Indeed, you should transpose your input to have rows as data points and columns as features: [coeff, score, latent, ~, explained] = pca(M'); The principal components are given by the columns of coeff in order of descending variance, so the first column holds the most important component. The variances for...

matlab,octave,pca,missing-data,least-squares

Thanks for all your help. I went through the references and was able to find their matlab code on the als algorithm from two of the references. For anybody wondering, the source code can be found in these two links: 1) http://research.ics.aalto.fi/bayes/software/index.shtml 2) https://www.cs.nyu.edu/~roweis/code.html...

matlab,machine-learning,pca,face-recognition,reduction

The reason is that "eigs" calculates the eigenvalues of the matrix, which includes SQRT in it... and I have negative values in Sb,Sw

The eigenvectors are the columns of the rotation matrix that prcomp returns. In order to rotate another data matrix, you just need to multiply it with the rotation matrix, and optionally scale it beforehand. In your case: result = scale(iris[, 1:4]) %*% pca1$rotation You can verify that this works using...

This behaviour is admittedly potentially weird, but it is nevertheless documented in the docstrings of the relevant functions. The class docstring of PCA says the following about whiten: whiten : bool, optional When True (False by default) the `components_` vectors are divided by n_samples times singular values to ensure uncorrelated...

c++,opengl,matrix,pca,eigenvector

I believe I found your problem. You need to check again the theory! As I may recall, you have the covariance defined in the theory as: C=1/M \Sum ( (p-pmean)*(p-pmean)^t ) Well, you may notice that C is a 3x3 matrix, NOT a value. Therefore, when you call Compute_EigenV and...

I suspect your reshape is wrong... When you read the image, it returns an 512*512*3 array I. When you reshape it with reshape(I,[],3), it becomes an 262144*3 array x. Now x*x' would yield an 262144*262144 array, which is too large for your memory. EDIT: Apparently this is the correct procedure...

scikit-learn,ipython-notebook,pca

It seems that sklearn will not run nicely on a 32 bit machine so when I ran this later on a 64 bit server it worked!!!!!

It's hard to say without knowing exactly what your data look like, but perhaps something like this would work: cols <- rainbow(17)[as.factor(gtex_pm$tissue)] plot(pc_gtex$x[,1], pc_gtex$x[,2], col=cols, main = "PCA", xlab = "PC1", ylab = "PC2") ...

Define a list of colors: col.list <- c("gray", "blue", "green", "red", "blue", "yellow", ...) plot(tab$EV2, tab$EV1, col=col.list[as.integer(tab$pop)], cex=1.2, pch=20, xlab="eigenvector 2", ylab="eigenvector 1") legend("topleft", legend=levels(tab$pop), cex=1,pch=20, col=1:nlevels(tab$pop)) ...

python,pandas,scikit-learn,pca,principal-components

You can try this. import pandas as pd import matplotlib.pyplot as plt from sklearn import decomposition # use your data file path here demo_df = pd.read_csv(file_path) demo_df.set_index('Unnamed: 0', inplace=True) target_names = demo_df.index.values tran_ne = demo_df.values pca = decomposition.PCA(n_components=4) pcomp = pca.fit_transform(tran_ne) pcomp1 = pcomp[:,0] fig, ax = plt.subplots() ax.scatter(x=pcomp1[0], y=0,...

The problem you encountered is because you have specified all of your variables as supplementary variables when you call PCA(). To illustrate with an example we can use the built in dataset USJudgeRatings. head(USJudgeRatings) CONT INTG DMNR DILG CFMG DECI PREP FAMI ORAL WRIT PHYS RTEN AARONSON,L.H. 5.7 7.9 7.7...

Let us recap what you are asking, to clarify : Find an eigenvector of a matrix mat This eigenvector should be associated with the largest eigenvalue of the matrix The matrix is the symmetric covariance matrix of a principal component analysis. In particular, it is symmetric. Your matrix is square...

axis,pca,point-cloud-library,ros,point-clouds

Orientations are represented by Quaternions in ROS, not by directional vectors. Quaternions can be a bit unintuitive, but fortunately there are some helper functions in the tf package, to generate quaternions, for example, from roll/pitch/yaw-angles. One way to fix the marker would therefore be, to convert the direction vector into...

r,pca,rotational-matrices,psych

I'd give manipulate a try - something in the veins of: library(psych) library(manipulate) l <- l_orig <- unclass(loadings(principal(Harman.5, 2, scores=TRUE))) manipulate( { if(rotateRight) l <<- factor.rotate(l, angle, 1, 2) if (rotateLeft) l <<- factor.rotate(l, -1*angle, 1, 2) plot(l, xlim = c(-1, 1), ylim = c(-1, 1), xlab = 1, ylab...

I think the problem is rather a PCA problem than an R problem. You multiply the original data with the rotation matrix and you wonder then why newData!=data. This would be only the case if the rotation matrix would be the identity matrix. What you probably were planning to do...

r,data-mining,text-mining,pca,text-extraction

I think that task is not for PCA. I would first try to introduce some kind of distance measure between 2 addresses. You can either use entire address as a single feature - then there're plenty of general-purpose string similarity measures, for example Levenshtein distance. There's a method in utils...

I think you should try theme(legend.position="none"). library(factoextra) plot(fviz_pca_biplot(pca, label="var", habillage=as.factor(kc$cluster)) + ggtitle("") + theme(text = element_text(size = 15), panel.background = element_blank(), panel.grid.major = element_blank(), panel.grid.minor = element_blank(), axis.line = element_line(colour = "black"), legend.position="none")) This is what I get: ...

matlab,machine-learning,data-mining,projection,pca

Oops, now I see my mistake. First of all, COEFF is orthogonal (not sure) so inv(COEFF) == COEFF' and the projection is found by proj = COEFF' * (x-m) ...

loadings(pca1) returns the PCA Loadings. unclass drops the class and converts it into a matrix. pca1$sdev^2 > 1 returns TRUE for columns where the eigenvalue > 1. [...,drop = F] selects the columns where the index is equals to TRUE while keeping the matrix structure even when only one column...

Principal Components do not necessarily have any correlation to classification accuracy. There could be a 2-variable situation where 99% of the variance corresponds to the first PC but that PC has no relation to the underlying classes in the data. Whereas the second PC (which only contributes to 1% of...

r,covariance,pca,eigenvalue,princomp

You explicitly told princomp to use correlation matrix in this line: test<-princomp(data,cor=T) If you omit the parameter and just use test <- printcomp(data), it will use covariance matrix and you'll get the results(roughly) you're expecting....

Adapting a previous answer, you can do perp.segment.coord <- function(x0, y0, a=0,b=1){ #finds endpoint for a perpendicular segment from the point (x0,y0) to the line # defined by lm.mod as y=a+b*x x1 <- (x0+b*y0-a*b)/(1+b^2) y1 <- a + b*x1 list(x0=x0, y0=y0, x1=x1, y1=y1) } ss<-perp.segment.coord(df$Person1, df$Person2,0,eigen$vectors.scaled[1,1]) g + geom_segment(data=as.data.frame(ss), aes(x...

this: cv::PCA pca(dst, cv::Mat(), CV_PCA_DATA_AS_ROW, 2); expects a cv::Mat with continuous data as input, which is not satisfied by a std::vector<std::vector<double> > tmpVec; try like: cv::Mat tmp; while(std::getline(file, numStream)) { ... cv::Mat m = cv::Mat(line).t(); // we need a row-vec tmp.push_back(m); } cv::PCA pca(tmp, cv::Mat(), CV_PCA_DATA_AS_ROW, 2); ...

python,python-2.7,numpy,scikit-learn,pca

You can simply drop the axis you want to keep from the data: mask = np.ones(data.shape[1], dtype=np.bool) mask[special_axis] = False data_new = data[:, mask] pca_transformed = PCA(n_components=14).fit_transform(data_new) This is the same as removing the projection along this feature. You can then stack the original axis with the PCA result if...

You could use bootstrapping on this. Simply re-sample your data with the bootstrapping package and record the principal components computed every time. Use the resulting empirical distribution to get your confidence intervals. The boot package makes this pretty easy. Here is an example calculating the Confidence Interval at 95% for...

machine-learning,cluster-analysis,pca,eigenvalue,eigenvector

As far as I can tell, you have mixed and shuffled aa number of approaches. No wonder it doesn't work... you could simply use jaccard distance (a simple inversion of jaccard similarity) + hierachical clustering you could do MDS to project you data, then k-means (probably what you are trying...

neural-network,classification,pca,pattern-recognition

Question: How many principal components should I use in pattern classification? Answer: As low as possible. When you apply PCA, you get number of principal components according to your data. Lets say you get 10 principal components from your data. You will control how much your variance are explained with...

Looking at car:::scatter3d.default shows that the coordinates are internally scaled by the min and max of each dimension; the following code scales before plotting: sc <- function(x,orig) { d <- diff(range(orig)) m <- min(orig) (x-m)/d } msc <- function(x) { sc(mean(x),x) } points3d(msc(data3d[,1]), msc(data3d[,2]), msc(data3d[,3]), col="red", size=20) ...

From what I see the only difference between the algorithms you list is the normalization by the standard deviation. It is a standard practice which ensures that values having different "range" are re-scaled to similar range. If your data is similarly scaled, this step is not strictly necessary. You can...

you will have to collect feature vectors from a lot of images, make a single pca from that (offline), and later use the mean & eigenvectors for the projection. // let's say, you have collected 10 feature vectors a 30 elements. // flatten them to a single row (reshape(1,1)) and...

python,machine-learning,scikit-learn,pca,feature-selection

The features that your PCA object has determined during fitting are in pca.components_. The vector space orthogonal to the one spanned by pca.components_ is discarded. Please note that PCA does not "discard" or "retain" any of your pre-defined features (encoded by the columns you specify). It mixes all of them...

I am not using Python, but I did something you need in C++ & opencv. Hope you succeed in converting it to whatever language. // choose how many eigenvectors you want: int nEigensOfInterest = 0; float sum = 0.0; for (int i = 0; i < mEiVal.rows; ++i) { sum...

eigen,face-recognition,pca,eigenvector

The accuracy would depend on the classifier you are using once you have the data in the PCA projected space. In the original Turk/Pentland eigenface paper http://www.face-rec.org/algorithms/PCA/jcn.pdf they just use kNN / Euclidean distance but a modern implementation might use SVMs e.g. with an rbf kernel as the classifier in...

KMeans().predict(X) ..docs here Predict the closest cluster each sample in X belongs to. In the vector quantization literature, cluster_centers_ is called the code book and each value returned by predict is the index of the closest code in the code book. Parameters: (New data to predict) X : {array-like, sparse...

I'm not sure how to do that with biplot, but if you work with the raw PCA output you can do essentially anything you want. Maybe something like: data <- replicate(100, rnorm(100)) pca <- prcomp(data) raw <- pca$x[,1:2] plot(raw[,1], raw[,2], col=rainbow(nrow(raw)), pch=20) ...will give you what you're looking for. Notice...

You can just choose some threshold value and replace cells: > threshold <- 0.2 > dd[abs(dd ) < threshold] <- NA > dd Comp.1 Comp.2 Comp.3 Comp.4 Sepal.Length 0.3613866 -0.6565888 -0.5820299 0.3154872 Sepal.Width NA -0.7301614 0.5979108 -0.3197231 Petal.Length 0.8566706 NA NA -0.4798390 Petal.Width 0.3582892 NA 0.5458314 0.7536574 ...

Yes, according to the pca help, "Rows of X correspond to observations and columns to variables." score just tells you the representation of M in the principal component space. You want the first column of coeff. numberOfDimensions = 5; coeff = pca(A); reducedDimension = coeff(:,1:numberOfDimensions); reducedData = A *...

Yes, as explained in the documentation, what normalize does, is scaling individual samples, independently to others: Normalization is the process of scaling individual samples to have unit norm. This is additionally explained in the documentation of the Normalizer class: Each sample (i.e. each row of the data matrix) with at...

Probably, you are running this from a wrong directory (i.e. you are calling something that MATLAB just can't find, so the problem is not the input).

python,matplotlib,scipy,scikit-learn,pca

import pandas as pd import numpy as np import matplotlib.pyplot as plt from sklearn.decomposition import PCA from sklearn.cluster import KMeans # read your data, replace 'stackoverflow.csv' with your file path df = pd.read_csv('stackoverflow.csv', usecols=[0, 2, 4], names=['freq', 'visit_length', 'conversion_cnt'], header=0).dropna() df.describe() Out[3]: freq visit_length conversion_cnt count 289705.0000 289705.0000 289705.0000 mean...

Pass the second, optional parameter to eigs, which controls how many eigenvectors are returned.

One possibility: Use xlim and ylim fuctnions library(ggbiplot) data(wine) wine.pca <- prcomp(wine, scale. = TRUE) p <- ggbiplot(wine.pca, obs.scale = 1, var.scale = 1, group=wine.class, varname.size = 8, labels.size=10, ellipse = TRUE, circle = TRUE) + scale_color_discrete(name = '') + geom_point(aes(colour=wine.class), size = 8) + theme(legend.direction ='horizontal', legend.position = 'top')...

The explained tells you how accurately you could represent the data by just using that principal component. In your case it means that just using the main principal component, you can describe very accurately (to a 99%) the data. Lets make a 2D example. Imagine you have data that is...

matlab,pca,dimensionality-reduction

You should take advantage of the built-in functions in Matlab, and use the pca function directly, or even the cov function if you want to compare eigs to pcaconv. Now to answer your question, both return the same eigenvectors but not in the same order. See the following example: >>...

r,machine-learning,data-mining,pca

The number of principal components can never exceed the number of samples. Perhaps too simply put, since you only have 5 samples you only need 5 variables to explain the variability.

python,machine-learning,scikit-learn,classification,pca

After you trained your LDA model with some data X, you may want to project some other data, Z. in this case what you should do is: lda = LDA(n_components=2) #creating a LDA object lda = lda.fit(X, y) #learning the projection matrix X_lda = lda.transform(X) #using the model to project...

Think of PCA as a transformation you apply to your data. You want two things to hold: Since the test set mimics a "real-world" situation where you get samples you didn't see before, you cannot use the test set for anything but evaluation of the classifier. You need to apply...

Yes, it holds that score = norm_ingredients * pc, where norm_ingredients is the normalized version of your input matrix so that its columns have zero mean, that is, norm_ingredients = ingredients - repmat(mean(ingredients), size(ingredients, 1), 1) ...

This was partly answered, but since it is my package, I will give a somewhat more complete answer. The summary table of the PCA or FA factor loadings tables is calculated in the print function. It it is returned (invisibly by print). However, it is available as the Vaccounted object....

I use the following function for reconstructing data from a prcomp object: #This function reconstructs a data set using a defined set of principal components. #arguments "pca" is the pca object from prcomp, "pcs" is a vector of principal components #to be used for reconstruction (default includes all pcs) prcomp.recon...

Is this what you mean, or am I misunderstanding: loadings <- data.frame(ir.pca$rotation, .names = row.names(ir.pca$rotation), names2 = c("(+)C" , "(-)C", "(*)C", "(%)C")) p + geom_text(data=loadings, mapping=aes(x = PC1, y = PC2, label = names2, colour = .names)) + coord_fixed(ratio=1) + labs(x = "PC1", y = "PC2", colour="Legend Title") UPDATE: Here's...

You can try this: pcaResult<-princomp(data) pc=pcaResult$scores ...