How to rebuild exactly the file in which there were changes?

Question

JavaScript

Good afternoon.
Probably someone knows how to compile only the file in which changes have occurred, for example, there are 20 files in the directory with the pug extension, there is a watcher that watches the change in the directory, when one file is changed, everyone is recompiled, mine is some kind of nonsense, when there are a lot of files, hellish pain begins, is it possible to monitor the file (it doesn’t matter which one) and if it has changed, compile it with its includes and not all 20.
I use gulp to build

// Pug task
    gulp.task('pug', function buildHTML(cb) {
      return gulp.src('pug/*.pug')
      .pipe(pug())
      .pipe(gulp.dest('./dist/'));
    });

    // Watch task
    gulp.task('watch',function(){
      global.isWatching = true;

      // Modules pug
      watch('pug/*.pug', function() {
        return runSequence('pug', browserSync.reload);
      });
    });

Answer 1

In galpa 4 version out of the box, you can do this:

gulp.task( 'pug', function () {
  return gulp.src( 'pug/*.pug', { since: gulp.lastRun( 'pug' ) } )
  .pipe( pug() )
  .pipe( gulp.dest( './dist/' ) );
});

There is also gulp-newer .