Changing letters to other letters with regexprep
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I'm trying to change letters in a string to different ones with one line of code. I simplified my code down to the following:
regexprep('ab',{'a','b'},{'b','c'});
Ideally, this would output 'bc'. Currently, regexprep runs through the string twice and ends up outputting 'cc'.
Is there any way to prevent this from happening? Is it possible with regexprep? Also, I would prefer to keep the letters as strings, as I'm planning on using the 'preservecase' condition.
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Stephen23
2018 年 9 月 28 日
編集済み: Stephen23
2018 年 9 月 28 日
>> str = 'a':'z'
str = abcdefghijklmnopqrstuvwxyz
>> old = 'bcd';
>> new = 'abc';
>> [X,Y] = ismember(str,old);
>> str(X) = new(Y(X))
str = aabcefghijklmnopqrstuvwxyz
Note that this works efficiently for any characters, not just a limited subset. In order to use regexprep you would probably need to use a dynamic replacement expression.
1 件のコメント
Walter Roberson
2018 年 9 月 28 日
Well, any characters representable in the first 65536 code points ;-) If you have codepoints beyond that, such as Linear B or CJK Unified Ideograms like ? then you would need more work.
その他の回答 (2 件)
OCDER
2018 年 9 月 28 日
Note that regexprep will replace things Sequentially.
regexprep('ab',{'a','b'},{'b','c'});
ab -> bb -> cc
To fix, reverse order like this:
regexprep('ab',{'b', 'a'},{'c', 'b'});
ab -> ac -> bc
2 件のコメント
Walter Roberson
2018 年 9 月 28 日
old = 'bcd';
new = 'abc';
lookup = char(0:255);
lookup(old+1) = new;
str = 'a':'z'
str = lookup(str+1)
You do not need to add 1 if you are willing to omit the possibility of char(0) being in the string.
2 件のコメント
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