Thread: word count problem

I think what you want to do is something like this:

Code:

#include <iostream>
#include <string>
#include <vector>

using namespace std;

int main() {
	vector<string> eg;
	string data;

	cout << "Enter all words then hit ctrl-D:\n";
	while (!cin.eof()) {
		cin >> data;
		eg.push_back(data);
	}

	int i, len = (int)eg.size();
	cout << endl;
	for (i=0;i<len;i++) {
		cout << (i+1) << ": " << eg[i] <<endl;
	}

	return 0;
}

Well, here is the output from that:

[root~/C++] ./a.out
Enter all words then hit ctrl-D:
one two and then
some more
etc

1: one
2: two
3: and
4: then
5: some
6: more
7: etc
8: etc

As you can see, for some reason "etc" is repeated twice. I'm pretty new to C++ too! I can maybe guess why that repetition is there tho; anyway, changing the while loop to this fixes that problem:

Code:

	cout << "Enter all words then hit ctrl-D:\n";
	while (cin >> data) {
		eg.push_back(data);
		if (cin.eof()) break;
	}

I'm not trying to spoil anything for you roelof, just you seem to be having a lot of trouble with this and I thought a working example might be helpful.

Looks good, the way words_count gets incremented is clever. Here's a way you could optimize it slightly by losing the "found" and "place" variables, so there is slightly less copying and testing:

Code:

    while (cin >> x )
    {
        size = words.size ();
        for (teller = 0; teller < size; teller ++)
        {
            if (x == words [teller])
            {
				words_count [teller] +=1 ;
				teller = -1;
				break;
            }
        }
		if (teller < 0) continue;
		words.push_back (x) ;
		words_count.push_back  (1) ;
    }

"Break" ends a loop (since it's done at that point) and "continue" proceeds to the next iteration, skipping the remaining code in the loop.

Yes, but the idea is that if teller is negative, then the word already exists in the vector, so you do not want to push it into the vector. Actually, instead of assigning -1 to teller, it would be simpler to make use of the fact that if teller == size, then the word was not found in the vector, hence you would replace the part of MK27's example that uses continue with:

Code:

if (teller == size)
{
    words.push_back(x);
    words_count.push_back(1);
}

By the way, later you will learn that it is easier to use a std::map container to solve this problem, but for now this is good practice

Originally Posted by laserlight

Actually, instead of assigning -1 to teller, it would be simpler to make use of the fact that if teller == size, then the word was not found in the vector, hence you would replace the part of MK27's example that uses continue with:

Correct. For some reason I was thinking that if the match were to the last element teller would == size anyway but of course the break prevents that.

Originally Posted by laserlight

By the way, later you will learn that it is easier to use a std::map container to solve this problem, but for now this is good practice

That's what I would have done, but I think in fact roelof's solution is more elegant and definitely more efficient -- sometimes less is more, syntax usage wise. A map would be a waste here.

Oke,

Maybe I could use a map but the book from which I try to learn c++ has not talked about maps.
So Im not allowed to use maps.

If I could maps , i would map.find to find out if a name is in the vector.

Roelof

Originally Posted by roelof

Maybe I could use a map but the book from which I try to learn c++ has not talked about maps.
So Im not allowed to use maps.

Yes, do not use them yet. It is just something to keep in mind for the future.

Originally Posted by MK27

I think in fact roelof's solution is more elegant and definitely more efficient

No, roelof's solution is less efficient in the long run: it performs a linear search of the existing words for each word, compared to the O(log n) search of the map for each word. As for elegant: that depends on your point of view, but the map solution is certainly more concise, even if one makes use of stuff like std::find instead of manually implementing the linear search.

Originally Posted by laserlight

No, roelof's solution is less efficient in the long run: it performs a linear search of the existing words for each word, compared to the O(log n) search of the map for each word. As for elegant: that depends on your point of view, but the map solution is certainly more concise, even if one makes use of stuff like std::find instead of manually implementing the linear search.

Okay, I thought you meant simply for the count, but obviously you would only need a single map, not a vector + map there. Silly me again!