-old),[new](#bu5rv7y-new)) replaces all occurrences of the substring old withnew. If old contains multiple substrings, then new either must be the same size as old, or must be a single substring.

example

Examples

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Replace placeholder text in a list of file names.

Create a string array.

str = ["\MyData\data.tar.gz"; "\MyScripts\cleandata.m"; "\MyScripts\preprocess.m"; "\MyScripts\publishResults.m"]

str = 4×1 string "\MyData\data.tar.gz" "\MyScripts\cleandata.m" "\MyScripts\preprocess.m" "\MyScripts\publishResults.m"

Replace <ROOT_DIR> with a string that is the name of a file path.

old = ""; new = "C:\MyProject"; newStr = replace(str,old,new)

newStr = 4×1 string "C:\MyProject\MyData\data.tar.gz" "C:\MyProject\MyScripts\cleandata.m" "C:\MyProject\MyScripts\preprocess.m" "C:\MyProject\MyScripts\publishResults.m"

Since R2020b

Create a string that includes a phone number.

str = "Hide the numbers in: (508) 555-1234"

str = "Hide the numbers in: (508) 555-1234"

Create a pattern that matches a digit using the digitsPattern function.

pat = pattern Matching:

digitsPattern(1)

Replace all digits with a "#" character.

newStr = replace(str,pat,"#")

newStr = "Hide the numbers in: (###) ###-####"

Create another pattern that matches only phone numbers.

pat = "(" + digitsPattern(3) + ") " + digitsPattern(3) + "-" + digitsPattern(4)

pat = pattern Matching:

"(" + digitsPattern(3) + ") " + digitsPattern(3) + "-" + digitsPattern(4)

Replace a phone number in a string that also has another number.

str = "12 calls made to: (508) 555-1234"; newStr = replace(str,pat,"(###) ###-####")

newStr = "12 calls made to: (###) ###-####"

For a list of functions that create pattern objects, see pattern.

Replace carriage returns with newline characters.

Create a string array.

str = ["Submission Date: 11/29/15\r"; "Acceptance Date: 1/20/16\r"; "Contact: john.smith@example.com\r\n"]

str = 3×1 string "Submission Date: 11/29/15\r" "Acceptance Date: 1/20/16\r" "Contact: john.smith@example.com\r\n"

Replace the carriage returns.

old = {'\r\n','\r'}; new = '\n'; newStr = replace(str,old,new)

newStr = 3×1 string "Submission Date: 11/29/15\n" "Acceptance Date: 1/20/16\n" "Contact: john.smith@example.com\n"

Input Arguments

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Input text, specified as a string array, character vector, or cell array of character vectors.

Substring to replace, specified as one of the following:

New substring, specified as a string array, character vector, or cell array of character vectors.

Tips

Extended Capabilities

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Thereplace function fully supports tall arrays. For more information, see Tall Arrays.

Usage notes and limitations:

Version History

Introduced in R2016b