Another treat! Early access to Intro To Machine Learning videos


(nok) #622

Interesting, I change the function parallel_trees with multiple-threads but not process. It works, not sure why though.

def parallel_trees(m, fn, n_jobs=8):
        return list(ProcessPoolExecutor(n_jobs).map(fn, m.estimators_))

to

def parallel_trees(m, fn, n_jobs=8):
        return list(ThreadPoolExecutor(n_jobs).map(fn, m.estimators_))

(Kevin Bird) #623

I’m working on building a random forest and I am getting really terrible results. I tried to just follow the lesson1 steps, but my data is giving me a -0.14 r^2. My question is what are the next steps at that point? Is it back to the drawing board or is there something I can do still? One thing I’m thinking is I might be able to do some sort of oversampling since I have 3.5% of my data equal to 1 and the rest is 0. Does that seem like a good next step with this terrible of a starting r^2? Or should I be looking at my data quality and make sure I am at least getting a decent r^2 out of the gate before trying to do anything extra?

Any advice on this would be much appreciated. Here is what my numbers look like:

m = RandomForestRegressor(n_jobs=-1)
m.fit(X_trn, y_trn)
print_score(m)
[0.08436980063282429, 0.192801180918272, 0.7956208204063915, -0.14288877498635988]

After looking into this further, I think the problem is with how r^2 is calculated compared to what I actually care about. Only 3.5% of my values are 1 so when I look at the predictions, I see a lot of Actual = 1 Predicted = 0.1.

This leads me to believe I should try to oversample to put the number of 1s up to 50% and then adjust that threshold later when I’m trying to decide what actually counts as a “1”.

On the bright side, I am getting a great r^2 for the RandomForestClassifier, but unfortunately the reason it’s good is because it is predicting almost all of them as 0.


(Arul Bharathi) #625

I noticed that somewhere in Lesson 2 or 3 that you have mentioned if we use set_rf_samples(), we should not be using oob_score. So that rule still should be followed, or that issue is fixed?


(Asutosh) #626

While going through this discussion forum, I came across a few discussions on the bootstrap argument of RandomForestRegressor() function and also about the set_rf_samples().

I also misunderstood it in the beginning and reading the conversations just got me more confused. So, I decided to dig a bit deep into fast.ai and sklearn source codes and came up with the following conclusions -->

n = no_of_rows_in_dataframe
if (bootstrap == false) {
  then all `n` rows are considered exactly once per tree for training
}
else if (set_rf_samples(k) is used) {
  then `k` rows are selected per tree for training & there might be some repetitions of rows
}
else {
  then `n` rows are selected per tree for training & there might be some repetitions of rows
}

Also, there were some ambiguity around the oob_score calculation. So, after exploring a bit, here’s what I concluded —>

/**************************************************************************************	
	for simplicity assuming output corresponding to each input is a single number.
	So, y.shape = (n, 1)
	y = actual outputs
	n = no_of_rows_in_data_frame
	
	For cases with a output vector, the oob_score can be calculated by simply taking average of oob_score 
    of each column of the vector.
****************************************************************************************/

total_prediction = zero_matrix of dimension (n x 1) /* used to accumulate total predictions for each row (by different trees in the forest) which will later be averaged */
  
no_of_predictions = zero_matrix of dimension (n x 1) /* total number of predictions for each row (which also represents total number of trees in which each row is Out-Of-Bag), used for averaging later */
  
for (tree in forest) {
	out_of_bag_samples = all_rows - set(rows used by `tree` for training)
    total_prediction += tree.predict(out_of_bag_samples)
    no_of_predictions = (increased by 1 for each row which was in out_of_bag_sample)
}

predictions = total_prediction / no_of_predictions
oob_score = r2_score(y, predictions)

For exact code of oob_score calculation, refer here


(ismail belghit) #627

great resources
thank you @jeremy sir
you rock :wink:


(Sanyam Bhutani) #628

I’ve just published the Lecture 1 Notes (With Jeremy’s permission).
Hope these are helpful and please do point out any points that could be corrected/improved.

I believe I’ll be able to share all the Notes before the End of this month.

Sanyam.


#629

Hello asutosh97

About the oob score I have a question:

if i understood what jeremy said (english is not my native language) the oob score allows you not to need a validation set to see how well the model works. For this reason it is also useful when we have little data.

my question is why in notebooks jeremy uses:

m = RandomForestRegressor(n_estimators=40, n_jobs=-1, oob_score=True)
m.fit(X_train, y_train)
print_score(m)

If i use the oob_score shouldn’t I use m.fit with the complete data and not only with the x_train and the y_train?


(Jeremy Howard) #630

This is looking great! Thanks for sharing :slight_smile:


(Igor Kasianenko) #631


In the 2nd lesson there intruduction to max_features of random forest. For me it looks similar to dropout of neural network from deep learning, is it correct intuition?

(Asutosh) #632

@Kasianenko
Yes, absolutely. Connecting the right dots.


(Asutosh) #633

Hello @fumpen,

as Jeremy says in one of his lectures, we can’t use any of the test data for calibration. Think of it as you don’t have it until you’ve trained your model completely. Else, you can’t get true results.


(Vishal ) #634

Hello Everyone,

In Lecture 2, @jeremy explains how a decision tree is formed by selecting a variable and a split-point, at each step, which yields the lowest MSE (as per the naive model). Can someone please explain why exactly is this the splitting methodology? From another source, decision tree splitting is done using the ‘Information Gain’. How are these two (MSE and Information Gain) connected?


(Asutosh) #635

hello @vahuja4, think it this way

Information Gain = MSE at Root Node - Avg. MSE of the childs after splitting

So, IG will be more when Avg. MSE drops the most. Both are basically indicating the same thing only.


(Vishal ) #636

I see. But why is this termed to be ‘Information Gain’? Also, would you know why is this the chosen methodology for splitting?


(Asutosh) #637

@vahuja4

  1. I think it is termed like that by convention because the more close your predictions come to actual values, you seem to have gained more information. And MSE basically denotes the gap between the actual values and model prediction. So, the closer the gap becomes(i.e. the more the MSE drops), it can be thought of as more information is gained.

  2. As you know in DecisionTreeRegressor, the pred`iction at a node is given by taking the average of all the data points belonging to it. So, our ultimate goal is to make this average as close to the actual value.
    So, we basically do a brute-force search of all possible splitting and check which one will give average closest to the actual values, and use MSE as a metric to measure the closeness.

I hope these answer your questions.


(Vishal ) #638

@asutosh97, thank you! Makes sense.


(Aditya) #639

Hey guys, if you want even a deeper understanding of your tree based models/xgb/sklearn etc, check this cool repos out,

What are your thoughts on this Jeremy?

(It’s really nice to interpret the Black Boxes properly…)

Both look Promising


(Asutosh) #640

looks interesting, where can I find documentations of using this?


(Aditya) #641

Probably the notebooks are there…

The plots looks amazing…

https://nbviewer.jupyter.org/github/slundberg/shap/tree/master/notebooks/


(Kofi Asiedu Brempong) #642

I’m trying out techniques learnt in lessons 1 and 2 on the house prices kaggle competition


the training set has 1460 rows, should i still split it into 2 to get a separate validation set or should I just rely on oob_score?