I have a bunch of strings which may of may not have random symbols and numbers in them. Some examples are:
contains(reserved[j])){
close();
i++){
letters[20]=word
I want to find any character that is NOT a letter, and replace it with a white space, so the above examples look like:
contains reserved j
close
i
letters word
What is the best way to do this?
It depends what you mean by "not a letter", but assuming you mean that letters are az or AZ then try this:
s = s.replaceAll("[^a-zA-Z]", " ");
If you want to collapse multiple symbols into a single space then add a plus at the end of the regular expression.
s = s.replaceAll("[^a-zA-Z]+", " ");
yourInputString = yourInputString.replaceAll("[^\\p{Alpha}]", " ");
^
denotes "all characters except"
\\p{Alpha}
denotes all alphabetic characters
See Pattern for details.
I want to find any character that is NOT a letter
That will be [^\\p{Alpha}]+
. The []
indicate a group. The \\p{Alpha}
matches any alphabetic character (both uppercase and lowercase, it does basically the same as \\p{Upper}\\p{Lower}
and a-zA-Z
. The ^
inside group inverses the matches. The +
indicates one-or-many matches in sequence.
and replace it with a white space
That will be " "
.
Summarized:
string = string.replaceAll("[^\\p{Alpha}]+", " ");
Also see the java.util.regex.Pattern
javadoc for a concise overview of available patterns. You can learn more about regexs at the great site http://regular-expression.info .
Use the regexp /[^a-zA-Z]/ which means, everything that is not in the az/AZ characters
In ruby I would do:
"contains(reserved[j]))".gsub(/[^a-zA-Z]/, " ")
=> "contains reserved j "
In Java should be something like:
import java.util.regex.*;
...
String inputStr = "contains(reserved[j])){";
String patternStr = "[^a-zA-Z]";
String replacementStr = " ";
// Compile regular expression
Pattern pattern = Pattern.compile(patternStr);
// Replace all occurrences of pattern in input
Matcher matcher = pattern.matcher(inputStr);
String output = matcher.replaceAll(replacementStr);
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