I know that in r-exam I can assign points to exercises eg. via expoints in the exercise' meta-data. However, I don't know to get the sum of the points across all exercises.
As a specific usecase: Consider a test that (by formal university requirements) must consist of (say) 90 points. So, I need to track the number of points that are already included via the exercises of the test.
I'm unaware which variable tracks this score (if any).
You are right, this information is not directly available, however it can be extracted from the metainformation contained in the output from any exams2xyz() interface. As a simple illustration consider:
library("exams")
set.seed(0)
exm <- exams2pdf(c("swisscapital.Rmd", "deriv.Rmd", "ttest.Rmd"),
n = 1, points = c(1, 17, 2))
Now exm is a list with only n = 1 exam, consisting of three exercises, each of which provides its metainform (among other details). So you can extract the points of the second exercise in the first (and only) exam via:
exm[[1]][[2]]$metainfo$points
## [1] 17
So to get the points from all exercises in the first exam:
sapply(exm[[1]], function(y) y$metainfo$points)
## exercise1 exercise2 exercise3
## 1 17 2
Of course, here it the points were explicitly set in exams2pdf() and were thus known. But the same approach can also be used if the points are set via the expoints tag in the individual exercises.
Related
I am just getting started with R so I am sorry if I say things that dont make sense.
I am trying to make a for loop which does the following,
l_dtest[[1]]<-vector()
l_dtest[[2]]<-vector()
l_dtest[[3]]<-vector()
l_dtest[[4]]<-vector()
l_dtest[[5]]<-vector()
all the way up till any number which will be assigned as n. for example, if n was chosen to be 100 then it would repeat this all the way to > l_dtest[[100]]<-vector().
I have tried multiple different attempts at doing this and here is one of them.
n<-4
p<-(1:n)
l_dtest<-list()
for(i in p){
print((l_dtest[i]<-vector())<-i)
}
Again I am VERY new to R so I don't know what I am doing or what is wrong with this loop.
The detailed background for why I need to do this is that I need to write an R function that receives as input the size of the population "n", runs a simulation of the model below with that population size, and returns the number of generations it took to reach a MRCA (most recent common ancestor).
Here is the model,
We assume the population size is constant at n. Generations are discrete and non-overlapping. The genealogy is formed by this random process: in each
generation, each individual chooses two parents at random from the previous generation. The choices are made randomly and equally likely over the n possibilities and each individual chooses twice. All choices are made independently. Thus, for example, it is possible that, when an individual chooses his two parents, he chooses the same individual twice, so that in
fact he ends up with just one parent; this happens with probability 1/n.
I don't understand the specific step at the begining of this post or why I need to do it but my teacher said I do. I don't know if this helps but the next step is choosing parents for the first person and then combining the lists from the step I posted with a previous step. It looks like this,
sample(1:5, 2, replace=T)
#[1] 1 2
l_dtemp[[1]]<-union(l_dtemp[[1]], l_d[[1]]) #To my understanding, l_dtem[[1]] is now receiving the listdescandants from l_d[[1]] bcs the ladder chose l_dtemp[[1]] as first parent
l_dtemp[[2]]<-union(l_dtemp[[2]], l_d[[1]]) #Same as ^^ but for l_d[[1]]'s 2nd choice which is l_dtemp[[2]]
sample(1:5, 2, replace=T)
#[1] 1 3
l_dtemp[[1]]<-union(l_dtemp[[1]], l_d[[2]])
l_dtemp[[3]]<-union(l_dtemp[[3]], l_d[[2]])
I am running a machine learning algorithm that uses CAT score for feature selection as
library(sda)
train1<- data.matrix(train, rownames.force = NA)
ranking.LDA = sda.ranking(train1[,1:lengthvar], train1[,lengthtrain], diagonal=FALSE)
topfs<-which(ranking.LDA[,"score"] >2)
My question is how to ask the CAT score to give me for example top 20 features? The only way I could extract features was setting a threshold, but this way, it gives me various number of features for different data set. What I want is always having eg. top 20 (or any other number) features.
Thanks in advance for your valuable contribution.
ranking.LDA gives a list of numbers.Hence we use a list function.
#As ranking.LDA gives a ranking of predictors we directly extract column names using this ranking.
colnames(train1[,ranking.LDA[1:20]])
Background:
On this combinatorics question, the issue is how to determine the sample space: the ways 8 different soccer teams can be paired up for the next round of competition. Two different answers have been advanced for that part of the problem: 28 (see comments OP) and 105 (see edit within OP and answer).
I'd like to do this manually to try to hone down on the mistake in whichever answer is incorrect.
What I have tried:
teams = 1:8
names(teams) = c("RM", "BCN", "SEV", "JUV", "ROM", "MC", "LIV", "BYN")
split(sample(teams), rep(1:(length(teams)/2), each=2))
Unfortunately, the output is a list, and I wanted a vector to be able to run something like:
unique(...,MARGIN=2)
Is there a way of doing this in an elegant manner?
After a now erased answer (thank you), I would go with
a <- replicate(1e5, unlist(split(sample(teams), rep(1:(length(teams)/2), each=2))))
to simulate 100,000 random samples, and later run
unique(a, MARGIN = 2).
But how can I account for the fact that the order of the 4 pairings of opponents doesn't matter, and that LIV-BYN and BYN-LIV, for example, is the same pairing (field advantage notwithstanding)?
> u = ncol(unique(replicate(1e6, unlist(split(sample(teams), rep(1:(length(teams)/2), each=2)))), MARGIN = 2))
> u / (factorial(4) * 2^4)
[1] 105
The idea of unlist is from #Song Zhengyi, and if his answer is un-deleted, I'll accept it. The complete answer is in the lines above.
u needs to be divided by 4! because
BCN-RM, BYN-SEV, JUV-ROM, LIV-MC
is exactly the same as
LIV-MC, BCN-RM, BYN-SEV, JUV-ROM
or
BCN-RM, LIV-MC, BYN-SEV, JUV-ROM
etc.
The term 2^4 is to avoid over-counting since for every possible unique draw, each one of the pairings can be flipped without loss (discarding field advantage): BCN-RM is the same as RM-BCN, and there are 4 pairs in each draw.
If field advantage is a consideration (real life)...
> u/factorial(4)
[1] 1680
we end up with 1,680 possible draws.
Lets say I have this data. My objective is to extraxt combinations of sequences.
I have one constraint, the time between two events may not be more than 5, lets call this maxGap.
User <- c(rep(1,3)) # One users
Event <- c("C","B","C") # Say this is random events could be anything from LETTERS[1:4]
Time <- c(c(1,12,13)) # This is a timeline
df <- data.frame(User=User,
Event=Event,
Time=Time)
If want to use these sequences as binary explanatory variables for analysis.
Given this dataframe the result should be like this.
res.df <- data.frame(User=1,
C=1,
B=1,
CB=0,
BC=1,
CBC=0)
(CB) and (CBC) will be 0 since the maxGap > 5.
I was trying to write a function for this using many for-loops, but it becomes very complex if the sequence becomes larger and the different number of evets also becomes larger. And also if the number of different User grows to 100 000.
Is it possible of doing this in TraMineR with the help of seqeconstraint?
Here is how you would do that with TraMineR
df.seqe <- seqecreate(id=df$User, timestamp=df$Time, event=df$Event)
constr <- seqeconstraint(maxGap=5)
subseq <- seqefsub(df.seqe, minSupport=0, constraint=constr)
(presence <- seqeapplysub(subseq, method="presence"))
which gives
(B) (B)-(C) (C)
1-(C)-11-(B)-1-(C) 1 1 1
presence is a table with a column for each subsequence that occurs at least once in the data set. So, if you have several individuals (event sequences), the table will have one row per individual and the columns will be the binary variable you are looking for. (See also TraMineR: Can I get the complete sequence if I give an event sub sequence? )
However, be aware that TraMineR works fine only with subsequences of length up to about 4 or 5. We suggest to set maxK=3 or 4 in seqefsub. The number of individuals should not be a problem, nor should the number of different possible events (the alphabet) as long as you restrict the maximal subsequence length you are looking for.
Hope this helps
For a text analysis program, I would like to analyze the co-occurrence of certain words in a text. For example, I would like to see that e.g. the words "Barack" and "Obama" appear more often together (i.e. have a positive correlation) than others.
This does not seem to be that difficult. However, to be honest, I only know how to calculate the correlation between two numbers, but not between two words in a text.
How can I best approach this problem?
How can I calculate the correlation between words?
I thought of using conditional probabilities, since e.g. Barack Obama is much more probable than Obama Barack; however, the problem I try to solve is much more fundamental and does not depend on the ordering of the words
The Ngram Statistics Package (NSP) is devoted precisely to this task. They have a paper online which describes the association measures they use. I haven't used the package myself, so I cannot comment on its reliability/requirements.
Well a simple way to solve your question is by shaping the data in a 2x2 matrix
obama | not obama
barack A B
not barack C D
and score all occuring bi-grams in the matrix. That way you can for instance use simple chi squared.
I don't know how this is commonly done, but I can think of one crude way to define a notion of correlation that captures word adjacency.
Suppose the text has length N, say it is an array
text[0], text[1], ..., text[N-1]
Suppose the following words appear in the text
word[0], word[1], ..., word[k]
For each word word[i], define a vector of length N-1
X[i] = array(); // of length N-1
as follows: the ith entry of the vector is 1 if the word is either the ith word or the (i+1)th word, and zero otherwise.
// compute the vector X[i]
for (j = 0:N-2){
if (text[j] == word[i] OR text[j+1] == word[i])
X[i][j] = 1;
else
X[i][j] = 0;
}
Then you can compute the correlation coefficient between word[a] and word[b] as the dot product between X[a] and X[b] (note that the dot product is the number of times these words are adjacent) divided by the lenghts (the length is the square root of the number of appearances of the word, well maybe twice that). Call this quantity COR(X[a],X[b]). Clearly COR(X[a],X[a]) = 1, and COR(X[a],X[b]) is larger if word[a], word[b] are often adjacent.
This can be generalized from "adjacent" to other notions of near - for example we could have chosen to use 3 word (or 4, 5, etc.) blocks instead. One can also add weights, probably do many more things as well if desired. One would have to experiment to see what is useful, if any of it is of use at all.
This problem sounds like a bigram, a sequence of two "tokens" in a larger body of text. See this Wikipedia entry, which has additional links to the more general n-gram problem.
If you want to do a full analysis, you'd most likely take any given pair of words and do a frequency analysis. E.g., the sentence "Barack Obama is the Democratic candidate for President," has 8 words, so there are 8 choose 2 = 28 possible pairs.
You can then ask statistical questions like, "in how many pairs does 'Obama' follow 'Barack', and in how many pairs does some other word (not 'Obama') follow 'Barack'? In this case, there are 7 pairs that include 'Barack' but in only one of them is it paired with 'Obama'.
Do the same for every possible word pair (e.g., "in how many pairs does 'candidate' follow 'the'?"), and you've got a basis for comparison.