I'm working with ranger, a fast implementation of Random Forests. The problem is I have no idea how to interpret the $forest component of the result. The document simply says
forest: Saved forest (If write.forest set to TRUE). Note that the variable IDs in the split.varIDs object do not necessarily
represent the column number in R.
Well, that isn't really helpful, so I tried inspecting its components myself, by their names are not self-explanatory.
> names(ranger(Species ~ ., data = iris)$forest)
[1] "dependent.varID" "num.trees"
[3] "child.nodeIDs" "split.varIDs"
[5] "split.values" "is.ordered"
[7] "class.values" "levels"
[9] "independent.variable.names" "treetype"
Some components like num.trees are trivial to understand, but things like child.nodeIDs are really mind-blowing.
> ranger(Species ~ ., data = iris)$forest$child.nodeIDs[[1]]
[[1]]
[1] 1 3 5 0 7 9 11 0 0 0 13 15 0 0 0 0 0
[[2]]
[1] 2 4 6 0 8 10 12 0 0 0 14 16 0 0 0 0 0
Is it documented somewhere?
See the documentation for the ranger::treeInfo function: https://www.rdocumentation.org/packages/ranger/versions/0.11.2/topics/treeInfo
Related
I am trying to train a model using bstTree method and print out the confusion matrix. adverse_effects is my class attribute.
set.seed(1234)
splitIndex <- createDataPartition(attended_num_new_bstTree$adverse_effects, p = .80, list = FALSE, times = 1)
trainSplit <- attended_num_new_bstTree[ splitIndex,]
testSplit <- attended_num_new_bstTree[-splitIndex,]
ctrl <- trainControl(method = "cv", number = 5)
model_bstTree <- train(adverse_effects ~ ., data = trainSplit, method = "bstTree", trControl = ctrl)
predictors <- names(trainSplit)[names(trainSplit) != 'adverse_effects']
pred_bstTree <- predict(model_bstTree$finalModel, testSplit[,predictors])
plot.roc(auc_bstTree)
conf_bstTree= confusionMatrix(pred_bstTree,testSplit$adverse_effects)
But I get the error 'Error in confusionMatrix.default(pred_bstTree, testSplit$adverse_effects) :
The data must contain some levels that overlap the reference.'
max(pred_bstTree)
[1] 1.03385
min(pred_bstTree)
[1] 1.011738
> unique(trainSplit$adverse_effects)
[1] 0 1
Levels: 0 1
How can I fix this issue?
> head(trainSplit)
type New_missed Therapytypename New_Diesease gender adverse_effects change_in_exposure other_reasons other_medication
5 2 1 14 13 2 0 0 0 0
7 2 0 14 13 2 0 0 0 0
8 2 0 14 13 2 0 0 0 0
9 2 0 14 13 2 1 0 0 0
11 2 1 14 13 2 0 0 0 0
12 2 0 14 13 2 0 0 0 0
uvb_puva_type missed_prev_dose skintypeA skintypeB Age DoseB DoseA
5 5 1 1 1 22 3.000 0
7 5 0 1 1 22 4.320 0
8 5 0 1 1 22 4.752 0
9 5 0 1 1 22 5.000 0
11 5 1 1 1 22 5.000 0
12 5 0 1 1 22 5.000 0
I had similar problem, which refers to this error. I used function confusionMatrix:
confusionMatrix(actual, predicted, cutoff = 0.5)
An I got the following error: Error in confusionMatrix.default(actual, predicted, cutoff = 0.5) : The data must contain some levels that overlap the reference.
I checked couple of things like:
class(actual) -> numeric
class(predicted) -> integer
unique(actual) -> plenty values, since it is probability
unique(predicted) -> 2 levels: 0 and 1
I concluded, that there is problem with applying cutoff part of the function, so I did it before by:
predicted<-ifelse(predicted> 0.5,1,0)
and run the confusionMatrix function, which works now just fine:
cm<- confusionMatrix(actual, predicted)
cm$table
which generated correct outcome.
One takeaway for your case, which might improve interpretation once you make code working:
you mixed input values for your confusion matrix(as per confusionMatrix package documetation), instead of:
conf_bstTree= confusionMatrix(pred_bstTree,testSplit$adverse_effects)
you should have written:
conf_bstTree= confusionMatrix(testSplit$adverse_effects,pred_bstTree)
As said it will most likely help you interpret confusion matrix, once you figure out way to make it work.
Hope it helps.
max(pred_bstTree) [1] 1.03385
min(pred_bstTree) [1] 1.011738
and errors tells it all. Plotting ROC is simply checking the effect of different threshold points. Based on threshold rounding happens e.g. 0.7 will be converted to 1 (TRUE class) and 0.3 will be go 0 (FALSE class); in case threshold is 0.5. Threshold values are in range of (0,1)
In your case regardless of threshold you will always get all observations into TRUE class as even minimum prediction is greater than 1. (Thats why #phiver was wondering if you are doing regression instead of classification) . Without any zero in prediction there is no level in 'prediction' which coincide with zero level in adverse_effects and hence this error.
PS: It will be difficult to tell root cause of error without you posting your data
When do text mining using R, after reprocessing text data, we need create a document-term matrix for further exploring. But in similar with Chinese, English also have some certain phases, such as "semantic distance", "machine learning", if you segment them into word, it have totally different meanings, I want to know how to segment document into phases but not word(term).
You can do this in R using the quanteda package, which can detect multi-word expressions as statistical collocates, which would be the multi-word expressions that you are probably referring to in English. To remove the collocations containing stop words, you would first tokenise the text, then remove the stop words leaving a "pad" in place to prevent false adjacencies in the results (two words that were not actually adjacent before the removal of stop words between them).
require(quanteda)
pres_tokens <-
tokens(data_corpus_inaugural) %>%
tokens_remove("\\p{P}", padding = TRUE, valuetype = "regex") %>%
tokens_remove(stopwords("english"), padding = TRUE)
pres_collocations <- textstat_collocations(pres_tokens, size = 2)
head(pres_collocations)
# collocation count count_nested length lambda z
# 1 united states 157 0 2 7.893307 41.19459
# 2 let us 97 0 2 6.291128 36.15520
# 3 fellow citizens 78 0 2 7.963336 32.93813
# 4 american people 40 0 2 4.426552 23.45052
# 5 years ago 26 0 2 7.896626 23.26935
# 6 federal government 32 0 2 5.312702 21.80328
# convert the corpus collocations into single tokens, for top 1,500 collocations
pres_compounded_tokens <- tokens_compound(pres_tokens, pres_collocations[1:1500])
tokens_select(pres_compounded_tokens[2], "*_*")
# tokens from 1 document.
# 1793-Washington :
# [1] "called_upon" "shall_endeavor" "high_sense" "official_act"
Using this "compounded" token set, we can now turn this into a document-feature matrix where the features consist of a mixture of original terms (those not found in a collocation) and the collocations. As can be seen below, "united" occurs alone and as part of the collocation "united_states".
pres_dfm <- dfm(pres_compounded_tokens)
head(pres_dfm[1:5, grep("united|states", featnames(pres_dfm))])
# Document-feature matrix of: 5 documents, 10 features (86% sparse).
# 5 x 10 sparse Matrix of class "dfm"
# features
# docs united states statesmen statesmanship reunited unitedly devastates statesman confederated_states united_action
# 1789-Washington 4 2 0 0 0 0 0 0 0 0
# 1793-Washington 1 0 0 0 0 0 0 0 0 0
# 1797-Adams 3 9 0 0 0 0 0 0 0 0
# 1801-Jefferson 0 0 0 0 0 0 0 0 0 0
# 1805-Jefferson 1 4 0 0 0 0 0 0 0 0
If you want a more brute-force approach, it's possible simply to create a document-by-bigram matrix this way:
# just form all bigrams
head(dfm(data_inaugural_corpus, ngrams = 2))
## Document-feature matrix of: 57 documents, 63,866 features.
## (showing first 6 documents and first 6 features)
## features
## docs fellow-citizens_of of_the the_senate senate_and and_of the_house
## 1789-Washington 1 20 1 1 2 2
## 1797-Adams 0 29 0 0 2 0
## 1793-Washington 0 4 0 0 1 0
## 1801-Jefferson 0 28 0 0 3 0
## 1805-Jefferson 0 17 0 0 1 0
## 1809-Madison 0 20 0 0 2 0
I have made some classification models where 1 means it is the same person, and 0 means they are different.
If I print the head of my predictions it looks the following way:
> head(PredictCTree)
[1] 0 0 0 0 0 0
Levels: 0 1
> head(PredictSVM)
1 1.1 1.2 1.3 1.7 1.14
0 0 0 0 0 0
Levels: 0 1
> head(PredictForest)
1.212 1.839 1.906 1.951 1.1011 1.1151
1 1 1 0 1 1
Levels: 0 1
So if I want to average them and add them up I have to make them numeric, but here is where I am struggling:
Example:
> PredictForest[1]
1.212
1
Levels: 0 1
basically I want to add 1 + 0 (for PredictForest and SVM)
as.numeric(PredictForest[1])
[1] 2
but I end up getting this answer:
> as.numeric(PredictForest[1]) + as.numeric(fitted.results[1] + as.numeric(PredictCTree[1] ))
[1] 4
Any suggestions?
My expected output would be:
> as.numeric(PredictForest[1]) + as.numeric(fitted.results[1] + as.numeric(PredictCTree[1] ))
[1] 1
So later on I could divide or give weights in order to test and get the most probable class.
Thank you!
If you try to convert a factor into a number, it'll give you the number of the level in the factor. To convert into numbers, you can first run as.character, which will safely turn it into a format that you can run as.numeric on.
test <- as.factor(c(0, 1))
as.numeric(test)
# [1] 1 2
as.numeric(as.character(test))
# [1] 0 1
The R FAQ recommends a different approach for speed
7.10 How do I convert factors to numeric?
It may happen that when reading numeric data into R (usually, when reading in a file), they come in as factors. If f is such a factor object, you can use
as.numeric(as.character(f))
to get the numbers back. More efficient, but harder to remember, is
as.numeric(levels(f))[as.integer(f)]
In any case, do not call as.numeric() or their likes directly for the task at hand (as as.numeric() or unclass() give the internal codes).
I am trying to use svm() to classify my data. A sample of my data is as follows:
ID call_YearWeek week WeekCount oc
x 2011W01 1 0 0
x 2011W02 2 1 1
x 2011W03 3 0 0
x 2011W04 4 0 0
x 2011W05 5 1 1
x 2011W06 6 0 0
x 2011W07 7 0 0
x 2011W08 8 1 1
x 2011W09 9 0 0
x 2011W10 10 0 0
x 2011W11 11 0 0
x 2011W12 12 1 1
x 2011W13 13 1 1
x 2011W14 14 1 1
x 2011W15 15 0 0
x 2011W16 16 2 1
x 2011W17 17 0 0
x 2011W18 18 0 0
x 2011W19 19 1 1
The third column shows week of the year. The 4th column shows number of calls in that week and the last column is a binary factor (if a call was received in that week or not). I used the following lines of code:
train <- data[1:105,]
test <- data[106:157,]
model <- svm(oc~week,data=train)
plot(model,train,week)
plot(model,train)
none of the last two lines work. they dont show any plots and they return no error. I wonder why this is happening.
Thanks
Seems like there are two problems here, first is that not all svm types are supported by plot.svm -- only the classification methods are, and not the regression methods. Because your response is numeric, svm() assumes you want to do regression so it chooses "eps-regression" by default. If you want to do classification, change your response to a factor
model <- svm(factor(oc)~week,data=train)
which will then use "C-classification" by default.
The second problem is that there does not seem to be a univariate predictor plot implemented. It seems to want two variables (one for x and one for y).
It may be better to take a step back and describe exactly what you want your plot to look like.
I have a sample code in R as follows:
library(igraph)
rm(list=ls())
dat=read.csv(file.choose(),header=TRUE,row.names=1,check.names=T) # read .csv file
m=as.matrix(dat)
net=graph.adjacency(adjmatrix=m,mode="undirected",weighted=TRUE,diag=FALSE)
where I used csv file as input which contain following data:
23732 23778 23824 23871 58009 58098 58256
23732 0 8 0 1 0 10 0
23778 8 0 1 15 0 1 0
23824 0 1 0 0 0 0 0
23871 1 15 0 0 1 5 0
58009 0 0 0 1 0 7 0
58098 10 1 0 5 7 0 1
58256 0 0 0 0 0 1 0
After this I used following command to check weight values:
E(net)$weight
Expected output is somewhat like this:
> E(net)$weight
[1] 8 1 10 1 15 1 1 5 7 1
But I'm getting weird values (and every time different):
> E(net)$weight
[1] 2.121996e-314 2.121996e-313 1.697597e-313 1.291034e-57 1.273197e-312 5.092790e-313 2.121996e-314 2.121996e-314 6.320627e-316 2.121996e-314 1.273197e-312 2.121996e-313
[13] 8.026755e-316 9.734900e-72 1.273197e-312 8.027076e-316 6.320491e-316 8.190221e-316 5.092790e-313 1.968065e-62 6.358638e-316
I'm unable to find where and what I am doing wrong?
Please help me to get the correct expected result and also please tell me why is this weird output and that too every time different when I run it.??
Thanks,
Nitin
Just a small working example below, much clearer than CSV input.
library('igraph');
adjm1<-matrix(sample(0:1,100,replace=TRUE,prob=c(0.9,01)),nc=10);
g1<-graph.adjacency(adjm1);
plot(g1)
P.s. ?graph.adjacency has a lot of good examples (remember to run library('igraph')).
Related threads
Creating co-occurrence matrix
Co-occurrence matrix using SAC?
The problem seems to be due to the data-type of the matrix elements. graph.adjacency expects elements of type numeric. Not sure if its a bug.
After you do,
m <- as.matrix(dat)
set its mode to numeric by:
mode(m) <- "numeric"
And then do:
net <- graph.adjacency(m, mode = "undirected", weighted = TRUE, diag = FALSE)
> E(net)$weight
[1] 8 1 10 1 15 1 1 5 7 1