Flutter, Dart Split sentence every one character to 2 characters size, for flutter custom tts project - flutter

Question

Example:
var sentence = (hello there, I am Bob.)
var result = [
'he',
'el',
'll',
'lo',
' ',
'th',
'he',
'er',
're',
',',
' ',
'I',
' ',
'am',
' ',
'bo',
'ob',
'.']
I've found here working example, though it is in Javascript and I don't really know how to adopt it for Dart, and not sure how will behave once white space and punctuation is added in. Punctation and white space I need always split on its own not in combination with letters, I need them as well, as I will use them to add pauses in between words and sentences.
Thank you
var a = 12345678;
a= a.toString();
var arr=[];
for (var i =0; i<a.length-1; i++) {
arr.push(Number(a[i]+''+a[i+1]));
}
console.log(arr);

Answer 1

You could use regular expressions to split the sentence. For example:
void main() {
var exp = RegExp('([A-Za-z]{1,2}|[,!.?\s ])');
var str = "hello there, I am Bob.";
var matches = exp.allMatches(str);
for (var m in matches) {
print(m.group(0));
}
}
This looks for letters (A-Z or a-z) in groups of either 1 or 2, or single punctuation characters (,!.?) \s represents a white space.
Running the above would produce:
he
ll
o
th
er
e
,
I
am
Bo
b
.

Answer 2

Another approach
void main() {
var a = "1234!5678";
a = a.toString();
var arr = [];
for (var i = 0; i < a.length - 1; i++) {
if (a[i + 1] == '!') {
continue;
}
if (a[i] == '!') {
arr.add(a[i]);
continue;
}
arr.add(a[i] + '' + a[i + 1]);
}
print(arr);
}
I don't know dart much but I wrote this simple algorithm on dartpad and it works

Answer 3

If someone is having same issue, this is how I solved it
void main(String string) {
var test = "I Hello there I am Bob 23!";
List<String> nameArray = test.split('');
for (int curIndex = 0; curIndex < nameArray.length; curIndex++) {
if (curIndex >= 1 && nameArray[curIndex].contains(new RegExp(r'[a-zA-Z]')) && nameArray[curIndex-1].contains(new RegExp(r'[a-zA-Z]'))) {
print(nameArray[curIndex-1] + nameArray[curIndex]); // checks if current curIndex and previous curIndex are letters, if so returns previous and curent letters joined
} else {
if (curIndex >= 1 && nameArray[curIndex].contains(new RegExp(r'[a-zA-Z]')) && nameArray[curIndex+1].contains(new RegExp(r'[a-zA-Z]'))) {
null; // checks if curIndex and next curIndex are letters, if so returns null
}else{
print(nameArray[curIndex]);
}
}
}
}
Which returns
I
He
el
ll
lo
th
he
er
re
I
am
Bo
ob
2
3
!