Mixing
Create words' stream using scanner
1
There is needed to return a stream of all words that have 3letters and more from a file. Is there better way then following, maybe using Stream.iterate:
private Stream<String> getWordsStream(String path){
Stream.Builder<String> wordsStream = Stream.builder();
FileInputStream inputStream = null;
try {
inputStream = new FileInputStream(path);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
Scanner s = new Scanner(inputStream);
s.useDelimiter("([^a-zA-Z])");
Pattern pattern = Pattern.compile("([a-zA-Z]{3,})");
while ((s.hasNext())){
if(s.hasNext(pattern)){
wordsStream.add(s.next().toUpperCase());
}
else {
s.next();
}
}
s.close();
return wordsStream.build();
}
java loops java-stream builder word
asked Nov 19 '18 at 21:43
a_chubenkoa_chubenko
4615
add a comment |
1
There is needed to return a stream of all words that have 3letters and more from a file. Is there better way then following, maybe using Stream.iterate:
private Stream<String> getWordsStream(String path){
Stream.Builder<String> wordsStream = Stream.builder();
FileInputStream inputStream = null;
try {
inputStream = new FileInputStream(path);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
Scanner s = new Scanner(inputStream);
s.useDelimiter("([^a-zA-Z])");
Pattern pattern = Pattern.compile("([a-zA-Z]{3,})");
while ((s.hasNext())){
if(s.hasNext(pattern)){
wordsStream.add(s.next().toUpperCase());
}
else {
s.next();
}
}
s.close();
return wordsStream.build();
}
java loops java-stream builder word
asked Nov 19 '18 at 21:43
a_chubenkoa_chubenko
4615
Which Java version?
– shmosel
Nov 19 '18 at 21:48
Did you mean to calls.next(pattern)?
– shmosel
Nov 19 '18 at 21:53
Maybe reading the entire stream as a string, then splitting it with a space (or whatever you're using), then checking each for their length.
– PhaseRush
Nov 19 '18 at 22:04
Java 9. I mean: is it possible to write this method more closely to stream style, without while loop at all
– a_chubenko
Nov 19 '18 at 22:06
add a comment |
1
1
1
There is needed to return a stream of all words that have 3letters and more from a file. Is there better way then following, maybe using Stream.iterate:
private Stream<String> getWordsStream(String path){
Stream.Builder<String> wordsStream = Stream.builder();
FileInputStream inputStream = null;
try {
inputStream = new FileInputStream(path);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
Scanner s = new Scanner(inputStream);
s.useDelimiter("([^a-zA-Z])");
Pattern pattern = Pattern.compile("([a-zA-Z]{3,})");
while ((s.hasNext())){
if(s.hasNext(pattern)){
wordsStream.add(s.next().toUpperCase());
}
else {
s.next();
}
}
s.close();
return wordsStream.build();
}
java loops java-stream builder word
asked Nov 19 '18 at 21:43
a_chubenkoa_chubenko
4615
There is needed to return a stream of all words that have 3letters and more from a file. Is there better way then following, maybe using Stream.iterate:
private Stream<String> getWordsStream(String path){
Stream.Builder<String> wordsStream = Stream.builder();
FileInputStream inputStream = null;
try {
inputStream = new FileInputStream(path);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
Scanner s = new Scanner(inputStream);
s.useDelimiter("([^a-zA-Z])");
Pattern pattern = Pattern.compile("([a-zA-Z]{3,})");
while ((s.hasNext())){
if(s.hasNext(pattern)){
wordsStream.add(s.next().toUpperCase());
}
else {
s.next();
}
}
s.close();
return wordsStream.build();
}
java loops java-stream builder word
java loops java-stream builder word
asked Nov 19 '18 at 21:43
a_chubenkoa_chubenko
4615
asked Nov 19 '18 at 21:43
a_chubenkoa_chubenko
4615
edited Nov 20 '18 at 15:49
a_chubenko
asked Nov 19 '18 at 21:43
a_chubenkoa_chubenko
4615
asked Nov 19 '18 at 21:43
a_chubenkoa_chubenko
4615
asked Nov 19 '18 at 21:43
a_chubenkoa_chubenko
4615
4615
Which Java version?
– shmosel
Nov 19 '18 at 21:48
Did you mean to calls.next(pattern)?
– shmosel
Nov 19 '18 at 21:53
Maybe reading the entire stream as a string, then splitting it with a space (or whatever you're using), then checking each for their length.
– PhaseRush
Nov 19 '18 at 22:04
Java 9. I mean: is it possible to write this method more closely to stream style, without while loop at all
– a_chubenko
Nov 19 '18 at 22:06
add a comment |
Which Java version?
– shmosel
Nov 19 '18 at 21:48
Did you mean to calls.next(pattern)?
– shmosel
Nov 19 '18 at 21:53
Maybe reading the entire stream as a string, then splitting it with a space (or whatever you're using), then checking each for their length.
– PhaseRush
Nov 19 '18 at 22:04
Java 9. I mean: is it possible to write this method more closely to stream style, without while loop at all
– a_chubenko
Nov 19 '18 at 22:06
Which Java version?
– shmosel
Nov 19 '18 at 21:48
Which Java version?
– shmosel
Nov 19 '18 at 21:48
Did you mean to call
s.next(pattern)?– shmosel
Nov 19 '18 at 21:53
Did you mean to call
s.next(pattern)?– shmosel
Nov 19 '18 at 21:53
Maybe reading the entire stream as a string, then splitting it with a space (or whatever you're using), then checking each for their length.
– PhaseRush
Nov 19 '18 at 22:04
Maybe reading the entire stream as a string, then splitting it with a space (or whatever you're using), then checking each for their length.
– PhaseRush
Nov 19 '18 at 22:04
Java 9. I mean: is it possible to write this method more closely to stream style, without while loop at all
– a_chubenko
Nov 19 '18 at 22:06
Java 9. I mean: is it possible to write this method more closely to stream style, without while loop at all
– a_chubenko
Nov 19 '18 at 22:06
add a comment |
3 Answers
3
active
oldest
votes
2
You can use Files.lines() and a Pattern:
private static final Pattern SPACES = Pattern.compile("[^a-zA-Z]+");
public static Stream<String> getWordStream(String path) throws IOException{
return Files.lines(Paths.get(path))
.flatMap(SPACES::splitAsStream)
.filter(word -> word.length() >= 3);
}
answered Nov 20 '18 at 7:23
LinoLino
7,88421936
There is tested for a book with 105 K words. This method is the fastest, took 0.29s.
– a_chubenko
Nov 20 '18 at 12:05
1
@Alex well, this method also skips the conversion to uppercase. Further, your original code was designed to process words consisting of ASCII letters only, whereas this code treats everything separated by a single space character as a word.
– Holger
Nov 20 '18 at 12:07
Pattern was changed to Pattern SPACES = Pattern.compile("([^a-zA-Z])");
– a_chubenko
Nov 20 '18 at 12:10
@Alex You should probably usePattern.compile("[^a-zA-Z]+")(notice the+at the end). So you won't get "empty" words, e.g a text like:"I have 100 dollars"would produce an array:["I", "have", "", "", "", "", "dollars"]with the current pattern
– Lino
Nov 20 '18 at 12:19
add a comment |
5
The worst part of your code is the following part
FileInputStream inputStream = null;
try {
inputStream = new FileInputStream(path);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
Scanner s = new Scanner(inputStream);
So when the file is absent, you will print the FileNotFoundException stack trace and proceed with a null input stream, leading to a NullPointerException. Instead of requiring the caller to deal with a spurious NullPointerException, you should declare the FileNotFoundException in the method signature. Otherwise, return an empty stream in the erroneous case.
But you don’t need to contruct a FileInputStream at all, as Scanner offers constructors accepting a File or Path. Combine this with the capability of returning a stream of matches (since Java 9) and you get:
private Stream<String> getWordsStream(String path) {
try {
Scanner s = new Scanner(Paths.get(path));
return s.findAll("([a-zA-Z]{3,})").map(mr -> mr.group().toUpperCase());
} catch(IOException ex) {
Logger.getLogger(Main.class.getName()).log(Level.SEVERE, null, ex);
return Stream.empty();
}
}
or preferably
private Stream<String> getWordsStream(String path) throws IOException {
Scanner s = new Scanner(Paths.get(path));
return s.findAll("([a-zA-Z]{3,})").map(mr -> mr.group().toUpperCase());
}
You don’t even need .useDelimiter("([^a-zA-Z])") here, as skipping all nonmatching stuff is the default behavior.
Closing the returned Stream will also close the Scanner.
So the caller should use it like this
try(Stream<String> s = getWordsStream("path/to/file")) {
s.forEach(System.out::println);
}
answered Nov 20 '18 at 8:03
HolgerHolger
163k23231438
There is tested for a book with 105 K words. This method took about 0.6s.
– a_chubenko
Nov 20 '18 at 12:06
add a comment |
0
Thre're much easier approach: read lines from file to the Stream and filter it with required condition (e.g. length >= 3). Files.lines() has lazy loading, so it does not ready all words from the file at the beginning, it does it every time when next word is required
public static void main(String... args) throws IOException {
getWordsStream(Paths.get("d:/words.txt")).forEach(System.out::println);
}
public static Stream<String> getWordsStream(Path path) throws IOException {
final Scanner scan = new Scanner(path);
return StreamSupport.stream(new Spliterators.AbstractSpliterator<String>(Long.MAX_VALUE,
Spliterator.DISTINCT | Spliterator.IMMUTABLE | Spliterator.NONNULL | Spliterator.ORDERED) {
@Override
public boolean tryAdvance(Consumer<? super String> action) {
while (scan.hasNext()) {
String word = scan.next();
// you can use RegExp if you have more complicated condition
if (word.length() < 3)
continue;
action.accept(word);
return true;
}
return false;
}
}, false).onClose(scan::close);
}
answered Nov 19 '18 at 22:04
oleg.cherednikoleg.cherednik
6,08221118
1
andlines()will return a stream of words? would be a very bad name for such method - maybe Java 9, but for sure in Java 11 it is returning a stream of lines
– Carlos Heuberger
Nov 19 '18 at 22:17
It returns lines that have more then 3 letter symbols in any Java. I have an unprepared text. So needs a loop again for finding all matches in a line.
– a_chubenko
Nov 19 '18 at 22:29
1
well, actuallylines()returns all lines, not only the ones with more than 3 letter. It'sgetWordsStream()that returns the lines with 3 or more letters. But question is about words with 3 or more letters, not about lines
– Carlos Heuberger
Nov 20 '18 at 1:05
With java-11, you need not to change the signature of that method and can useFiles.lines(Path.of(path))instead.
– nullpointer
Nov 20 '18 at 2:12
Fixed. Same lazy load approach. Pease a cake.
– oleg.cherednik
Nov 20 '18 at 5:22
|
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
You can use Files.lines() and a Pattern:
private static final Pattern SPACES = Pattern.compile("[^a-zA-Z]+");
public static Stream<String> getWordStream(String path) throws IOException{
return Files.lines(Paths.get(path))
.flatMap(SPACES::splitAsStream)
.filter(word -> word.length() >= 3);
}
answered Nov 20 '18 at 7:23
LinoLino
7,88421936
There is tested for a book with 105 K words. This method is the fastest, took 0.29s.
– a_chubenko
Nov 20 '18 at 12:05
1
@Alex well, this method also skips the conversion to uppercase. Further, your original code was designed to process words consisting of ASCII letters only, whereas this code treats everything separated by a single space character as a word.
– Holger
Nov 20 '18 at 12:07
Pattern was changed to Pattern SPACES = Pattern.compile("([^a-zA-Z])");
– a_chubenko
Nov 20 '18 at 12:10
@Alex You should probably usePattern.compile("[^a-zA-Z]+")(notice the+at the end). So you won't get "empty" words, e.g a text like:"I have 100 dollars"would produce an array:["I", "have", "", "", "", "", "dollars"]with the current pattern
– Lino
Nov 20 '18 at 12:19
add a comment |
2
You can use Files.lines() and a Pattern:
private static final Pattern SPACES = Pattern.compile("[^a-zA-Z]+");
public static Stream<String> getWordStream(String path) throws IOException{
return Files.lines(Paths.get(path))
.flatMap(SPACES::splitAsStream)
.filter(word -> word.length() >= 3);
}
answered Nov 20 '18 at 7:23
LinoLino
7,88421936
There is tested for a book with 105 K words. This method is the fastest, took 0.29s.
– a_chubenko
Nov 20 '18 at 12:05
1
@Alex well, this method also skips the conversion to uppercase. Further, your original code was designed to process words consisting of ASCII letters only, whereas this code treats everything separated by a single space character as a word.
– Holger
Nov 20 '18 at 12:07
Pattern was changed to Pattern SPACES = Pattern.compile("([^a-zA-Z])");
– a_chubenko
Nov 20 '18 at 12:10
@Alex You should probably usePattern.compile("[^a-zA-Z]+")(notice the+at the end). So you won't get "empty" words, e.g a text like:"I have 100 dollars"would produce an array:["I", "have", "", "", "", "", "dollars"]with the current pattern
– Lino
Nov 20 '18 at 12:19
add a comment |
2
2
2
You can use Files.lines() and a Pattern:
private static final Pattern SPACES = Pattern.compile("[^a-zA-Z]+");
public static Stream<String> getWordStream(String path) throws IOException{
return Files.lines(Paths.get(path))
.flatMap(SPACES::splitAsStream)
.filter(word -> word.length() >= 3);
}
answered Nov 20 '18 at 7:23
LinoLino
7,88421936
You can use Files.lines() and a Pattern:
private static final Pattern SPACES = Pattern.compile("[^a-zA-Z]+");
public static Stream<String> getWordStream(String path) throws IOException{
return Files.lines(Paths.get(path))
.flatMap(SPACES::splitAsStream)
.filter(word -> word.length() >= 3);
}
answered Nov 20 '18 at 7:23
LinoLino
7,88421936
edited Nov 20 '18 at 12:43
answered Nov 20 '18 at 7:23
LinoLino
7,88421936
answered Nov 20 '18 at 7:23
LinoLino
7,88421936
answered Nov 20 '18 at 7:23
LinoLino
7,88421936
7,88421936
There is tested for a book with 105 K words. This method is the fastest, took 0.29s.
– a_chubenko
Nov 20 '18 at 12:05
1
@Alex well, this method also skips the conversion to uppercase. Further, your original code was designed to process words consisting of ASCII letters only, whereas this code treats everything separated by a single space character as a word.
– Holger
Nov 20 '18 at 12:07
Pattern was changed to Pattern SPACES = Pattern.compile("([^a-zA-Z])");
– a_chubenko
Nov 20 '18 at 12:10
@Alex You should probably usePattern.compile("[^a-zA-Z]+")(notice the+at the end). So you won't get "empty" words, e.g a text like:"I have 100 dollars"would produce an array:["I", "have", "", "", "", "", "dollars"]with the current pattern
– Lino
Nov 20 '18 at 12:19
add a comment |
There is tested for a book with 105 K words. This method is the fastest, took 0.29s.
– a_chubenko
Nov 20 '18 at 12:05
1
@Alex well, this method also skips the conversion to uppercase. Further, your original code was designed to process words consisting of ASCII letters only, whereas this code treats everything separated by a single space character as a word.
– Holger
Nov 20 '18 at 12:07
Pattern was changed to Pattern SPACES = Pattern.compile("([^a-zA-Z])");
– a_chubenko
Nov 20 '18 at 12:10
@Alex You should probably usePattern.compile("[^a-zA-Z]+")(notice the+at the end). So you won't get "empty" words, e.g a text like:"I have 100 dollars"would produce an array:["I", "have", "", "", "", "", "dollars"]with the current pattern
– Lino
Nov 20 '18 at 12:19
There is tested for a book with 105 K words. This method is the fastest, took 0.29s.
– a_chubenko
Nov 20 '18 at 12:05
There is tested for a book with 105 K words. This method is the fastest, took 0.29s.
– a_chubenko
Nov 20 '18 at 12:05
1
1
@Alex well, this method also skips the conversion to uppercase. Further, your original code was designed to process words consisting of ASCII letters only, whereas this code treats everything separated by a single space character as a word.
– Holger
Nov 20 '18 at 12:07
@Alex well, this method also skips the conversion to uppercase. Further, your original code was designed to process words consisting of ASCII letters only, whereas this code treats everything separated by a single space character as a word.
– Holger
Nov 20 '18 at 12:07
Pattern was changed to Pattern SPACES = Pattern.compile("([^a-zA-Z])");
– a_chubenko
Nov 20 '18 at 12:10
Pattern was changed to Pattern SPACES = Pattern.compile("([^a-zA-Z])");
– a_chubenko
Nov 20 '18 at 12:10
@Alex You should probably use
Pattern.compile("[^a-zA-Z]+") (notice the + at the end). So you won't get "empty" words, e.g a text like: "I have 100 dollars" would produce an array: ["I", "have", "", "", "", "", "dollars"] with the current pattern– Lino
Nov 20 '18 at 12:19
@Alex You should probably use
Pattern.compile("[^a-zA-Z]+") (notice the + at the end). So you won't get "empty" words, e.g a text like: "I have 100 dollars" would produce an array: ["I", "have", "", "", "", "", "dollars"] with the current pattern– Lino
Nov 20 '18 at 12:19
add a comment |
5
The worst part of your code is the following part
FileInputStream inputStream = null;
try {
inputStream = new FileInputStream(path);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
Scanner s = new Scanner(inputStream);
So when the file is absent, you will print the FileNotFoundException stack trace and proceed with a null input stream, leading to a NullPointerException. Instead of requiring the caller to deal with a spurious NullPointerException, you should declare the FileNotFoundException in the method signature. Otherwise, return an empty stream in the erroneous case.
But you don’t need to contruct a FileInputStream at all, as Scanner offers constructors accepting a File or Path. Combine this with the capability of returning a stream of matches (since Java 9) and you get:
private Stream<String> getWordsStream(String path) {
try {
Scanner s = new Scanner(Paths.get(path));
return s.findAll("([a-zA-Z]{3,})").map(mr -> mr.group().toUpperCase());
} catch(IOException ex) {
Logger.getLogger(Main.class.getName()).log(Level.SEVERE, null, ex);
return Stream.empty();
}
}
or preferably
private Stream<String> getWordsStream(String path) throws IOException {
Scanner s = new Scanner(Paths.get(path));
return s.findAll("([a-zA-Z]{3,})").map(mr -> mr.group().toUpperCase());
}
You don’t even need .useDelimiter("([^a-zA-Z])") here, as skipping all nonmatching stuff is the default behavior.
Closing the returned Stream will also close the Scanner.
So the caller should use it like this
try(Stream<String> s = getWordsStream("path/to/file")) {
s.forEach(System.out::println);
}
answered Nov 20 '18 at 8:03
HolgerHolger
163k23231438
There is tested for a book with 105 K words. This method took about 0.6s.
– a_chubenko
Nov 20 '18 at 12:06
add a comment |
5
The worst part of your code is the following part
FileInputStream inputStream = null;
try {
inputStream = new FileInputStream(path);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
Scanner s = new Scanner(inputStream);
So when the file is absent, you will print the FileNotFoundException stack trace and proceed with a null input stream, leading to a NullPointerException. Instead of requiring the caller to deal with a spurious NullPointerException, you should declare the FileNotFoundException in the method signature. Otherwise, return an empty stream in the erroneous case.
But you don’t need to contruct a FileInputStream at all, as Scanner offers constructors accepting a File or Path. Combine this with the capability of returning a stream of matches (since Java 9) and you get:
private Stream<String> getWordsStream(String path) {
try {
Scanner s = new Scanner(Paths.get(path));
return s.findAll("([a-zA-Z]{3,})").map(mr -> mr.group().toUpperCase());
} catch(IOException ex) {
Logger.getLogger(Main.class.getName()).log(Level.SEVERE, null, ex);
return Stream.empty();
}
}
or preferably
private Stream<String> getWordsStream(String path) throws IOException {
Scanner s = new Scanner(Paths.get(path));
return s.findAll("([a-zA-Z]{3,})").map(mr -> mr.group().toUpperCase());
}
You don’t even need .useDelimiter("([^a-zA-Z])") here, as skipping all nonmatching stuff is the default behavior.
Closing the returned Stream will also close the Scanner.
So the caller should use it like this
try(Stream<String> s = getWordsStream("path/to/file")) {
s.forEach(System.out::println);
}
answered Nov 20 '18 at 8:03
HolgerHolger
163k23231438
There is tested for a book with 105 K words. This method took about 0.6s.
– a_chubenko
Nov 20 '18 at 12:06
add a comment |
5
5
5
The worst part of your code is the following part
FileInputStream inputStream = null;
try {
inputStream = new FileInputStream(path);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
Scanner s = new Scanner(inputStream);
So when the file is absent, you will print the FileNotFoundException stack trace and proceed with a null input stream, leading to a NullPointerException. Instead of requiring the caller to deal with a spurious NullPointerException, you should declare the FileNotFoundException in the method signature. Otherwise, return an empty stream in the erroneous case.
But you don’t need to contruct a FileInputStream at all, as Scanner offers constructors accepting a File or Path. Combine this with the capability of returning a stream of matches (since Java 9) and you get:
private Stream<String> getWordsStream(String path) {
try {
Scanner s = new Scanner(Paths.get(path));
return s.findAll("([a-zA-Z]{3,})").map(mr -> mr.group().toUpperCase());
} catch(IOException ex) {
Logger.getLogger(Main.class.getName()).log(Level.SEVERE, null, ex);
return Stream.empty();
}
}
or preferably
private Stream<String> getWordsStream(String path) throws IOException {
Scanner s = new Scanner(Paths.get(path));
return s.findAll("([a-zA-Z]{3,})").map(mr -> mr.group().toUpperCase());
}
You don’t even need .useDelimiter("([^a-zA-Z])") here, as skipping all nonmatching stuff is the default behavior.
Closing the returned Stream will also close the Scanner.
So the caller should use it like this
try(Stream<String> s = getWordsStream("path/to/file")) {
s.forEach(System.out::println);
}
answered Nov 20 '18 at 8:03
HolgerHolger
163k23231438
The worst part of your code is the following part
FileInputStream inputStream = null;
try {
inputStream = new FileInputStream(path);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
Scanner s = new Scanner(inputStream);
So when the file is absent, you will print the FileNotFoundException stack trace and proceed with a null input stream, leading to a NullPointerException. Instead of requiring the caller to deal with a spurious NullPointerException, you should declare the FileNotFoundException in the method signature. Otherwise, return an empty stream in the erroneous case.
But you don’t need to contruct a FileInputStream at all, as Scanner offers constructors accepting a File or Path. Combine this with the capability of returning a stream of matches (since Java 9) and you get:
private Stream<String> getWordsStream(String path) {
try {
Scanner s = new Scanner(Paths.get(path));
return s.findAll("([a-zA-Z]{3,})").map(mr -> mr.group().toUpperCase());
} catch(IOException ex) {
Logger.getLogger(Main.class.getName()).log(Level.SEVERE, null, ex);
return Stream.empty();
}
}
or preferably
private Stream<String> getWordsStream(String path) throws IOException {
Scanner s = new Scanner(Paths.get(path));
return s.findAll("([a-zA-Z]{3,})").map(mr -> mr.group().toUpperCase());
}
You don’t even need .useDelimiter("([^a-zA-Z])") here, as skipping all nonmatching stuff is the default behavior.
Closing the returned Stream will also close the Scanner.
So the caller should use it like this
try(Stream<String> s = getWordsStream("path/to/file")) {
s.forEach(System.out::println);
}
answered Nov 20 '18 at 8:03
HolgerHolger
163k23231438
edited Nov 20 '18 at 8:19
answered Nov 20 '18 at 8:03
HolgerHolger
163k23231438
answered Nov 20 '18 at 8:03
HolgerHolger
163k23231438
answered Nov 20 '18 at 8:03
HolgerHolger
163k23231438
163k23231438
There is tested for a book with 105 K words. This method took about 0.6s.
– a_chubenko
Nov 20 '18 at 12:06
add a comment |
There is tested for a book with 105 K words. This method took about 0.6s.
– a_chubenko
Nov 20 '18 at 12:06
There is tested for a book with 105 K words. This method took about 0.6s.
– a_chubenko
Nov 20 '18 at 12:06
There is tested for a book with 105 K words. This method took about 0.6s.
– a_chubenko
Nov 20 '18 at 12:06
add a comment |
0
Thre're much easier approach: read lines from file to the Stream and filter it with required condition (e.g. length >= 3). Files.lines() has lazy loading, so it does not ready all words from the file at the beginning, it does it every time when next word is required
public static void main(String... args) throws IOException {
getWordsStream(Paths.get("d:/words.txt")).forEach(System.out::println);
}
public static Stream<String> getWordsStream(Path path) throws IOException {
final Scanner scan = new Scanner(path);
return StreamSupport.stream(new Spliterators.AbstractSpliterator<String>(Long.MAX_VALUE,
Spliterator.DISTINCT | Spliterator.IMMUTABLE | Spliterator.NONNULL | Spliterator.ORDERED) {
@Override
public boolean tryAdvance(Consumer<? super String> action) {
while (scan.hasNext()) {
String word = scan.next();
// you can use RegExp if you have more complicated condition
if (word.length() < 3)
continue;
action.accept(word);
return true;
}
return false;
}
}, false).onClose(scan::close);
}
answered Nov 19 '18 at 22:04
oleg.cherednikoleg.cherednik
6,08221118
1
andlines()will return a stream of words? would be a very bad name for such method - maybe Java 9, but for sure in Java 11 it is returning a stream of lines
– Carlos Heuberger
Nov 19 '18 at 22:17
It returns lines that have more then 3 letter symbols in any Java. I have an unprepared text. So needs a loop again for finding all matches in a line.
– a_chubenko
Nov 19 '18 at 22:29
1
well, actuallylines()returns all lines, not only the ones with more than 3 letter. It'sgetWordsStream()that returns the lines with 3 or more letters. But question is about words with 3 or more letters, not about lines
– Carlos Heuberger
Nov 20 '18 at 1:05
With java-11, you need not to change the signature of that method and can useFiles.lines(Path.of(path))instead.
– nullpointer
Nov 20 '18 at 2:12
Fixed. Same lazy load approach. Pease a cake.
– oleg.cherednik
Nov 20 '18 at 5:22
|
show 3 more comments
0
Thre're much easier approach: read lines from file to the Stream and filter it with required condition (e.g. length >= 3). Files.lines() has lazy loading, so it does not ready all words from the file at the beginning, it does it every time when next word is required
public static void main(String... args) throws IOException {
getWordsStream(Paths.get("d:/words.txt")).forEach(System.out::println);
}
public static Stream<String> getWordsStream(Path path) throws IOException {
final Scanner scan = new Scanner(path);
return StreamSupport.stream(new Spliterators.AbstractSpliterator<String>(Long.MAX_VALUE,
Spliterator.DISTINCT | Spliterator.IMMUTABLE | Spliterator.NONNULL | Spliterator.ORDERED) {
@Override
public boolean tryAdvance(Consumer<? super String> action) {
while (scan.hasNext()) {
String word = scan.next();
// you can use RegExp if you have more complicated condition
if (word.length() < 3)
continue;
action.accept(word);
return true;
}
return false;
}
}, false).onClose(scan::close);
}
answered Nov 19 '18 at 22:04
oleg.cherednikoleg.cherednik
6,08221118
1
andlines()will return a stream of words? would be a very bad name for such method - maybe Java 9, but for sure in Java 11 it is returning a stream of lines
– Carlos Heuberger
Nov 19 '18 at 22:17
It returns lines that have more then 3 letter symbols in any Java. I have an unprepared text. So needs a loop again for finding all matches in a line.
– a_chubenko
Nov 19 '18 at 22:29
1
well, actuallylines()returns all lines, not only the ones with more than 3 letter. It'sgetWordsStream()that returns the lines with 3 or more letters. But question is about words with 3 or more letters, not about lines
– Carlos Heuberger
Nov 20 '18 at 1:05
With java-11, you need not to change the signature of that method and can useFiles.lines(Path.of(path))instead.
– nullpointer
Nov 20 '18 at 2:12
Fixed. Same lazy load approach. Pease a cake.
– oleg.cherednik
Nov 20 '18 at 5:22
|
show 3 more comments
0
0
0
Thre're much easier approach: read lines from file to the Stream and filter it with required condition (e.g. length >= 3). Files.lines() has lazy loading, so it does not ready all words from the file at the beginning, it does it every time when next word is required
public static void main(String... args) throws IOException {
getWordsStream(Paths.get("d:/words.txt")).forEach(System.out::println);
}
public static Stream<String> getWordsStream(Path path) throws IOException {
final Scanner scan = new Scanner(path);
return StreamSupport.stream(new Spliterators.AbstractSpliterator<String>(Long.MAX_VALUE,
Spliterator.DISTINCT | Spliterator.IMMUTABLE | Spliterator.NONNULL | Spliterator.ORDERED) {
@Override
public boolean tryAdvance(Consumer<? super String> action) {
while (scan.hasNext()) {
String word = scan.next();
// you can use RegExp if you have more complicated condition
if (word.length() < 3)
continue;
action.accept(word);
return true;
}
return false;
}
}, false).onClose(scan::close);
}
answered Nov 19 '18 at 22:04
oleg.cherednikoleg.cherednik
6,08221118
Thre're much easier approach: read lines from file to the Stream and filter it with required condition (e.g. length >= 3). Files.lines() has lazy loading, so it does not ready all words from the file at the beginning, it does it every time when next word is required
public static void main(String... args) throws IOException {
getWordsStream(Paths.get("d:/words.txt")).forEach(System.out::println);
}
public static Stream<String> getWordsStream(Path path) throws IOException {
final Scanner scan = new Scanner(path);
return StreamSupport.stream(new Spliterators.AbstractSpliterator<String>(Long.MAX_VALUE,
Spliterator.DISTINCT | Spliterator.IMMUTABLE | Spliterator.NONNULL | Spliterator.ORDERED) {
@Override
public boolean tryAdvance(Consumer<? super String> action) {
while (scan.hasNext()) {
String word = scan.next();
// you can use RegExp if you have more complicated condition
if (word.length() < 3)
continue;
action.accept(word);
return true;
}
return false;
}
}, false).onClose(scan::close);
}
answered Nov 19 '18 at 22:04
oleg.cherednikoleg.cherednik
6,08221118
edited Nov 20 '18 at 8:34
answered Nov 19 '18 at 22:04
oleg.cherednikoleg.cherednik
6,08221118
answered Nov 19 '18 at 22:04
oleg.cherednikoleg.cherednik
6,08221118
answered Nov 19 '18 at 22:04
oleg.cherednikoleg.cherednik
6,08221118
6,08221118
1
andlines()will return a stream of words? would be a very bad name for such method - maybe Java 9, but for sure in Java 11 it is returning a stream of lines
– Carlos Heuberger
Nov 19 '18 at 22:17
It returns lines that have more then 3 letter symbols in any Java. I have an unprepared text. So needs a loop again for finding all matches in a line.
– a_chubenko
Nov 19 '18 at 22:29
1
well, actuallylines()returns all lines, not only the ones with more than 3 letter. It'sgetWordsStream()that returns the lines with 3 or more letters. But question is about words with 3 or more letters, not about lines
– Carlos Heuberger
Nov 20 '18 at 1:05
With java-11, you need not to change the signature of that method and can useFiles.lines(Path.of(path))instead.
– nullpointer
Nov 20 '18 at 2:12
Fixed. Same lazy load approach. Pease a cake.
– oleg.cherednik
Nov 20 '18 at 5:22
|
show 3 more comments
1
andlines()will return a stream of words? would be a very bad name for such method - maybe Java 9, but for sure in Java 11 it is returning a stream of lines
– Carlos Heuberger
Nov 19 '18 at 22:17
It returns lines that have more then 3 letter symbols in any Java. I have an unprepared text. So needs a loop again for finding all matches in a line.
– a_chubenko
Nov 19 '18 at 22:29
1
well, actuallylines()returns all lines, not only the ones with more than 3 letter. It'sgetWordsStream()that returns the lines with 3 or more letters. But question is about words with 3 or more letters, not about lines
– Carlos Heuberger
Nov 20 '18 at 1:05
With java-11, you need not to change the signature of that method and can useFiles.lines(Path.of(path))instead.
– nullpointer
Nov 20 '18 at 2:12
Fixed. Same lazy load approach. Pease a cake.
– oleg.cherednik
Nov 20 '18 at 5:22
1
1
and
lines() will return a stream of words? would be a very bad name for such method - maybe Java 9, but for sure in Java 11 it is returning a stream of lines– Carlos Heuberger
Nov 19 '18 at 22:17
and
lines() will return a stream of words? would be a very bad name for such method - maybe Java 9, but for sure in Java 11 it is returning a stream of lines– Carlos Heuberger
Nov 19 '18 at 22:17
It returns lines that have more then 3 letter symbols in any Java. I have an unprepared text. So needs a loop again for finding all matches in a line.
– a_chubenko
Nov 19 '18 at 22:29
It returns lines that have more then 3 letter symbols in any Java. I have an unprepared text. So needs a loop again for finding all matches in a line.
– a_chubenko
Nov 19 '18 at 22:29
1
1
well, actually
lines() returns all lines, not only the ones with more than 3 letter. It's getWordsStream() that returns the lines with 3 or more letters. But question is about words with 3 or more letters, not about lines– Carlos Heuberger
Nov 20 '18 at 1:05
well, actually
lines() returns all lines, not only the ones with more than 3 letter. It's getWordsStream() that returns the lines with 3 or more letters. But question is about words with 3 or more letters, not about lines– Carlos Heuberger
Nov 20 '18 at 1:05
With java-11, you need not to change the signature of that method and can use
Files.lines(Path.of(path)) instead.– nullpointer
Nov 20 '18 at 2:12
With java-11, you need not to change the signature of that method and can use
Files.lines(Path.of(path)) instead.– nullpointer
Nov 20 '18 at 2:12
Fixed. Same lazy load approach. Pease a cake.
– oleg.cherednik
Nov 20 '18 at 5:22
Fixed. Same lazy load approach. Pease a cake.
– oleg.cherednik
Nov 20 '18 at 5:22
|
show 3 more comments
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Which Java version?
– shmosel
Nov 19 '18 at 21:48
Did you mean to call
s.next(pattern)?– shmosel
Nov 19 '18 at 21:53
Maybe reading the entire stream as a string, then splitting it with a space (or whatever you're using), then checking each for their length.
– PhaseRush
Nov 19 '18 at 22:04
Java 9. I mean: is it possible to write this method more closely to stream style, without while loop at all
– a_chubenko
Nov 19 '18 at 22:06