Add a break iterator that returns indices into the string instead of tokens in the string

src/calibre/spell/break_iterator.py

@@ -22,6 +22,16 @@ def split_into_words(text, lang='en'):
         it.set_text(text)
         return it.split()
 
+
+def split_into_words_and_positions(text, lang='en'):
+    with _lock:
+        it = _iterators.get(lang, None)
+        if it is None:
+            it = _iterators[lang] = _icu.BreakIterator(_icu.UBRK_WORD, lang_as_iso639_1(lang) or lang)
+        it.set_text(text)
+        return it.split2()
+
+
 def index_of(needle, haystack, lang='en'):
     with _lock:
         it = _iterators.get(lang, None)

src/calibre/utils/icu.c

@@ -666,6 +666,45 @@ end:
 
 } // }}}
 
+// BreakIterator.split2 {{{
+static PyObject *
+icu_BreakIterator_split2(icu_BreakIterator *self, PyObject *args, PyObject *kwargs) {
+#if PY_VERSION_HEX >= 0x03030000 
+#error Not implemented for python >= 3.3
+#endif
+
+    int32_t prev = 0, p = 0, sz = 0;
+    PyObject *ans = NULL, *temp = NULL;
+  
+    ans = PyList_New(0);
+    if (ans == NULL) return PyErr_NoMemory();
+
+    p = ubrk_first(self->break_iterator);
+    while (p != UBRK_DONE) {
+        prev = p; p = ubrk_next(self->break_iterator);
+        if (self->type == UBRK_WORD && ubrk_getRuleStatus(self->break_iterator) == UBRK_WORD_NONE) 
+            continue;  // We are not at the start of a word
+        sz = (p == UBRK_DONE) ? self->text_len - prev : p - prev;
+        if (sz > 0) {
+#ifdef Py_UNICODE_WIDE
+            sz = u_countChar32(self->text + prev, sz);
+            prev = u_countChar32(self->text, prev);
+#endif
+            temp = Py_BuildValue("II", prev, sz); 
+            if (temp == NULL) {
+                Py_DECREF(ans); ans = NULL; break; 
+            } 
+            if (PyList_Append(ans, temp) != 0) {
+                Py_DECREF(temp); Py_DECREF(ans); ans = NULL; break; 
+            }
+            Py_DECREF(temp);
+        }
+    }
+
+    return ans;
+
+} // }}}
+
 static PyMethodDef icu_BreakIterator_methods[] = {
     {"set_text", (PyCFunction)icu_BreakIterator_set_text, METH_VARARGS,
      "set_text(unicode object) -> Set the text this iterator will operate on"
@@ -675,6 +714,10 @@ static PyMethodDef icu_BreakIterator_methods[] = {
      "split() -> Split the current text into tokens, returning a list of tokens"
     },
 
+    {"split2", (PyCFunction)icu_BreakIterator_split2, METH_VARARGS,
+     "split2() -> Split the current text into tokens, returning a list of 2-tuples of the form (position of token, length of token). The numbers are suitable for indexing python strings regardless of narrow/wide builds."
+    },
+
     {"index", (PyCFunction)icu_BreakIterator_index, METH_VARARGS,
      "index(token) -> Find the index of the first match for token. Useful to find, for example, words that could also be a part of a larger word. For example, index('i') in 'string i' will be 7 not 3. Returns -1 if not found."
     },

src/calibre/utils/icu_test.py

@@ -150,10 +150,11 @@ class TestICU(unittest.TestCase):
 
     def test_break_iterator(self):
         ' Test the break iterator '
-        from calibre.spell.break_iterator import split_into_words as split, index_of
+        from calibre.spell.break_iterator import split_into_words as split, index_of, split_into_words_and_positions
         for q in ('one two three', ' one two three', 'one\ntwo  three ', 'one-two,three'):
             self.ae(split(unicode(q)), ['one', 'two', 'three'], 'Failed to split: %r' % q)
         self.ae(split(u'I I\'m'), ['I', "I'm"])
+        self.ae(split_into_words_and_positions('one \U0001f431 three'), [(0, 3), (6 if sys.maxunicode >= 0x10ffff else 7, 5)])
         self.ae(0, index_of('i', 'i'))
         self.ae(4, index_of('i', 'six i'))
         self.ae(-1, index_of('i', ''))