So i have the following problem:
I have to tokenize a string using String.split() and the tokens must be in the form 07dd ddd ddd, where d is a digit. I thought of using the following regex : ^(07\\d{2}\\s\\d{3}\\d{3}) and pass it as an argument to String.split(). But for some reason, although i do have substrings under that form, it outputs the whole initial string and doesn't tokenize it.
I initially thought that it was using an empty string as a splitter, as an empty string indeed matches that regex, but even after I added & (.)+ to the regex in order to assure that the splitter hasn't got length 0, it still outputs the whole initial string.
I know that i could have used Pattern's and Matchers to solve it much faster, but i have to use String.split(). Any ideas why this happens?
A Few Pointers
Your pattern ^(07\d{2}\s\d{3}\d{3}) is missing a space between the two last groups of digits
The reason you get the whole string back is that this pattern was never found in the first place: there is no split
If you split on this pattern (once fixed), the resulting array will be strings that are in-between this pattern (these tokens are actually removed)
If you want to use this pattern (once fixed), you need a Match All not a Split. This will look like arrayOfMatches = yourString.match(/pattern/g);
If you want to split, you need to use a delimiter that is present between the token (this delimiter could in fact just be a zero-width position asserted by the 07 about to follow)
Further Reading
Match All and Split are Two Sides of the Same Coin
Related
I am trying to do the follow:
I have some strings that I need to separate, they have this form:
node:info:sequence(id:ASDF,LMD)
node:info:sequence:id:QWES
Those are the possible individual string formats...
Now I have to separate them when come concatenated by comma... like this
node:info:sequence(id:ASDF,LMD),node:info:sequence:id:QWES
So I tried
entries.split(",node");
Which... kinda works but of course I cut the "node" part from the previous string, is there anyway I can detect that , followed by node but split it by the comma , only?
You may use
s.split(",(?=node\\b)")
See the regex demo
The positive lookahead (?=node\b) will make sure only those commas are matched that are followed with a whole word node (as the \b is a word boundary).
I have a following String:
MYLMFILLAAGCSKMYLLFINNAARPFASSTKAASTVVTPHHSYTSKPHHSTTSHCKSSD
I want to split such a string every time a K or R is encountered, except when followed by a P.
Therefore, I want the following output:
MYLMFILLAAGCSK
MYLLFINNAARPFASSTK
AASTVVTPHHSYTSKPHHSTTSHCK
SSD
At first, I tried using simple .split() function in java but I couldn't get the desired result. Because I really don't know how to mention it in the .split() function not to split if there is a P right after K or R.
I've looked at other similar questions and they suggest to use Pattern matching but I don't know how to use it in this context.
You can use split:
String[] parts = str.split("(?<=[KR])(?!P)");
Because you want to keep the input you're splitting on, you must use a look behind, which asserts without consuming. There are two look arounds:
(?<=[KR]) means "the previous char is either K or R"
(?!P) means "the next char is not a P"
This regex matches between characters where you want to split.
Some test code:
String str = "MYLMFILLAAGCSKMYLLFINNAARPFASSTKAASTVVTPHHSYTSKPHHSTTSHCKSSD";
Arrays.stream(str.split("(?<=[KR])(?!P)")).forEach(System.out::println);
Output:
MYLMFILLAAGCSK
MYLLFINNAARPFASSTK
AASTVVTPHHSYTSKPHHSTTSHCK
SSD
Just try this regexp:
(K)([^P]|$)
and substitute each matching by
\1\n\2
as ilustrated in the following demo. No negative lookahead needed. But you cannot use it with split, as it should eliminate the not P character after the K also.
You can do a first transform like the one above, and then .split("\n");
so it should be:
"MYLMFILLAAGCSKMYLLFINNAARPFASSTKAASTVVTPHHSYTSKPHHSTTSHCKSSDK"
.subst("(K)([^P]|$)", "\1\n\2").split("\n");
I'm practicing reading input and then tokenizing it.
For example, if I have [882,337] I want to just get the numbers 882 and 337. I tried using the following code:
String test = "[882,337]";
String[] tokens = test.split("\\[|\\]|,");
System.out.println(tokens[0]);
System.out.println(tokens[1]);
System.out.println(tokens[2]);
It kind of works, the output is:
(blank line)
882
337
What I don't understand is why token[0] is empty? I would expect there to only be two tokens where token[0] = 882 and token[1] = 337.
I checked out some links but didn't find the answer.
Thanks for the help!
Split splits the given String. If you split "[882,337]" on "[" or "," or "]" then you actually have:
nothing
882
337
nothing
But, as you have called String.split(delimiter), this calls String.split(delimiter, limit) with a limit of zero.
From the documentation:
The limit parameter controls the number of times the pattern is applied and therefore affects the length of the resulting array. If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array's length will be no greater than n, and the array's last entry will contain all input beyond the last matched delimiter. If n is non-positive then the pattern will be applied as many times as possible and the array can have any length. If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded.
(emphasis mine)
So in this configuration the final, empty, strings are discarded. You are therefore left with exactly what you have.
Usually, to tokenize something like this, one would go for a combination of replaceAll and split:
final String[] tokens = input.replaceAll("^\\[|\\]$").split(",");
This will first strip off the start (^[) and end (]$) brackets and then split on ,. This way you don't have to have somewhat obtuse program logic where you start looping from an arbitrary index.
As an alternative, for more complex tokenizations, one can use Pattern - might be overkill here, but worth bearing in mind before you get into writing multiple replaceAll chains.
First we need to define, in Regex, the tokens we want (rather than those we're splitting on) - in this case it's simple, it's just digits so \d.
So, in order to extract all digit only (no thousands/decimal separators) values from an arbitrary String on would do the following:
final List<Integer> tokens = new ArrayList<>(); <-- to hold the tokens
final Pattern pattern = Pattern.compile("\\d++"); <-- the compiled regex
final Matcher matcher = pattern.matcher(input); <-- the matcher on input
while(matcher.find()) { <-- for each matched token
tokens.add(Integer.parseInt(matcher.group())); <-- parse and `int` and store
}
N.B: I have used a possessive regex pattern for efficiency
So, you see, the above code is somewhat more complex than the simple replaceAll().split(), but it is much more extensible. You can use arbitrary complex regex to token almost any input.
The symbols where the string is split are here:
String test = "[882,337]";
^ ^ ^
Because The first char matches your delimiter, everything left from it will be the first result. Well, left from the first letter is nothing, so the result is the empty string.
One could expect the same behaviour for the end, since the last symbol also matches the delimiter. But:
Trailing empty strings are therefore not included in the resulting array.
See Javadoc.
Splitting creates two (or more) things from one thing. For instance if you split a,b by , you will get a and b.
But in case of ",b" you will get "" and "b". You can think of it this way:
"" exists at start, end and even in-between all characters of string:
""+","+"b" -> ",b" so if we split on this "," we are getting left and right part: "" and "b"
Similar things happens in case of "a," and at first result array is ["a",""] but here split method removes trailing empty strings and returns only ["a"] (you can turn off this clearing mechanism by using split(",", -1)).
So in case of
String test = "[882,337]";
String[] tokens = test.split("\\[|\\]|,");
you are splitting:
""+"["+"882"+","+"337"+"]"+""
here: ^ ^ ^
which at first creates array ["", "882", "337", ""] but then trailing empty string is removed and finally you are receiving:
["", "882", "337"]
Only case where empty string is removed from start of result array is when
you are using Java 8 (or newer) and splitting on regex which is zero-length like split("") or lets say before each x with split("(?=x)") (more info at: Why in Java 8 split sometimes removes empty strings at start of result array?)
and when this empty string was result of split method. For instance "".split("") will not remove "", more info here: https://stackoverflow.com/a/25058091/1393766
That's because each delimiter has a "before" and "after" result, even if it is empty. Consider
882,337
You expect that to produce two results.
Similarly, you expect
882,337,
to produce three, with the last one being empty (assuming your limit is big enough, or assuming you're using almost any other language / implementation of split()). Extending that logically,
,882,337,
must produce four, with the first and last results being empty. This is exactly the case you have, except you have multiple delimiters.
I'm stucked for a while with a regex that does me the following:
split my sentences with this: "[\W+]"
but if it finds a word like this: "aaa-aa" (not "aaa - aa" or "aaa--aaa-aa"), the word isnt splitted, but the whole word.
Basically, i want to split a sentece per words, but also considering "aaa-aa" is a word. I'have sucessfully done that by creating two separate functions, one for spliting with \w, and other to find words like "aaa-aa". Finally, i then add both, and subctract each compound word.
For example, the sentence:
"Hello my-name is Richard"
First i collect {Hello, my, name, is, Richard}
then i collect {my-name}
then i add {my-name} to {Hello, my, name, is, Richard}
then i take out {my} and {name} in here {Hello, my, name, is, Richard}.
result: {Hello, my-name, is, Richard}
this approach does what i need, but for parsing large files, this becomes too heavy, because for each sentence there's too many copies needed. So my question is, there is anything i can do to include everything in one pattern? Like:
"split me the text using this pattern "[\W+], but if you find a word like this "aaa-aa", consider it a word and not two words.
If you want to use a split() rather than explicitly matching the words you are interested in, the following should do what you want: [\s-]{2,}|\s To break that down, you first split on two or more whitespaces and/or hyphens - so a single '-' won't match so 'one-two' will be left alone but something like 'one--two', 'one - two' or even 'one - --- - two' will be split into 'one' and 'two'. That still leaves the 'normal' case of a single whitespace - 'one two' - unmatched, so we add an or ('|') followed by a single whitespace (\s). Note that the order of the alternatives is important - RE subexpressions separated by '|' are evaluated left-to-right so we need to put the spaces-and-hyphens alternative first. If we did it the other way around, when presented with something like 'one -two' we'd match on the first whitespace and return 'one', '-two'.
If you want to interactively play around with Java REs I can thoroughly recommend http://myregexp.com/signedJar.html which allows you to edit the RE and see it matching against a sample string as you edit the RE.
Why not to use pattern \\s+? This does exactly what you want without any tricks: splits text by words separated by whitespace.
Your description isn't clear enough, but why not just split it up by spaces?
I am not sure whether this pattern would work, because I don't have developer tools for Java, you might try it though, it uses character class substraction, which is supported only in Java regex as far as I know:
[\W&&[^-]]+
it means match characters if they are [\W] and [^-], that is characters are [\W] and not [-].
Almost the same regular expression as in your previous question:
String sentence = "Hello my-name is Richard";
Pattern pattern = Pattern.compile("(?<!\\w)\\w+(-\\w+)?(?!\\w)");
Matcher matcher = pattern.matcher(sentence);
while (matcher.find()) {
System.out.println(matcher.group());
}
Just added the option (...)? to also match non-hypened words.
I need to be able to split an input String by commas, semi-colons or white-space (or a mix of the three). I would also like to treat multiple consecutive delimiters in the input as a single delimiter. Here's what I have so far:
String regex = "[,;\\s]+";
return input.split(regex);
This works, except for when the input string starts with one of the delimiter characters, in which case the first element of the result array is an empty String. I do not want my result to have empty Strings, so that something like, ",,,,ZERO; , ;;ONE ,TWO;," returns just a three element array containing the capitalized Strings.
Is there a better way to do this than stripping out any leading characters that match my reg-ex prior to invoking String.split?
Thanks in advance!
No, there isn't. You can only ignore trailing delimiters by providing 0 as a second parameter to String's split() method:
return input.split(regex, 0);
but for leading delimiters, you'll have to strip them first:
return input.replaceFirst("^"+regex, "").split(regex, 0);
If by "better" you mean higher performance then you might want to try creating a regular expression that matches what you want to match and using Matcher.find in a loop and pulling out the matches as you find them. This saves modifying the string first. But measure it for yourself to see which is faster for your data.
If by "better" you mean simpler, then no I don't think there is a simpler way than the way you suggested: removing the leading separators before applying the split.
Pretty much all splitting facilities built into the JDK are broken one way or another. You'd be better off using a third-party class such as Splitter, which is both flexible and correct in how it handles empty tokens and whitespaces:
Splitter.on(CharMatcher.anyOf(";,").or(CharMatcher.WHITESPACE))
.omitEmptyStrings()
.split(",,,ZERO;,ONE TWO");
will yield an Iterable<String> containing "ZERO", "ONE", "TWO"
You could also potentially use StringTokenizer to build the list, depending what you need to do with it:
StringTokenizer st = new StringTokenizer(",,,ZERO;,ONE TWO", ",; ", false);
while(st.hasMoreTokens()) {
String str = st.nextToken();
//add to list, process, etc...
}
As a caveat, however, you'll need to define each potential whitespace character separately in the second argument to the constructor.