Suppose we have 2 labels: 0 and 1.
The data number with label 0 is 1000 but the data with label 1 is just 100.
In this situation, the training of classification will be bias to the result of label 0.
What can be done in this scenario?
Can we generate samples manually corresponding to label 1?
If we can do so, how to validate that the generated samples possess the same properties/characteristics as the original data?
See this aricle.
It's about a method called SMOTE which stands for Synthetic Minority Over-sampling Technique.
Basically if you have data distributed like this (small number of red dots, larger number of green dots):
You synthesize new samples around the existing ones:
This method is one of the commonly used ones and it is described in greater detail in the article linked above. There are other simpler methods like removing some datapoints from the majority class or duplicating some of the ones in the minority class.
The images have been taken from the article.
Related
I am using a GLMM model to determine differences in soil compaction across 3 locations and 2 seasons in undisturbed and disturbed sites. I used location and seas as random effects. My teacher says to use the compaction reading divided by its upper bound as the Y value against the different sites (fixed effect). (I was previously using disturbed and undisturbed sites as 1,0 as Y against the compaction reading - so the opposite way around.) The random variables are minimal. I was using both glmer (glmer to determine AIC and therefore best model fit (but this cannot be done in glmmPQL)) while glmmPQL provides all amounts of variation which glmer does not. So while these outcomes are very similar when using disturbed and undisturbed as Y (as well as matching the graphs) only glmmPQL is similar to the graphs when using proportion of compaction reading. glmer using proportions is totally different. Additionally my teacher says I need to validate my model choice with a chi-squared value and if over-dispersed use a quasi binomial. But I cannot find any way to do this in glmmPQL and with glmer showing strange results using proportions as Y I am unsure if this is correct. I also cannot use quasi binomial in either glmer or glmmPQL.
My response was the compaction reading which is measured from 0 to 6 (kg per cm squared) inclusive. The explanatory variable was Type (diff soil either disturbed and not disturbed = 4 categories because some were artificially disturbed to pull out differences). All compaction readings were divided by 6 to make them a proportion and so a continuous variable bounded by 0 and 1 but containing values of both 0 and 1. (I also tried the reverse and coded disturbed as 1 and undisturbed as 0 and compared these groups separately across all Types (due to 4 Types) and left compaction readings as original). Using glmer with code:
model1 <- glmer(comp/6 ~ Type +(1|Loc/Seas), data=mydata,
family = "binomial")
model2 <- glmer(comp/6~Type +(1|Loc) , data=mydata, family="binomial")
and using glmmPQL:
mod1 <-glmmPQL(comp/6~Type, random=~1|Loc, family = binomial, data=mydata)
mod2 <- glmmPQL(comp/6~Type, random=~1|Loc/Seas, family = binomial, data=mydata)
I could compare models in glmer but not in glmmPQL but the latter gave me the variance for all random effects plus residual variance whereas glmer did not provide the residual variance (so was left wondering what was the total amount and what proportion were these random effects responsible for)
When I used glmer this way, the results were completely different to glmmPQL as in no there was no sig difference at all in glmer but very much a sig diff in glmmPQL. (However if I do the reverse and code by disturbed and undisturbed these do provide similar results between glmer and glmmPQL and what is suggested by the graphs - but my supervisor said this is not strictly correct (eg: mod3 <- glmmPQL(Status~compaction, random=~1|Loc/Seas, family = binomial, data=mydata) where Status is 1 or 0 according to disturbed or undisturbed) plus my supervisor would like me to provide a chi squared goodness of fit for the model chosen - so can only use glmer here ?). Additionally, the random effects variance is minimal, and glmer model choice removes these as non significant (although keeping one in provides a smaller AIC). Removing them (as suggested by the chi-squared test (but not AIC) and running as only a glm is consistent to both results from glmmPQL and what is observed on the graph. Sorry if this seems very pedantic, but I am trying to do what is correct for my supervisor and for the species I am researching. I know there are differences.. they are seen, observed, eyeballing the data suggests so and so do the graphs.. Maybe I should just run the glm ? Thank you for answering me. I will find some output to post.
I have a design for an experiment that I would like to code in psychtoolbox using MATLAB. I understand the basics of how to use this toolbox however it would be extremely helpful if someone has designed a similar experiment before and could provide me with some code that could help me to carry out the following:
The experimental procedure will be composed of 80 trials divided into 5 blocks with 16 trials in each block. The experiment consists of the participant selecting a number from the screen. There is a desirable number (target number) and an associated less desirable number (lure number). I won't go into further detail about the reasoning behind this experiment as it is not relevant to my question.
I have attached an image that shows 1 block of trials (16 trials). The other 4 blocks are the same as this block.
Target and lure numbers will be presented on the screen to choose from (an example can be seen in image below).
In some of the trials as can be seen from the trials table only one target number and one lure number are presented for the participant to choose from (instead of two targets and two lures).
The lure(s) that appears with each target(s) should not always be the same. I want the lure that is shown with the target to be randomly selected in each trial (as can be seen in the trials image there is more than one possible lure). In the trials image I have attached the trial numbers are presented just for clarity, in each block the presentation of the targets needs to be randomized.
You can use the BalanceTrials function that comes with psychtoolbox. You use all the possible lures and targets as inputs and it returns a random order of all possible combinations. You can also specify a minimum length of the list it returns, but if there are more combinations it will make the list longer to make it balanced. Here is an example:
numberOfTrials = 80;
targetNumbers = {'3','4','5','6','7','4 5','4 6','4 7'};
lureNumbers = {'3','4','5','6','7','4 7'};
[targets, lures] = BalanceTrials(numberOfTrials, 1, targetNumbers, lureNumbers);
You can split this up into 5 blocks or you do it each time for each block.
I am have built a small text analysis model that is classifying small text files as either good, bad, or neutral. I was using a Support-Vector Machine as my classifier. However, I was wondering if instead of classifying all three I could classify into either Good or Bad but if the support for that text file is below .7 or some user specified threshold it would classify that text file as neutral. I know this isn't looked at as the best way of doing this, I am just trying to see what would happen if I took a different approach.
The operator Drop Uncertain Predictions might be what you want.
After you have applied your model to some test data, the resulting example set will have a prediction and two new attributes called confidence(Good) and confidence(Bad). These confidences are between 0 and 1 and for the two class case they will sum to 1 for each example within the example set. The highest confidence dictates the value of the prediction.
The Drop Uncertain Predictions operator requires a min confidence parameter and will set the prediction to missing if the maximum confidence it finds is below this value (you can also have different confidences for different class values for more advanced investigations).
You could then use the Replace Missing Values operator to change all missing predictions to be a text value of your choice.
Hashing reduces dimensionality while one-hot-encoding essentially blows up the feature space by transforming multi-categorical variables into many binary variables. So it seems like they have opposite effects. My questions are:
What is the benefit of doing both on the same dataset? I read something about capturing interactions but not in detail - can somebody elaborate on this?
Which one comes first and why?
Binary one-hot-encoding is needed for feeding categorical data to linear models and SVMs with the standard kernels.
For example, you might have a feature which is a day of a week. Then you create a one-hot-encoding for each of them.
1000000 Sunday
0100000 Monday
0010000 Tuesday
...
0000001 Saturday
Feature-hashing is mostly used to allow for significant storage compression for parameter vectors: one hashes the high dimensional input vectors into a lower dimensional feature space. Now the parameter vector of a resulting classifier can therefore live in the lower-dimensional space instead of in the original input space. This can be used as a method of dimension reduction thus usually you expect to trade a bit of decreasing of performance with significant storage benefit.
The example in wikipedia is a good one. Suppose your have three documents:
John likes to watch movies.
Mary likes movies too.
John also likes football.
Using a bag-of-words model, you first create below document to words model. (each row is a document, each entry in the matrix indicates whether a word appears in the document).
The problem with this process is that such dictionaries take up a large amount of storage space, and grow in size as the training set grows.
Instead of maintaining a dictionary, a feature vectorizer that uses the hashing trick can build a vector of a pre-defined length by applying a hash function h to the features (e.g., words) in the items under consideration, then using the hash values directly as feature indices and updating the resulting vector at those indices.
Suppose you generate below hashed features with 3 buckets. (you apply k different hash functions to the original features and count how many times the hashed value hit a bucket).
bucket1 bucket2 bucket3
doc1: 3 2 0
doc2: 2 2 0
doc3: 1 0 2
Now you successfully transformed the features in 9-dimensions to 3-dimensions.
A more interesting application of feature hashing is to do personalization. The original paper of feature hashing contains a nice example.
Imagine you want to design a spam filter but customized to each user. The naive way of doing this is to train a separate classifier for each user, which are unfeasible regarding either training (to train and update the personalized model) or serving (to hold all classifiers in memory). A smart way is illustrated below:
Each token is duplicated and one copy is individualized by concatenating each word with a unique user id. (See USER123_NEU and USER123_Votre).
The bag of words model now holds the common keywords and also use-specific keywords.
All words are then hashed into a low dimensioanl feature space where the document is trained and classified.
Now to answer your questions:
Yes. one-hot-encoding should come first since it is transforming a categorical feature to binary feature to make it consumable by linear models.
You can apply both on the same dataset for sure as long as there is benefit to use the compressed feature-space. Note if you can tolerate the original feature dimension, feature-hashing is not required. For example, in a common digit recognition problem, e.g., MINST, the image is represented by 28x28 binary pixels. The input dimension is only 784. For sure feature hashing won't have any benefit in this case.
Why is Shannon's Entropy measure used in Decision Tree branching?
Entropy(S) = - p(+)log( p(+) ) - p(-)log( p(-) )
I know it is a measure of the no. of bits needed to encode information; the more uniform the distribution, the more the entropy. But I don't see why it is so frequently applied in creating decision trees (choosing a branch point).
Because you want to ask the question that will give you the most information. The goal is to minimize the number of decisions/questions/branches in the tree, so you start with the question that will give you the most information and then use the following questions to fill in the details.
For the sake of decision trees, forget about the number of bits and just focus on the formula itself. Consider a binary (+/-) classification task where you have an equal number of + and - examples in your training data. Initially, the entropy will be 1 since p(+) = p(-) = 0.5. You want to split the data on an attribute that most decreases the entropy (i.e., makes the distribution of classes least random). If you choose an attribute, A1, that is completely unrelated to the classes, then the entropy will still be 1 after splitting the data by the values of A1, so there is no reduction in entropy. Now suppose another attribute, A2, perfectly separates the classes (e.g., the class is always + for A2="yes" and always - for A2="no". In this case, the entropy is zero, which is the ideal case.
In practical cases, attributes don't typically perfectly categorize the data (the entropy is greater than zero). So you choose the attribute that "best" categorizes the data (provides the greatest reduction in entropy). Once the data are separated in this manner, another attribute is selected for each of the branches from the first split in a similar manner to further reduce the entropy along that branch. This process is continued to construct the tree.
You seem to have an understanding of the math behind the method, but here is a simple example that might give you some intuition behind why this method is used: Imagine you are in a classroom that is occupied by 100 students. Each student is sitting at a desk, and the desks are organized such there are 10 rows and 10 columns. 1 out of the 100 students has a prize that you can have, but you must guess which student it is to get the prize. The catch is that everytime you guess, the prize is decremented in value. You could start by asking each student individually whether or not they have the prize. However, initially, you only have a 1/100 chance of guessing correctly, and it is likely that by the time you find the prize it will be worthless (think of every guess as a branch in your decision tree). Instead, you could ask broad questions that dramatically reduce the search space with each question. For example "Is the student somewhere in rows 1 though 5?" Whether the answer is "Yes" or "No" you have reduced the number of potential branches in your tree by half.