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Correct way of normalizing and scaling the MNIST dataset

I've looked everywhere but couldn't quite find what I want. Basically the MNIST dataset has images with pixel values in the range [0, 255]. People say that in general, it is good to do the following:
Scale the data to the [0,1] range.
Normalize the data to have zero mean and unit standard deviation (data - mean) / std.
Unfortunately, no one ever shows how to do both of these things. They all subtract a mean of 0.1307 and divide by a standard deviation of 0.3081. These values are basically the mean and the standard deviation of the dataset divided by 255:
from torchvision.datasets import MNIST
import torchvision.transforms as transforms
trainset = torchvision.datasets.MNIST(root='./data', train=True, download=True)
print('Min Pixel Value: {} \nMax Pixel Value: {}'.format(trainset.data.min(), trainset.data.max()))
print('Mean Pixel Value {} \nPixel Values Std: {}'.format(trainset.data.float().mean(), trainset.data.float().std()))
print('Scaled Mean Pixel Value {} \nScaled Pixel Values Std: {}'.format(trainset.data.float().mean() / 255, trainset.data.float().std() / 255))
This outputs the following
Min Pixel Value: 0
Max Pixel Value: 255
Mean Pixel Value 33.31002426147461
Pixel Values Std: 78.56748962402344
Scaled Mean: 0.13062754273414612
Scaled Std: 0.30810779333114624
However clearly this does none of the above! The resulting data 1) will not be between [0, 1] and will not have mean 0 or std 1. In fact this is what we are doing:
[data - (mean / 255)] / (std / 255)
which is very different from this
[(scaled_data) - (mean/255)] / (std/255)
where scaled_data is just data / 255.

Euler_Salter
I may have stumbled upon this a little too late, but hopefully I can help a little bit.
Assuming that you are using torchvision.Transform, the following code can be used to normalize the MNIST dataset.
train_loader = torch.utils.data.DataLoader(
datasets.MNIST('./data', train=True
transform=transforms.Compose([
transforms.ToTensor(),
transforms.Normalize((0.1307,), (0.3081,))
])),
Usually, 'transforms.ToTensor()' is used to turn the input data in the range of [0,255] to a 3-dimensional Tensor. This function automatically scales the input data to the range of [0,1]. (This is equivalent to scaling the data down to 0,1)
Therefore, it makes sense that the mean and std used in the 'transforms.Normalize(...)' will be 0.1307 and 0.3081, respectively. (This is equivalent to normalizing zero mean and unit standard deviation.)
Please refer to the link below for better explanation.
https://pytorch.org/vision/stable/transforms.html

I think you misunderstand one critical concept: these are two different, and inconsistent, scaling operations. You can have only one of the two:
mean = 0, stdev = 1
data range [0,1]
Think about it, considering the [0,1] range: if the data are all small positive values, with min=0 and max=1, then the sum of the data must be positive, giving a positive, non-zero mean. Similarly, the stdev cannot be 1 when none of the data can possibly be as much as 1.0 different from the mean.
Conversely, if you have mean=0, then some of the data must be negative.
You use only one of the two transformations. Which one you use depends on the characteristics of your data set, and -- ultimately -- which one works better for your model.
For the [0,1] scaling, you simply divide by 255.
For the mean=0, stdev=1 scaling, you perform the simple linear transformation you already know:
new_val = (old_val - old_mean) / old_stdev
Does that clarify it for you, or have I entirely missed your point of confusion?

Purpose
Two of the most important reasons for features scaling are:
You scale features to make them all of the same magnitude (i.e. importance or weight).
Example:
Dataset with two features: Age and Weight. The ages in years and the weights in grams! Now a fella in the 20th of his age and weights only 60Kg would translate to a vector = [20 yrs, 60000g], and so on for the whole dataset. The Weight Attribute will dominate during the training process. How is that, depends on the type of the algorithm you are using - Some are more sensitive than others: E.g. Neural Network where the Learning Rate for Gradient Descent get affected by the magnitude of the Neural Network Thetas (i.e. Weights), and the latter varies in correlation to the input (i.e. features) during the training process; also Feature Scaling improves Convergence. Another example is the K-Mean Clustering Algorithm requires Features of the same magnitude since it is isotropic in all directions of space. INTERESTING LIST.
You scale features to speed up execution time.
This is straightforward: All these matrices multiplications and parameters summation would be faster with small numbers compared to very large number (or very large number produced from multiplying features by some other parameters..etc)
Types
The most popular types of Feature Scalers can be summarized as follows:
StandardScaler: usually your first option, it's very commonly used. It works via standardizing the data (i.e. centering them), that's to bring them to a STD=1 and Mean=0. It gets affected by outliers, and should only be used if your data have Gaussian-Like Distribution.
MinMaxScaler: usually used when you want to bring all your data point into a specific range (e.g. [0-1]). It heavily gets affected by outliers simply because it uses the Range.
RobustScaler: It's "robust" against outliers because it scales the data according to the quantile range. However, you should know that outliers will still exist in the scaled data.
MaxAbsScaler: Mainly used for sparse data.
Unit Normalization: It basically scales the vector for each sample to have unit norm, independently of the distribution of the samples.
Which One & How Many
You need to get to know your dataset first. As per mentioned above, there are things you need to look at before, such as: the Distribution of the Data, the Existence of Outliers, and the Algorithm being utilized.
Anyhow, you need one scaler per dataset, unless there is a specific requirement, such that if there exist an algorithm that works only if data are within certain range and has mean of zero and standard deviation of 1 - all together. Nevertheless, I have never come across such case.
Key Takeaways
There are different types of Feature Scalers that are used based on some rules of thumb mentioned above.
You pick one Scaler based on the requirements, not randomly.
You scale data for a purpose, for example, in the Random Forest Algorithm you do NOT usually need to scale.

Well the data gets scaled to [0,1] using torchvision.transforms.ToTensor() and then the normalization (0.1306,0.3081) is applied.
You can look about it in the Pytorch documentation : https://pytorch.org/vision/stable/transforms.html.
Hope that answers your question.

partially define initial centroid for scikit-learn K-Means clustering

Scikit documentation states that:
Method for initialization:
‘k-means++’ : selects initial cluster centers for k-mean clustering in a smart way to speed up convergence. See section Notes in k_init for more details.
If an ndarray is passed, it should be of shape (n_clusters, n_features) and gives the initial centers.
My data has 10 (predicted) clusters and 7 features. However, I would like to pass array of 10 by 6 shape, i.e. I want 6 dimensions of centroid of be predefined by me, but 7th dimension to be iterated freely using k-mean++.(In another word, I do not want to specify initial centroid, but rather control 6 dimension and only leave one dimension to vary for initial cluster)
I tried to pass 10x6 dimension, in hope it would work, but it just throw up the error.

Sklearn does not allow you to perform this kind of fine operations.
The only possibility is to provide a 7th feature value that is random, or similar to what Kmeans++ would have achieved.
So basically you can estimate a good value for this as follows:
import numpy as np
from sklearn.cluster import KMeans
nb_clust = 10
# your data
X = np.random.randn(7*1000).reshape( (1000,7) )
# your 6col centroids
cent_6cols = np.random.randn(6*nb_clust).reshape( (nb_clust,6) )
# artificially fix your centroids
km = KMeans( n_clusters=10 )
km.cluster_centers_ = cent_6cols
# find the points laying on each cluster given your initialization
initial_prediction = km.predict(X[:,0:6])
# For the 7th column you'll provide the average value
# of the points laying on the cluster given by your partial centroids
cent_7cols = np.zeros( (nb_clust,7) )
cent_7cols[:,0:6] = cent_6cols
for i in range(nb_clust):
init_7th = X[ np.where( initial_prediction == i ), 6].mean()
cent_7cols[i,6] = init_7th
# now you have initialized the 7th column with a Kmeans ++ alike
# So now you can use the cent_7cols as your centroids
truekm = KMeans( n_clusters=10, init=cent_7cols )

That is a very nonstandard variation of k-means. So you cannot expect sklearn to be prepared for every exotic variation. That would make sklearn slower for everybody else.
In fact, your approach is more like certain regression approaches (predicting the last value of the cluster centers) rather than clustering. I also doubt the results will be much better than simply setting the last value to the average of all points assigned to the cluster center using the other 6 dimensions only. Try partitioning your data based on the nearest center (ignoring the last column) and then setting the last column to be the arithmetic mean of the assigned data.
However, sklearn is open source.
So get the source code, and modify k-means. Initialize the last component randomly, and while running k-means only update the last column. It's easy to modify it this way - but it's very hard to design an efficient API to allow such customizations through trivial parameters - use the source code to customize at his level.

Sklearn LDA-analysis won't generate 2 dimensions

I'm trying to plot a 3-feature dataset with a binary classification on a matplotlib plot. This worked with an example dataset provided in a guide (http://www.apnorton.com/blog/2016/12/19/Visualizing-Multidimensional-Data-in-Python/) but when I try to instead insert my own dataset, the LinearDiscriminantAnalysis will only output a one-dimensional series, no matter what number I put in "n_components". Why would this not work with my own code?
Data = pd.read_csv("DataFrame.csv", sep=";")
x = Data.iloc[:, [3, 5, 7]]
y = Data.iloc[:, 8]
lda = LDA(n_components=2)
lda_transformed = pd.DataFrame(lda.fit_transform(x, y))
plt.scatter(lda_transformed[y==0][0], lda_transformed[y==0][1], label='Loss', c='red')
plt.scatter(lda_transformed[y==1][0], lda_transformed[y==1][1], label='Win', c='blue')
plt.legend()
plt.show()

In the case when the number of different class labels, C, is less than the number of observations (almost always), then linear discriminant analysis will always produce C - 1 discriminating components. Using n_components from the sklearn API is only a means to choose possibly fewer components, e.g. in the case when you know what dimensionality you'd like to reduce down to. But you could never use n_components to get more components.
This is discussed in the Wikipedia section on Multiclass LDA. The definition of the between-class scatter is given as
\Sigma_{b} = (1 / C) \sum_{i}^{C}( (\mu_{i} - mu)(\mu_{i} - mu)^{T}
which is the empirical covariance matrix among the population of class means. By definition, such a covariance matrix has rank at most C - 1.
... the variability between features will be contained in the subspace spanned by the eigenvectors corresponding to the C − 1 largest eigenvalues ...
So because LDA uses a decomposition of the class mean covariance matrix, it means the dimensionality reduction it can provide is based on the number of class labels, and not on the sample size nor the feature dimensionality.
In the example you linked, it doesn't matter how many features there are. The point is that the example uses 3 simulated cluster centers, so there are 3 class labels. This means linear discriminant analysis could produce projection of the data onto either 1-dimensional or 2-dimensional discriminating subspaces.
But in your data, you start out with only 2 class labels, a binary problem. This means the dimensionality of the linear discriminant model can be at most 1-dimensional, literally a line that forms the decision boundary between the two classes. Dimensionality reduction with LDA in this case would simply be the projection of data points onto a particular normal vector of that separating line.
If you want to specifically reduce down to two dimensions, you can try many of the other algorithms that sklearn provides: t-SNE, ISOMAP, PCA and kernel PCA, random projection, multi-dimensional scaling, among others. Many of these allow you to choose the dimensionality of the projected space, up to the original feature dimensionality, or sometimes you can even project into larger spaces, like with kernel PCA.

In the example you give, dimension reduction by LDA reduces the data from 13 features to 2 features, however in your example it reduces from 3 to 1 (even though you wanted to get 2 features), thus it is not possible to plot in 2D.
If you really want to select 2 features out of 3, you can use feature_selection.SelectKBest to choose 2 best features and there won't be any problems plotting in 2D.
For more information, please read this fantastic answer for PCA:
https://stats.stackexchange.com/questions/2691/making-sense-of-principal-component-analysis-eigenvectors-eigenvalues

Probably it's because of sklearn implementation that won't allowed you to do so if you only have 2 class. The problem has been stated in here, https://github.com/scikit-learn/scikit-learn/issues/1967.

How to change parameters of a scikit learn function dynamically i.e. find best parameter

I am trying to do dimensionality reduction using PCA function of sklearn, specifically
from sklearn.decomposition import PCA
def mypca(X,comp):
pca = PCA(n_components=comp)
pca.fit(X)
PCA(copy=True, n_components=comp, whiten=False)
Xpca = pca.fit_transform(X)
return Xpca
for n_comp in range(10,1000,20):
Xpca = mypca(X,n_comp) # X is a 2 dimensional array
print Xpca
I am calling mypca function from a loop with different values for comp. I am doing this in order to find the best value of comp for the problem I am trying to solve. But mypca function always returns the same value i.e. Xpca irrespective of value of comp.
The value it returns is correct for first value of comp I send from the loop i.e. Xpca value which it sends each time is correct for comp = 10 in my case.
What should I do in order to find best value of comp?

You use PCA to reduce the dimension.
From your code:
for n_comp in range(10,1000,20):
Xpca = mypca(X,n_comp) # X is a 2 dimensional array
print Xpca
Your input dataset X is only a 2 dimensional array, the minimum n_comp is 10, so the PCA try to find the 10 best dimension for you. Since 10 > 2, you will always get the same answer. :)

It looks like you're trying to pass different values for number of components, and re-fit with each. A great thing about PCA is that it's actually not necessary to do this. You can fit the full number of components (even as many components as dimensions in your dataset), then simply discard the components you don't want (i.e. those with small variance). This is equivalent to re-fitting the entire model with fewer components. Saves a lot of computation.
How to do it:
# x = input data, size(<points>, <dimensions>)
# fit the full model
max_components = x.shape[1] # as many components as input dimensions
pca = PCA(n_components=max_components)
pca.fit(x)
# transform the data (contains all components)
y_all = pca.transform(x)
# keep only the top k components (with greatest variance)
k = 2
y = y_all[:, 0:k]
In terms of how to select the number of components, it depends what you want to do. One standard way of choosing the number of components k is to look at the fraction of variance explained (R^2) by each choice of k. If your data is distributed near a low-dimensional linear subspace, then when you plot R^2 vs. k, the curve will have an 'elbow' shape. The elbow will be located at the dimensionality of the subspace. It's good practice to look at this curve because it helps understand the data. Even if there's no clean elbow, it's common to choose a threshold value for R^2, e.g. to preserve 95% of the variance.
Here's how to do it (this should be done on the model with max_components components):
# Calculate fraction of variance explained
# for each choice of number of components
r2 = pca.explained_variance_.cumsum() / x.var(0).sum()
Another way you might want to proceed is to take the PCA-transformed data and feed it to a downstream algorithm (e.g. classifier/regression), then select your number of components based on the performance (e.g. using cross validation).
Side note: Maybe just a formatting issue, but your code block in mypca() should be indented, or it won't be interpreted as part of the function.

Extracting PCA components with sklearn

I am using sklearn's PCA for dimensionality reduction on a large set of images. Once the PCA is fitted, I would like to see what the components look like.
One can do so by looking at the components_ attribute. Not realizing that was available, I did something else instead:
each_component = np.eye(total_components)
component_im_array = pca.inverse_transform(each_component)
for i in range(num_components):
component_im = component_im_array[i, :].reshape(height, width)
# do something with component_im
In other words, I create an image in the PCA space that has all features but 1 set to 0. By inversely transforming them, I should then get the image in the original space which, once transformed, can be expressed solely with that PCA component.
The following image shows the results. On the left is the component calculated using my method. On the right is pca.components_[i] directly. Additionally, with my method, most images are very similar (but they are different) while by accessing the components_ the images are very different as I would have expected
Is there a conceptual problem in my method? Clearly the components from pca.components_[i] are correct (or at least more correct) than the ones I'm getting. Thanks!

Components and inverse transform are two different things. The inverse transform maps the components back to the original image space
#Create a PCA model with two principal components
pca = PCA(2)
pca.fit(data)
#Get the components from transforming the original data.
scores = pca.transform(data)
# Reconstruct from the 2 dimensional scores
reconstruct = pca.inverse_transform(scores )
#The residual is the amount not explained by the first two components
residual=data-reconstruct
Thus you are inverse transforming the original data and not the components, and thus they are completely different. You almost never inverse_transform the orginal data. pca.components_ are the actual vectors representing the underlying axis used to project the data to the pca space.

The difference between grabbing the components_ and doing an inverse_transform on the identity matrix is that the latter adds in the empirical mean of each feature. I.e.:
def inverse_transform(self, X):
return np.dot(X, self.components_) + self.mean_
where self.mean_ was estimated from the training set.