Rachel Rachel - 5 months ago 23
Java Question

HashMap : Adding values with common keys and printing them out

I have file which has String in the form

pair like people and count, example would be

"Reggy, 15"
"Jenny, 20"
"Reggy, 4"
"Jenny, 5"

and in the output I should have summed up all count values based on key so for our example output would be

"Reggy, 19"
"Jenny, 25"

Here is my approach:

  1. Read each line and for each line get key and count using scanner and having
    as delimiter

  2. Now see if key is already present before if then just add currentValues to previousValues if not then take currentValue as value of HashMap.

Sample Implementation:

public static void main(final String[] argv) {
final File file = new File("C:\\Users\\rachel\\Desktop\\keyCount.txt");

try {
final Scanner scanner = new Scanner(file);

while (scanner.hasNextLine()) {
if (scanner.hasNext(".*,")) {
String key;
final String value;

key = scanner.next(".*,").trim();

if (!(scanner.hasNext())) {
// pick a better exception to throw
throw new Error("Missing value for key: " + key);

key = key.substring(0, key.length() - 1);
value = scanner.next();

System.out.println("key = " + key + " value = " + value);
} catch (final FileNotFoundException ex) {

Part I am not clear about is how to divide key/value pair while reading them in and creating HashMap based on that.

Also is the approach am suggestion an optimal one or is there a way to enhance the performance more.

Answer Source

Since this is almost certainly a learning exercise, I'll stay away from writing code, letting you have all the fun.

Create a HashMap<String,Integer>. Every time that you see a key/value pair, check if the hash map has a value for the key (use 'containsKey(key)'). If it does, get that old value using get(key), add the new value, and store the result back using put(key, newValue). If the key is not there yet, add a new one - again, using put. Don't forget to make an int out if the String value (use Integer.valueOf(value) for that).

As far as optimizing goes, any optimization at this point would be premature: it does not even work! However, it's hard to get much faster than a single loop that you have, which is also rather straightforward.