Knowing the specific terms that you want to tell in a particular text, you can solve as follows:
Create a method that takes as arguments the text you want
and the list of elements that must be counted in the text. THE
idea focuses on:
1.1 Divide the text by the spaces contained in it (using the split function);
1.2 For each element resulting from the division of the text in point 1.1, check that it
it is part of the elements that are intended to be accounted for;
1.2.1 If there is, check if in HashMap there is already some element with this key
(in this case the element of the text);
1.2.1.1 If already exists, increment the value of the element on the map;
1.2.1.2 If not, insert this element into HashMap with the initial value of 1;
1.3 The% w / w will have, at the end, the amount in the text of each element.
Note: I recommend creating a method to check whether or not any element exists in the list of pre-determined elements to be counted.
Example:
//método fundamental (1.)
public void coontarElementosNumTexto(String texto, List<String> elementos)
{
HashMap<String,Integer> resMap = new HashMap();
String[] elementosTexto = texto.split(" ");
for (String var : elementosTexto )
{
if (existeElementoLista(var,params)){
if (resMap .get(var) != null){
resMap.put(var, resMap.get(var) + 1);
}
else{
resMap.put(var, 1);
}
}
}
System.out.println("O resultado final :\n" + resMap.toString());
}
The method I mentioned in the note might look like this:
//método para verificar se existe um elementos na lista dos elementos pré-definidos
private static boolean existe(String var, List<String> params) {
for (String param : params){
//Aqui eu ignoro se é maiúscula ou minúscula :)
if (var.equalsIgnoreCase(param))
return true;
}
return false;
}
Usage:
public static void main(String[] args) {
String a = "";
List<String> elementosContabeis = new ArrayList<>();
elementosContabeis.add("Stack");
elementosContabeis.add("Overflow");
elementosContabeis.add("Pt");
count("Finalmente o Stack Overflow em Pt foi lançado. Parabéns a todos que tornaram isso possível, poder ter uma versão em Pt é muito fixe!", elementosContabeis);
}
Result:
O resultado:
{Stack=1, Overflow=1, Pt=2}
