Java String replaceAll regex删除除数字，点和空格之外的所有内容

Question

I've browsed a lot of regex topics but none of these lead me to success in my particular situation.

(Using java) I have some charsequence which I then convert to array and extract numbers to array of doubles.

asdsad 59 asdf .2 asdf 56 89 .a 2.4 wef 95 asdf.

then I want to use regex to remove the extra part and compose the following string

59 2 56 89 2.4 95

so then I could just use .split(" ") and put them all to an array of doubles.

Until this moment I used the following expression

[^0-9.\s]

but it leaves the extra dots and therefore not reliable. Now I'm trying something like

[^0-9.\s]|([^0-9]\.[^0-9])

but it's not working at all, I'm not really good with regex, so could you explain me why the last expression is not working and how to fix it.

Answer 1

Have you tried string.replaceAll("[^\\\\d\\\\. ]","") ?

You can see the results here: https://regex101.com/r/X6gLaY/2

String string = "asdsad 59 asdf 2 asdf 56 89 .a 2.4 wef 95 asdf.";
String regex = "[^\\d\\. ]| \\.|\\.$";
System.out.println(string.replaceAll(regex,""));

Java example: http://ideone.com/w4BWOZ

Outputs: 59 2 56 89 2.4 95

Answer 2

Add an alternative to remove dots that are not preceded with a digit that is followed with a dot and a digit:

[^\d\s.]+|(?<!\d\.\d)\.

See this regex demo .

Details :

• [^\\d\\s.]+ - 1+ chars other than digits, whitespaces and dots
• | - or
• (?<!\\d\\.\\d) - a location not preceded with a digit, dot, digit
• \\. - a dot.

Sample code:

String re = "[^\\d\\s.]+|(?<!\\d\\.\\d)\\.";
System.out.println("asdsad 59 asdf 2 asdf 56 89 .a 2.4 wef 95 asdf.".replaceAll(re, ""));
System.out.println("asdsad 59 asdf .2 asdf 56 89 .a 2.4 wef 95 asdf.".replaceAll(re, ""));

Java demo

Answer 3

I've played with regex for half a day until I come up with this.

Apparently it really matters what the order of the expression is. I assume it's because it iterates over each condition and always uses the data that's left after execution of previous condition, so I changed the regular expression to:

1. exclude all dots followed by non-digit

2. exclude all non-digits followed by dot

3. exclude all left-over non-digits

[^0-9]\.|\.[^0-9]|[^0-9.\s]

Now it works like a charm. Hope it helps someone. :)

Answer 4

You can split directly on this regex:

([^\d.]|\B\.|\.\B)+

ie

String[] parts = str.split("([^\\d.]|\\B\\.|\\.\\B)+");

Although this (may) leave a leading blank in the array.

To go directly to a double[] in one line (handling the leading blank):

double[] numbers = Arrays.stream(str.split("([^\\d.]|\\B\\.|\\.\\B)+"))
    .filter(s -> !s.isEmpty())
    .mapToDouble(Double::parseDouble)
    .toArray();

See live demo .