How can I control the number of configurations being evaluated during hyperband tuning in mlr3? I noticed that when I tune 6 parameters in xgboost(), the code evaluates about 9 configurations. When I tune the same number of parameters in catboost(), the code starts with evaluating 729 configurations. I am using eta = 3 in both cases.
The number of sampled configurations in hyperband is defined by the lower and upper bound of the budget hyperparameter and eta. You can get a preview of the schedule and number of configurations:
mlr3hyperband::hyperband_schedule(r_min = 1, r_max = 81, eta = 3)
#> bracket stage budget n
#> 1: 4 0 1 81
#> 2: 4 1 3 27
#> 3: 4 2 9 9
#> 4: 4 3 27 3
#> 5: 4 4 81 1
#> 6: 3 0 3 34
#> 7: 3 1 9 11
#> 8: 3 2 27 3
#> 9: 3 3 81 1
#> 10: 2 0 9 15
#> 11: 2 1 27 5
#> 12: 2 2 81 1
#> 13: 1 0 27 8
#> 14: 1 1 81 2
#> 15: 0 0 81 5
mlr3hyperband::hyperband_n_configs(r_min = 1, r_max = 81, eta = 3)
#> 143
If you want to evaluate more configurations, increase the repetition parameter of hyperband. Hyperband will start again after evaluating the last bracket. Install the latest gh version for this devtools::install_github("mlr-org/mlr3hyperband").
Related
According to https://cran.r-project.org/web/packages/randomForest/randomForest.pdf, classification trees are fully grown, meaning node size = 1.
However, if trees are really grown to a maximum, then shouldn't each terminal node contain a single case (data point, species, etc)?
If I run:
library(randomForest)
data(iris) #150 cases
set.seed(352)
rf <- randomForest(Species ~ ., iris)
hist(treesize(rf),main ="number of nodes")
I can see that most "fully grown" trees only have about 10 nodes, meaning node size can't be equal to 1...Right?
for example, (-1) below represents a terminal node for the 134th tree in the forest. Only 8 terminal nodes!?
> getTree(rf,134)
left daughter right daughter split var split point status prediction
1 2 3 3 2.50 1 0
2 0 0 0 0.00 -1 1
3 4 5 4 1.75 1 0
4 6 7 3 4.95 1 0
5 8 9 3 4.85 1 0
6 10 11 4 1.60 1 0
7 12 13 1 6.50 1 0
8 14 15 1 5.95 1 0
9 0 0 0 0.00 -1 3
10 0 0 0 0.00 -1 2
11 0 0 0 0.00 -1 3
12 0 0 0 0.00 -1 3
13 0 0 0 0.00 -1 2
14 0 0 0 0.00 -1 2
15 0 0 0 0.00 -1 3
I would be greatful if someone can explain
"Fully grown" -> "Nothing left to split". A (node of a-) decision tree is fully grown, if all data records assigned to it hold/make the same prediction.
In the iris dataset case, once you reach a node with 50 setosa data records in it, it doesn't make sense to split it into two child nodes with 25 and 25 setosas each.
I am attempting to read Aaron Hsu's thesis on A data parallel compiler hosted on the GPU, where I have landed at some APL code I am unable to fix. I've attached both a screenshot of the offending page (page number 74 as per the thesis numbering on the bottom):
The transcribed code is as follows:
d ← 0 1 2 3 1 2 3 3 4 1 2 3 4 5 6 5 5 6 3 4 5 6 5 5 6 3 4
This makes sense: create an array named d.
⍳≢d
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
This too makes sense. Count the number of elements in d and create a sequence of
that length.
⍉↑d,¨⍳≢d
0 1 2 3 1 2 3 3 4 1 2 3 4 5 6 5 5 6 3 4 5 6 5 5 6 3 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
This is slightly challenging, but let me break it down:
zip the sequence ⍳≢d = 1..27 with the d array using the ,¨ idiom, which zips the two arrays using a catenation.
Then, split into two rows using ↑ and transpose to get columns using ⍉
Now the biggie:
(⍳≢d)#(d,¨⍳≢d)⊢7 27⍴' '
INDEX ERROR
(⍳≢d)#(d,¨⍳≢d)⊢7 27⍴' '
Attempting to break it down:
⍳≢d counts number of elements in d
(d,¨⍳≢d) creates an array of pairs (d, index of d)
7 27⍴' ' creates a 7 x 27 grid: presumably 7 because that's the max value of d + 1, for indexing reasons.
Now I'm flummoxed about how the use of ⊢ works: as far as I know, it just ignores everything to the left! So I'm missing something about the parsing of this expression.
I presume it is parsed as:
(⍳≢d)#((d,¨⍳≢d)⊢(7 27⍴' '))
which according to me should be evaluated as:
(⍳≢d)#((d,¨⍳≢d)⊢(7 27⍴' '))
= (⍳≢d)#((7 27⍴' ')) [using a⊢b = b]
= not the right thing
As I was writing this down, I managed to fix the bug by sheer luck: if we increment d to be d + 1 so we are 1-indexed, the bug no longer manifests:
d ← d + 1
d
1 2 3 4 2 3 4 4 5 2 3 4 5 6 7 6 6 7 4 5 6 7 6 6 7 4 5
then:
(⍳≢d)#(d,¨⍳≢d)⊢7 27⍴' '
1
2 5 10
3 6 11
4 7 8 12 19 26
9 13 20 27
14 16 17 21 23 24
15 18 22 25
However, I still don't understand how this works! I presume the context will be useful
for others attempting to leave the thesis, so I'm going to leave the rest of it up.
Please explain what (⍳≢d)#(d,¨⍳≢d)⊢7 27⍴' ' does!
I've attached the raw screenshot to make sure I didn't miss something:
I'm happy to see that you found the the off-by-one error. It stems from Aaron Hsu working with index origin 0. If you set ⎕IO←0 then his code will work.
Some dyadic operators can take an array operand, giving the sequence OPERATOR operand argument, e.g. in -#(1 2 3)(4 5 6 7). This poses a problem because both the operand and the argument are arrays, and juxtaposition of arrays forms a new array with those arrays as elements by a process known as stranding. Compare:
(1 2 3)(4 5 6 7)
┌─────┬───┐
│1 2 3│4 5│
└─────┴───┘
However, in the case of the operator with its array operand, we want to "break" this strand so the left part can act as operand while the right part acts as argument. One way to break the stranding up is by applying a function to the argument, giving the sequence OPERATOR operand Function argument. Now, we don't actually need any transformation of the argument, so an identity function will do: -#(1 2 3)⊢(4 5 6 7).
As for what (⍳≢d)#(d,¨⍳≢d)⊢7 27⍴' ' actually does:
7 27⍴' ' creates a blank matrix.
(⍳≢d) are indices to insert into specified slots in the matrix.
#(d,¨⍳≢d) indicates at which locations in the matrix the above should replace the existing values
⊢ serves solely to separate (d,¨⍳≢d) from 7 27⍴' '. The code could also have been written as ((⍳≢d)#(d,¨⍳≢d))7 27⍴' ' with parentheses serving to "bind" the operand to the operator.
I have been trying to run Cox PH model on a sample data set of 10k customers (randomly taken from 32 million customer base) for predicting probability of survival in time t (which is month in my case). I am using recurrent event survival analysis using counting process for e-commerce. For this...
1. Observation starting point: right after a customer makes first purchase
2. Start/Stop times: Months of two consecutive purchases (as in the data)
I have a few independent variables as in the sample data below:
id start stop status tenure orders revenue Quantity
A 0 20 0 0 1 $89.0 1
B 0 17 0 0 1 $556.0 2
C 0 17 0 0 1 $900.0 2
D 32 33 0 1679 9 $357.8 9
D 26 32 1 1497 7 $326.8 7
D 23 26 1 1405 4 $142.9 4
D 17 23 1 1219 3 $63.9 3
D 9 17 1 978 2 $50.0 2
D 0 9 1 694 1 $35.0 1
E 0 15 0 28 2 $156.0 2
F 0 15 0 0 1 $348.0 1
F 12 14 0 375 2 $216.8 3
F 0 12 1 0 1 $67.8 2
G 9 15 0 277 2 $419.0 2
G 0 9 1 0 1 $359.0 1
While running cox PH using the following code:
fit10=coxph(Surv(start,stop,status)~orders+tenure+Quantity+revenue,data=test)
I keep getting the following error:
Warning: X matrix deemed to be singular; variable 1 2 3 4
I tried searching for the same error online but the answers I found said this could be because of interacting independent variables, whereas my variables are individual and continuous.
I have the following codes:
ItemSimilarity itemSimilarity = new UncenteredCosineSimilarity(dataModel);
recommender = new GenericItemBasedRecommender(dataModel,itemSimilarity);
List<RecommendedItem> items = recommender.mostSimilarItems(10, 5);
my datamodel is like this:
uid itemid socre
userid itemid score
1 6 5
1 10 3
1 11 5
1 12 4
1 13 5
2 2 3
2 6 5
2 10 3
2 12 5
when I run the code above,the result is just like this:
13
6
11
2
12
I debug the code,and find that the List items = recommender.mostSimilarItems(10, 5); return the items has the same score,that is one!
so,I have a problem.in my opinion,I think the mostsimilaritem should consider the item co-occurrence matrix：
2 6 10 11 12 13
2 0 1 1 0 1 0
6 1 0 2 1 2 1
10 1 2 0 1 2 1
11 0 1 1 0 1 1
12 1 2 2 1 0 1
13 0 1 1 1 1 0
in the matrix above ,the item 12's most similar should be [6,12,11,13,2],because the item 1 and item 12 is more similar than the other items,isn't it?
now,anyone who can explain this for me?thanks!
In your matrix you have much more data than in your input. In particular you seem to be imputing 0 values that are not in the data. That is why you are likely getting answers different from what you expect.
Mahout expects your IDs to be contiguous Integers starting from 0. This is true of your row and column ids. Your matrix looks like it has missing ids. Just having Integers is not enough.
Could this be the problem? Not sure what Mahout would do with the input above.
I always keep a dictionary to map Mahout IDs to/from my own.
HI complete newbie question here: I have a table consisting of two columns. First column belongs to "bins" that are coded by where a the fruit flies live. The second column is either 0 or 1, neutral vs really like sugar, respectively. I have two question?
1) if I suspect that there is a single variable, something about where they live that is determining whether how much they like sugar. Is there a way that I can have the computer to group into just 2 clusters? All the bins that like sugar vs neutral. That way we can do further experiment to determine what is it about the bins.
2) automatically determine how many clusters there might be that is driving this behavior? For example may be there is 4 variables (4 clusters) that can determine the outcome of sugar preference.
Apologies if this is trivial. The table is listed below. thanks!
Bin sugar
1 1
1 1
1 0
1 0
2 1
2 0
2 0
3 1
3 0
3 1
3 1
4 1
4 1
4 1
5 1
5 0
5 1
6 0
6 0
6 0
7 0
7 1
7 1
8 1
8 0
8 1
9 1
9 0
9 0
9 0
10 0
10 0
10 0
11 1
11 1
11 1
12 0
12 0
12 0
12 0
13 0
13 0
13 1
13 0
13 0
14 0
14 0
14 0
14 0
15 1
15 0
15 0
16 1
16 1
17 1
17 1
18 0
18 1
18 1
17 1
19 1
20 1
20 0
20 0
20 1
21 0
21 0
21 1
21 0
22 1
22 0
22 1
22 1
23 1
23 1
24 1
24 0
25 0
25 1
25 0
26 1
26 1
27 1
27 1
Okay, assuming I understood what you meant, one approach to problem 1) should be addressed using bayes filtering.
Say event L is "a fly likes sugar", event B is "a fly is in bin B".
So what you have is:
number of flies = 84
size of each bins = (eg size of bin 1: 4)
probability that a fly likes sugar:
P(L) = flies that like sugar / total number of flies = 43/84
probability that a fly doesn't like sugar:
P(notL) = 1 - P(L) = 41/84
probability that a fly is in a given bin:
P(B) = size of the bin / sum of the sizes of all bins = 4/84 (for bin 1)
probability that a fly isn't in a given bin:
P(notB) = 1 - P(B) = 80/84 (for bin 1)
probability that a fly likes sugar, knowing that's in bin B:
P(L|B) = flies that like sugar in a bin / size of the bin
(eg for bin 1 is 2/4 = 1/2)
probability that a fly likes sugar, knowing that it's not in bin B:
P(L|notB) = (total flies that like sugar - flies that like sugar in the bin)/(size of bins - size of the bin)) = 41/80
You want to know the probability that a fly is in a given bin B knowing that likes sugar, which you can obtain with:
P(B|L) = (P(L|B) * P(B)) / (P(L|B) * P(B) + P(L|notB) * P(notB))
If you compute P(B|L) and P(B|notL) for each bin, then you know which of the bins have the highest probability of containing flies that like sugar. Then you can further study those bins.
Hope i was clear, my statistics is a bit rusty and I'm not even sure I am doing everything correctly. Take it as a hint to point you in the right direction to address the problem.
You can refer here to get more accurate reasoning and results.
As for problem 2)... I have to think about it a bit more.