I'm doing some very basic Linear Algebra and I'm probably missing the point completely here.
Let's say I have the following matrices:
v1 = [5, 8]
v2 = [3, 4]
v3 = [4, 4]
v4 = [2, 1]
Expected output:
M1 = [5 - 3, 8 - 4] = [2, 4]
M2 = [4 - 2, 4 - 1] = [2, 3]
Actual output:
0 0
0 0
2 4
-1 0
2 3
Here is the code:
std::vector<double> calculate(std::vector<double> values, std::vector<double> values2)
{
std::vector<double> vel(2, 0);
for(unsigned i=0; (i < values.size()); i++)
{
vel[i] = values[i] - values2[i];
}
return vel;
}
std::vector<std::vector<double> > values = { {5,8}, {3, 4}, {4, 4}, {2, 1}};
std::vector<std::vector<double> > v;
v.resize(2);
for(unsigned i=0; (i < values.size()-1); i++)
{
v[i].resize(2);
v.push_back(calculate(values[i], values[i + 1]));
//v[i] = calculate(values[i], values[i + 1]);
}
for(unsigned i=0; (i < v.size()); i++)
{
for(unsigned j=0; (j < v[i].size()); j++)
{
std::cout << v[i][j] << " ";
}
std::cout << std::endl;
}
The problem being is that, the following should iterate over 4 times, calculating the 4 matrices, and the final resulting 2D vector should only contain 2 values.
I'm probably missing something stupid.
v.resize(2); // now it contains `{{} {}}`.
for(unsigned i=0; (i < values.size()-1); i++) //for each input except the last (3 iterations)
//starting with {5,8} and {3,4}
v[i].resize(2); //resize one of the vectors already in v to 2
//now v contains {{0,0}, {}
v.push_back(calculate(values[i], values[i + 1])); //push back the calculations
//now v contains {{0,0}, {}, {2,4}}
for(............. (i < values.size()-1); i++)
//next the middle pair of inputs {3,4} and {4,4}
v[i].resize(2); //resize one of the vectors already in v to 2
//now v contains {{0,0}, {0,0}, {2,4}}
v.push_back(calculate(values[i], values[i + 1])); //push back the calculations
//now v contains {{0,0}, {0,0}, {2,4}, {-1,0}}
for(............. (i < values.size()-1); i++)
//finally the last pair of inputs {4,4} and {2,1}
v[i].resize(2); //resize the {2,4} to 2, but it was already two
//now v contains {{0,0}, {0,0}, {2,4}, {-1,0}}
v.push_back(calculate(values[i], values[i + 1])); //push back the calculations
//now v contains {{0,0}, {0,0}, {2,4}, {-1,0}, {2,3}}
for(............. (i < values.size()-1); ....) //done iterating
Do you have a debugger? It's really important to learn how to step through code like this, have someone show you, or find a tutorial. Stepping through this code in a debugger would have made it obvious what was going on to you.
Luckily, the code is REALLY easy to fix:
std::vector<std::vector<double> > v;
for(unsigned i=0; i<values.size()-1; i+=2) //NOTE: i+=2!!!
{
v.push_back(calculate(values[i], values[i + 1]));
}
for(unsigned i=0; (i < values.size()-1); i++)
{
v[i].resize(2);
v.push_back(calculate(values[i], values[i + 1]));
//v[i] = calculate(values[i], values[i + 1]);
}
This loop operates on two elements of values
, so the counter needs to increase by 2 per iteration. otherwise you would subtract v1-v2
, v2-v3
, v3-v4
and so on. Also there is no need to determine the vector size prior to the push_back (and it actually dangerous, because the index is likely out-of-bounds before the element is pushed back).
for(unsigned i=0; (i < values.size()-1); i+=2)
{
//v[i].resize(2);
v.push_back(calculate(values[i], values[i + 1]));
//v[i] = calculate(values[i], values[i + 1]);
}
As pointed out by others v.resize(2);
shouldn't be there either because it just adds two empty elements initially, which you don't want.
v.resize(2);
I don't know what you were expecting the above to do, but this (coupled with the code below) is what is responsible for those first two lines of output.
for(unsigned i=0; (i < values.size()-1); i++)
{
v[i].resize(2);
I don't know what you were expecting the above to do, but this (coupled with the code above) is what is responsible for those first two lines of output.
Here's what's happening: Your first resize (the resize outside the loop) populates v
with two empty vectors. The first two passes through the loop populate v[0]
and v[1]
as vectors of doubles with two elements that are both zero.
v.push_back(calculate(values[i], values[i + 1]));
This will ultimately add three more elements to v
, one being {5,8}-{3, 4}={2,4}, the next being {3,4}-{4,4}={-1,0}, and the last being {4,4}-{2,1}={2,3}. So your output should be
0 0
0 0
2 4
-1 0
2 3
To get rid of those first two lines, simply delete the two calls to resize
. To get rid of the penultimate line (that -1 0
that is between the two desired lines of output), change your loop increment from one ( i++
) to two ( i += 2
).
This output
0 0
0 0
2 4
-1 0
2 3
can be explained very simply
At first using method resize
v.resize(2);
you added two empty std::vector in vector v.
Then in loop
for(unsigned i=0; (i < values.size()-1); i++)
{
v[i].resize(2);
v.push_back(calculate(values[i], values[i + 1]));
//v[i] = calculate(values[i], values[i + 1]);
}
you resized each of them. So now v[0] and v[1] are
0 0
0 0
And then you get subtraction of
values[0] - values[1]
values[1] - values[2]
values[2] - values[3]
After that vector v was appended with
2 4
-1 0
2 3
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