I am doing some clustering of documents using cosine similarity between each document. This is fine. However my problem is a little strange in that I only want to cluster certain documents with others, not all of the documents against each other. Here's an example...
I have two spreadsheets with 3 labels apiece. I want to cluster the labels that are similar to each other BETWEEN the documents but not in the internal of the document, so for instance
Doc1: has labels: sex and gender, tobacco use years, current age
Doc2: has labels: gender, age now, time of use
I want to cluster the labels between the two documents but not inside the document, so I've created a similarity matrix that looks like this:
d1_l1 d1_l2 d1_l3 d2_l1 d2_l2 d2_l3
d1_l1 1.0000000 NA NA 0.5773503 0.0 0.0000000
d1_l2 NA 1.0000000 NA 0.0000000 0.0 0.3333333
d1_l3 NA NA 1.0 0.0000000 0.5 0.0000000
d2_l1 0.5773503 0.0000000 0.0 1.0000000 NA NA
d2_l2 0.0000000 0.0000000 0.5 NA 1.0 NA
d2_l3 0.0000000 0.3333333 0.0 NA NA 1.0000000
where the cosine similarity between labels in the same document is set as NA. The problem is that agnes and other hierarchical clustering methods don't accept NA values. So what should I do? Am I thinking about this the wrong way?
Related
I have an sf object containing the following columns of data:
HR60 HR70 HR80 HR90 HC60 HC70 HC80 HC90
0.000000 0.000000 8.855827 0.000000 0.0000000 0.0000000 0.3333333 0.0000000
0.000000 0.000000 17.208742 15.885624 0.0000000 0.0000000 1.0000000 1.0000000
1.863863 1.915158 3.450775 6.462453 0.3333333 0.3333333 1.0000000 2.0000000
...
How can I calculate the median of HR60 to HR90 columns for all observations and place it in a different column, let's say HR-median? I tried to use apply(), but this kind of works for the whole dataset only and I need only these 4 columns to be considered.
We can select those columns
df1$HR_median <- apply(subset(df1, select = HR60:HR90), 1, median)
I have a dataset with five categorical variables. And I ran a multinomial logistic regression with the function multinom in package nnet, and then derived the p values from the coefficients. But I do not know how to interpret the results.
The p values were derived according to UCLA's tutorial: https://stats.idre.ucla.edu/r/dae/multinomial-logistic-regression/ .
Just like this:
z <- summary(test)$coefficients/summary(test)$standard.errors
p <- (1 - pnorm(abs(z), 0, 1)) * 2
p
And I got this:
(Intercept) Age1 Age2 Age3 Age4 Unit1 Unit2 Unit3 Unit4 Unit5 Level1 Level2 Area1 Area2
Not severe 0.7388029 9.094373e-01 0 0.000000e+00 0.000000e+00 0 0.75159758 0 0 0.0000000 0.8977727 0.9333862 0.6285447 0.4457171
Very severe 0.0000000 1.218272e-09 0 6.599380e-06 7.811761e-04 0 0.00000000 0 0 0.0000000 0.7658748 0.6209889 0.0000000 0.0000000
Severe 0.0000000 8.744405e-08 0 1.052835e-06 3.299770e-04 0 0.00000000 0 0 0.0000000 0.8843606 0.4862364 0.0000000 0.0000000
Just so so 0.0000000 1.685045e-07 0 5.507560e-03 2.973261e-06 0 0.08427447 0 NaN 0.3010429 0.5552963 0.7291180 0.0000000 0.0000000
Not severe at all 0.0000000 0.000000e+00 0 0.000000e+00 0.000000e+00 0 NaN NaN 0 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
But how should I interpret these p values? Age3 is significantly related to Very severe? I am green to statistics and have no idea. Help me understand the results please. Thank you in advance.
I suggest using stargazer package to display coefficients and p-values (I believe that it is a more convenient and common way)
Regarding the interpretation of the results, in a multinomial model you can say: keeping all other variables constant, if Age3 is higher by one unit, the log odds for Very Severe relative to the reference category is higher/lower by that amount indicated by the value of the coefficient. The p-value just shows you whether the association between these two variables (predictor and response) is significant or not. Interpretation is the same that of other models.
Note: in case of p-value the null hypothesis is always that the coefficient is equal to zero (no effect at all). When p-value is less than 0.05, you can safely reject the null hypothesis and state that the predictor has an effect on the response variable.
I hope I could give you some hints
I am having some troubles regarding a cluster analysis that I am trying to do with the pvclust package.
Specifically, I have a data matrix composed by species (rows) and sampling stations (columns). I want to perform a CA in order to group my sampling stations according to my species abundance (which I have previously log(x+1) transformed).
Once having prepared adequately my matrix,I've tried to run a CA according to the pvclust package, using Ward's clustering method and Bray-Curtis as distance index. However, every time I get the following error message:
''Error in hclust(distance, method = method.hclust) :
invalid clustering method''
I then tried to perform the same analysis using another cluster method, and I had no problem. I also tried to perform the same analysis using the hclust function from the vegan package, and I had no problem at all, too. The analysis run without any problems.
To better understand my problem, I'll display part of my matrix and the
script that I used to perfrom the analysis:
P1 P2 P3 P4 P5 P6
1 10.8750000 3.2888889 2.0769231 1.4166667 3.2395833 5.333333
3 0.3645833 0.3027778 0.3212038 0.7671958 0.4993676 0.000000
4 0.0000000 0.0000000 2.3500000 0.0000000 0.0000000 0.264000
5 0.0000000 0.7333333 0.2692308 0.0000000 0.2343750 0.000000
6 0.0000000 0.9277778 0.0000000 0.2936508 0.7291667 0.000000
7 0.4166667 6.3500000 1.0925463 0.5476190 0.1885169 0.000000
8 1.6250000 0.0000000 0.0000000 0.0000000 5.2187500 0.000000
9 0.0000000 0.8111111 0.0000000 0.0000000 0.0000000 0.000000
10 2.6770833 0.6666667 2.3304890 4.5906085 2.9652778 0.000000
15 1.8020833 0.9666667 1.4807137 3.3878968 0.1666667 0.000000
16 17.8750000 4.9555556 1.4615385 6.5000000 7.8593750 7.666667
19 4.5312500 1.0555556 3.5766941 6.7248677 2.3196181 0.000000
20 0.0000000 0.6777778 0.5384615 0.0000000 0.0000000 0.000000
21 0.0000000 0.9777778 0.0000000 0.2500000 0.0000000 0.000000
24 1.2500000 3.0583333 0.1923077 0.0000000 4.9583333 0.000000
25 0.0000000 0.0000000 2.5699634 0.0000000 0.0000000 0.000000
26 6.6666667 2.2333333 24.8730020 55.9980159 17.6239583 0.000000
Where P1-P6 are my sampling stations, and the leftmost row numbers are my different species. I'll denote this example matrix just as ''platforms''.
Afterwards, I've used the following code lines:
dist <- function(x, ...){
vegdist(x, ...)
}
result<-pvclust(platforms,method.dist = "bray",method.hclust = "ward")
It is noteworthy that I run the three first codelines, since the bray-curtis index isn't originally available in the pvclust package. Thus, running these codelines allowed me to specify the bray-curtis index in the pvclust function
Does anyone know why it doesn't work with the pvclust package?
Any help will be much appreciated.
Kind regards,
Marie
There are two related issues:
When calling method.hclust you need to pass hclust compatible methods. In theory pvclust checks for ward and converts to ward.D, but you probably want to pass the (correct) names of either ward.D or ward.D2.
You cannot over-write dist in that fashion. However, you can pass a custom function to pvclust.
For instance, this should work:
library(vegan)
library(pvclust)
sample.data <- "P1 P2 P3 P4 P5 P6
10.8750000 3.2888889 2.0769231 1.4166667 3.2395833 5.3333330
0.3645833 0.3027778 0.3212038 0.7671958 0.4993676 0.0000000
0.0000000 0.0000000 2.3500000 0.0000000 0.0000000 0.2640000
0.0000000 0.7333333 0.2692308 0.0000000 0.2343750 0.0000000
0.0000000 0.9277778 0.0000000 0.2936508 0.7291667 0.0000000
0.4166667 6.3500000 1.0925463 0.5476190 0.1885169 0.0000000
1.6250000 0.0000000 0.0000000 0.0000000 5.2187500 0.0000000
0.0000000 0.8111111 0.0000000 0.0000000 0.0000000 0.0000000
2.6770833 0.6666667 2.3304890 4.5906085 2.9652778 0.0000000
1.8020833 0.9666667 1.4807137 3.3878968 0.1666667 0.0000000
17.8750000 4.9555556 1.4615385 6.5000000 7.8593750 7.6666670
4.5312500 1.0555556 3.5766941 6.7248677 2.3196181 0.0000000
0.0000000 0.6777778 0.5384615 0.0000000 0.0000000 0.0000000
0.0000000 0.9777778 0.0000000 0.2500000 0.0000000 0.0000000
1.2500000 3.0583333 0.1923077 0.0000000 4.9583333 0.0000000
0.0000000 0.0000000 2.5699634 0.0000000 0.0000000 0.0000000
6.6666667 2.2333333 24.8730020 55.9980159 17.6239583 0.0000000"
platforms <- read.table(text = sample.data, header = TRUE)
result <- pvclust(platforms,
method.dist = function(x){
vegdist(x, "bray")
},
method.hclust = "ward.D")
To begin, I have been able to put together a nested for loop to create the object I am after, and it works OK for a small toy data set, but the data I will be working with in general will be larger and I am trying to determine if a package exists in R with a built in function to accomplish this task.
The final object is a data frame or matrix that shows the conditional percentage in a column given the reference row. Here is the code for a toy data and my nested for loop that generates the final output object.
mylist <- list(
ID001=c("apple","orange","grape"),
ID002=c("banana","grape"),
ID003=c("apple","pineapple"),
ID004=c("orange","apple"),
ID005=c("orange","grape", "apple"))
dat <- reshape2:::melt(mylist)
names(dat) <- c("fruit","id")
dat <- dat[,c(2,1)]
theFruit <- unique(dat$fruit)
n=length(theFruit)
final.df <- data.frame(matrix(nrow=n,ncol=n, dimnames=list(theFruit,theFruit)))
for(i in theFruit){
for(j in theFruit){
tempid1 <- dat[dat$fruit==i,]$id
tempid2 <- dat[dat$fruit==j,]$id
final.df[i,j] <- round(length(which(tempid1%in%tempid2))/length(tempid1),2)
}
}
final.df
apple orange grape banana pineapple
apple 1.00 0.75 0.50 0.00 0.25
orange 1.00 1.00 0.67 0.00 0.00
grape 0.67 0.67 1.00 0.33 0.00
banana 0.00 0.00 1.00 1.00 0.00
pineapple 1.00 0.00 0.00 0.00 1.00
Reading the output we see that, given a person ate an apple (apple row), 75% also ate an orange (orange column). Similarly, given a person ate an orange (orange row) 100% also ate an apple (apple column). This is not intended to be symmetric with intersections of the two fruits eaten, it is column conditioned on row.
This seems to be akin to a market basket analysis application and I have been working with the arules package the past few days to get at this. In the vernacular of the arules package, I would say the name of the percentages populating the data frame are support values but I have not been able to generate a matrix or data frame of all of the support percentages from arules.
The data I will be working with will have a couple million IDs but only about 150 "products" so the output matrix would only be about 150x150. I can use arules to identify the compelling pairwise relationships but there is interest in seeing ALL of the conditionals.
Does anyone know if arules or another package can accomplish this?
You are looking for the confidence values (Wikipedia). You get a similar output to yours like this with arules:
library(arules)
library(reshape2)
trans <- as(mylist, "transactions")
rules <- apriori(trans, parameter = list(supp = 0, conf = 0, minlen=2, maxlen=2))
df <- inspect(rules)[, c("lhs", "rhs", "confidence")]
dcast(df, lhs~rhs, value.var="confidence", fill=1)
# lhs {apple} {banana} {grape} {orange} {pineapple}
# 1 {apple} 1.0000000 0.0000000 0.5000000 0.7500000 0.25
# 2 {banana} 0.0000000 1.0000000 1.0000000 0.0000000 0.00
# 3 {grape} 0.6666667 0.3333333 1.0000000 0.6666667 0.00
# 4 {orange} 1.0000000 0.0000000 0.6666667 1.0000000 0.00
# 5 {pineapple} 1.0000000 0.0000000 0.0000000 0.0000000 1.00
Of course you can make the first column to row names and convert the data frame to a matrix later on. I leave it up to you.
When using the knn() function in package class in R, there is an argument called "prob". If I make this true, I get the probability of that particular value being classified to whatever it is classified as.
I have a dataset where the classifier has 9 levels. Is there any way in which I can get the probability of a particular observation for all the 9 levels?
As far as I know the knn() function in class only returns the highest probability.
However, you can use the knnflex package which allows you to return all probability levels using knn.probability (see here, page 9-10).
This question still require proper answer.
If the probability for the most probable class is needed then the class package will be still suited. The clue is to set the argument prob to TRUE and k to higher than default 1 - class::knn(tran, test, cl, k = 5, prob = TRUE). The k has to be higher than default 1 to not get always 100% probability for each observation.
However if you want to get probabilities for each of the classes I will recommend the caret::knn3 function with predict one.
data(iris3)
train <- rbind(iris3[1:25,,1], iris3[1:25,,2], iris3[1:25,,3])
test <- rbind(iris3[26:50,,1], iris3[26:50,,2], iris3[26:50,,3])
cl <- factor(c(rep("s",25), rep("c",25), rep("v",25)))
# class package
# take into account k higher than 1 and prob equal TRUE
model <- class::knn(train, test, cl, k = 5, prob = TRUE)
tail(attributes(model)$prob, 10)
#> [1] 1.0 1.0 1.0 1.0 1.0 1.0 0.8 1.0 1.0 0.8
# caret package
model2 <- predict(caret::knn3(train, cl, k = 3), test)
tail(model2, 10)
#> c s v
#> [66,] 0.0000000 0 1.0000000
#> [67,] 0.0000000 0 1.0000000
#> [68,] 0.0000000 0 1.0000000
#> [69,] 0.0000000 0 1.0000000
#> [70,] 0.0000000 0 1.0000000
#> [71,] 0.0000000 0 1.0000000
#> [72,] 0.3333333 0 0.6666667
#> [73,] 0.0000000 0 1.0000000
#> [74,] 0.0000000 0 1.0000000
#> [75,] 0.3333333 0 0.6666667
Created on 2021-07-20 by the reprex package (v2.0.0)
I know there is an answer already marked here, but this is possible to complete without utilizing another function or package.
What you can do instead is build your knn model knn_model and check out it's attributes for the "prob" output, as such.
attributes(knn_model)$prob