Mongo: count the number of word occurrences in a set of documents
MapReduce might be a good fit that can process the documents on the server without doing manipulation on the client (as there isn't a feature to split a string on the DB server (open issue).
Start with the map
function. In the example below (which likely needs to be more robust), each document is passed to the map
function (as this
). The code looks for the summary
field and if it's there, lowercases it, splits on a space, and then emits a 1
for each word found.
var map = function() {
var summary = this.summary;
if (summary) {
// quick lowercase to normalize per your requirements
summary = summary.toLowerCase().split(" ");
for (var i = summary.length - 1; i >= 0; i--) {
// might want to remove punctuation, etc. here
if (summary[i]) { // make sure there's something
emit(summary[i], 1); // store a 1 for each word
}
}
}
};
Then, in the reduce
function, it sums all of the results found by the map
function and returns a discrete total for each word that was emit
ted above.
var reduce = function( key, values ) {
var count = 0;
values.forEach(function(v) {
count +=v;
});
return count;
}
Finally, execute the mapReduce:
> db.so.mapReduce(map, reduce, {out: "word_count"})
The results with your sample data:
> db.word_count.find().sort({value:-1})
{ "_id" : "is", "value" : 3 }
{ "_id" : "bad", "value" : 2 }
{ "_id" : "good", "value" : 2 }
{ "_id" : "this", "value" : 2 }
{ "_id" : "neither", "value" : 1 }
{ "_id" : "or", "value" : 1 }
{ "_id" : "something", "value" : 1 }
{ "_id" : "that", "value" : 1 }
A basic MapReduce example
var m = function() {
var words = this.summary.split(" ");
if (words) {
for(var i=0; i<words.length; i++) {
emit(words[i].toLowerCase(), 1);
}
}
}
var r = function(k, v) {
return v.length;
};
db.collection.mapReduce(
m, r, { out: { merge: "words_count" } }
)
This will insert word counts into a collection name words_count which you can sort (and index)
Note that it doesn't use stemming, omit punctuation, handles stop words etc.
Also note you can optimize the map function by accumulating repeating word(s) occurrences and emitting the count, not just 1