我使用巨大的数据文件,有时我只需要知道这些文件中的行数,通常我打开它们,一行一行地读取它们,直到我到达文件的末尾
我在想有没有更聪明的办法
我使用巨大的数据文件,有时我只需要知道这些文件中的行数,通常我打开它们,一行一行地读取它们,直到我到达文件的末尾
我在想有没有更聪明的办法
当前回答
上面的count()方法给出的答案是,如果文件末尾没有换行符,就会出现行数错误——它无法计算文件中的最后一行。
这个方法更适合我:
public int countLines(String filename) throws IOException {
LineNumberReader reader = new LineNumberReader(new FileReader(filename));
int cnt = 0;
String lineRead = "";
while ((lineRead = reader.readLine()) != null) {}
cnt = reader.getLineNumber();
reader.close();
return cnt;
}
其他回答
EOF处没有换行符('\n')的多行文件的最佳优化代码。
/**
*
* @param filename
* @return
* @throws IOException
*/
public static int countLines(String filename) throws IOException {
int count = 0;
boolean empty = true;
FileInputStream fis = null;
InputStream is = null;
try {
fis = new FileInputStream(filename);
is = new BufferedInputStream(fis);
byte[] c = new byte[1024];
int readChars = 0;
boolean isLine = false;
while ((readChars = is.read(c)) != -1) {
empty = false;
for (int i = 0; i < readChars; ++i) {
if ( c[i] == '\n' ) {
isLine = false;
++count;
}else if(!isLine && c[i] != '\n' && c[i] != '\r'){ //Case to handle line count where no New Line character present at EOF
isLine = true;
}
}
}
if(isLine){
++count;
}
}catch(IOException e){
e.printStackTrace();
}finally {
if(is != null){
is.close();
}
if(fis != null){
fis.close();
}
}
LOG.info("count: "+count);
return (count == 0 && !empty) ? 1 : count;
}
/**
* Count file rows.
*
* @param file file
* @return file row count
* @throws IOException
*/
public static long getLineCount(File file) throws IOException {
try (Stream<String> lines = Files.lines(file.toPath())) {
return lines.count();
}
}
在JDK8_u31上测试。但与此方法相比,性能确实较慢:
/**
* Count file rows.
*
* @param file file
* @return file row count
* @throws IOException
*/
public static long getLineCount(File file) throws IOException {
try (BufferedInputStream is = new BufferedInputStream(new FileInputStream(file), 1024)) {
byte[] c = new byte[1024];
boolean empty = true,
lastEmpty = false;
long count = 0;
int read;
while ((read = is.read(c)) != -1) {
for (int i = 0; i < read; i++) {
if (c[i] == '\n') {
count++;
lastEmpty = true;
} else if (lastEmpty) {
lastEmpty = false;
}
}
empty = false;
}
if (!empty) {
if (count == 0) {
count = 1;
} else if (!lastEmpty) {
count++;
}
}
return count;
}
}
经过测试,非常快。
如果你用这个
public int countLines(String filename) throws IOException {
LineNumberReader reader = new LineNumberReader(new FileReader(filename));
int cnt = 0;
String lineRead = "";
while ((lineRead = reader.readLine()) != null) {}
cnt = reader.getLineNumber();
reader.close();
return cnt;
}
你不能运行到大num行,比如100K行,因为从读取器返回。getLineNumber是int。你需要长类型的数据来处理最多的行。
我知道这是一个老问题,但公认的解决方案并不完全符合我所需要的。因此,我将其改进为接受各种行结束符(而不仅仅是换行)并使用指定的字符编码(而不是ISO-8859-n)。所有在一个方法(适当重构):
public static long getLinesCount(String fileName, String encodingName) throws IOException {
long linesCount = 0;
File file = new File(fileName);
FileInputStream fileIn = new FileInputStream(file);
try {
Charset encoding = Charset.forName(encodingName);
Reader fileReader = new InputStreamReader(fileIn, encoding);
int bufferSize = 4096;
Reader reader = new BufferedReader(fileReader, bufferSize);
char[] buffer = new char[bufferSize];
int prevChar = -1;
int readCount = reader.read(buffer);
while (readCount != -1) {
for (int i = 0; i < readCount; i++) {
int nextChar = buffer[i];
switch (nextChar) {
case '\r': {
// The current line is terminated by a carriage return or by a carriage return immediately followed by a line feed.
linesCount++;
break;
}
case '\n': {
if (prevChar == '\r') {
// The current line is terminated by a carriage return immediately followed by a line feed.
// The line has already been counted.
} else {
// The current line is terminated by a line feed.
linesCount++;
}
break;
}
}
prevChar = nextChar;
}
readCount = reader.read(buffer);
}
if (prevCh != -1) {
switch (prevCh) {
case '\r':
case '\n': {
// The last line is terminated by a line terminator.
// The last line has already been counted.
break;
}
default: {
// The last line is terminated by end-of-file.
linesCount++;
}
}
}
} finally {
fileIn.close();
}
return linesCount;
}
这个解决方案在速度上与公认的解决方案相当,在我的测试中大约慢了4%(尽管Java中的计时测试是出了名的不可靠)。
扫描与regex:
public int getLineCount() {
Scanner fileScanner = null;
int lineCount = 0;
Pattern lineEndPattern = Pattern.compile("(?m)$");
try {
fileScanner = new Scanner(new File(filename)).useDelimiter(lineEndPattern);
while (fileScanner.hasNext()) {
fileScanner.next();
++lineCount;
}
}catch(FileNotFoundException e) {
e.printStackTrace();
return lineCount;
}
fileScanner.close();
return lineCount;
}
还没记。