# Use all cv::Mat in std::vector and change their value efficiently

I have a vector of matrices with the same number of columns, i.e. `std::vector<cv::Mat> v`

and I want to apply a `cv::PCA`

(without dimensionality reduction) using all of them. Finally, I want that the result is applied to original matrices too.

Code speaking, this is the only solution that I came so far:

```
std::vector<cv::Mat> v;
//fill v
cv::Mat bigMat;
for(size_t i=0; i<v.size(); i++)
bigMat.push_back(v[i]);
cv::PCA pca(bigMat, cv::Mat(), CV_PCA_DATA_AS_ROW);
for(size_t i=0; i<v.size(); i++)
bigMat = pca.project(bigMat);
```

But this doesn't effect `v[i]`

. How can I do this efficiently? Notice that I have to use the projected version of `bigMat`

later. This is the only solution that came to my mind:

```
cv::PCA pca(bigMat, cv::Mat(), CV_PCA_DATA_AS_ROW);
bigMat.release();
for (size_t i=0; i<v.size(); i++){
v[i] = pca.project(v[i]);
bigMat.push_back(v[i]);
}
```

But this is kinda ugly and I don't know how efficient could be. Is there any better solution (where efficiency is the top priority)?

If you wander, `v[i]`

is the set of descriptors of the `i-th`

image. If you ask yourself why there is no dimensionality reduction in my PCA, give a look at this paper.

btw, your paper

doesmention dimensionality reduction (from 128 to 64)then, this is for sure wrong:

as you're recursively projecting the same Mat, while a plain:

would do, imho.

then, please clarify: why dou you need

boththe mat vector and bigMat ? this seems excessive.@berak thanks for your answer. The paper use multiple types of PCA, and some of them are about rotating the descriptors (without dimensionality reduction) to improve the encoding precision combined with power-law. I need

`bigMat`

because later I'm going to use it for k-means on the rotated descriptors using the vlfeat implementation, and so I pass as a pointer`bigMat.data()`

.ok, still, why do you need to have the projections in a

`vector<Mat>`

?you could project bigMat

once, and access each row vector like:@berak Thanks, following your tip I found an efficient workaround. Could you please give a look at this question?