problem with plotting several plots with mcmc_areas() function - r
So, I wanted to create several plots of my posterior sample using bayesplot.
My data contains several draws divided into features (the column "correlation" contains the posterior values):
head(cors_feat[cors_feat$feature==1,])
feature correlation
1 1 -0.5002517
2 1 0.1964202
3 1 -0.8603918
4 1 -0.7111672
5 1 0.1760033
6 1 0.2788863
head(cors_feat[cors_feat$feature==2,])
feature correlation
40001 2 -0.7149123
40002 2 -0.7666210
40003 2 -0.7542937
40004 2 -0.2094619
40005 2 0.4089077
40006 2 -0.8550481
I loop through the features, and in the loop I use the function mcmc_areas() to create the plots:
lst <- list()
for (i in 1:length(unique(cors_feat$feature))){
lst <- c(lst,mcmc_areas(cors_feat[cors_feat$feature==i,], pars = c("correlation"), prob = 0.95))
}
When I try to use the plot_grid function from from the "cowplot" library I get the following error:
plot_grid(plotlist = lst)
There were 50 or more warnings (use warnings() to see the first 50)
And most importantly, I cannot see any plot.
How can I solve this issue?
Related
Run DBSCAN against grouped coordinates
I'm attempting to run DBSCAN against some grouped coordinates in order to get sub-clusters. I've clustered some spacial data and I'd now like to further divide these regions according to the density of points within them. I think DBSCAN is probably the best way to do this.
My issue is that I can't figure out how to run DBSCAN against each cluster seperately and then output the cluster assignment as a new column. Here's some sample data:
library(dplyr)
library(dbscan)
# Create sample data
df <- data.frame(
"ID"=1:200,
"X"=c(1.0083,1.3166,1.3072,1.1311,1.2984,1.2842,1.1856,1.3451,1.1932,1.0926,1.2464,1.3197,1.2331,1.2996,1.3482,
1.1944,1.2800,1.3051,1.4471,0.9068,1.3150,1.1846,1.0232,1.0005,1.0640,1.3177,1.1015,0.9598,1.0354,1.2203,
0.8388,0.8655,1.3387,1.0133,1.0106,1.1753,1.3200,1.0139,1.1511,1.3508,1.2747,1.3681,1.1074,1.2735,1.2245,
0.9695,1.3250,1.0537,1.2020,1.3093,0.9268,1.3244,1.2626,1.3123,1.2819,1.1063,0.8759,1.0063,1.0173,1.0187,
1.2396,1.0241,1.2619,1.2682,1.0008,1.0827,1.3639,1.3099,1.0004,0.8886,1.2359,1.1370,1.2783,1.0803,1.1918,
1.1156,1.3313,1.1205,1.0776,1.3895,1.3559,0.8518,1.1315,1.3521,1.2281,1.2589,0.9974,1.1487,1.4204,0.9998,
1.0154,1.0098,0.8851,1.0252,0.9331,1.2197,1.0084,1.2303,1.2808,1.3125,0.5500,0.6694,0.3301,0.3787,0.6492,
0.6568,0.6773,0.3769,0.6237,0.7265,0.5509,0.3579,0.7201,0.2631,0.3881,0.7596,0.3343,0.7049,0.3430,0.2951,
0.5483,0.7699,0.3806,0.6555,0.2524,0.4030,0.6329,0.5006,0.2701,0.0822,0.5442,0.5233,0.7105,0.5660,0.3962,
0.3187,0.3143,0.5673,0.3731,0.7310,0.6376,0.4864,0.8865,0.3352,0.7540,0.0690,0.7983,0.6990,0.4090,0.5658,
0.5636,0.5420,0.7223,0.6146,0.5648,0.2711,0.3422,0.7214,0.2196,0.2848,0.6496,0.7907,0.7418,0.7825,0.4550,
0.4361,0.7417,0.2661,0.8978,0.7875,0.2343,0.3853,0.6874,0.7761,0.2905,0.6092,0.5329,0.6189,0.0684,0.5726,
0.5740,0.7060,0.4609,0.3568,0.7037,0.2874,0.6200,0.7149,0.5100,0.7059,0.2520,0.3105,0.6870,0.7888,0.3674,
0.6514,0.7271,0.6679,0.3752,0.7067),
"Y"=c(-1.2547,-1.1499,-1.1803,-1.0626,-1.2877,-1.1151,-1.0958,-1.1339,-1.0808,-1.5461,-1.0775,-1.1431,-1.0499,
-1.1521,-1.1675,-1.0963,-1.1407,-1.1916,-1.1229,-1.2297,-1.1308,-1.0341,-1.3071,-1.2370,-1.5043,-1.1154,
-1.5452,-1.0349,-1.5412,-1.0348,-1.3620,-1.3776,-1.1830,-1.2552,-1.2354,-1.0838,-1.1214,-1.2396,-1.4937,
-1.0793,-1.1857,-1.0679,-1.5425,-1.1633,-1.1620,-1.0838,-1.0750,-1.3493,-1.4155,-1.1354,-1.0615,-1.1494,
-1.1620,-1.1582,-1.1800,-1.5230,-1.3019,-1.2484,-1.5490,-1.2435,-1.0487,-1.2330,-1.1234,-1.0924,-1.0702,
-1.0446,-1.1077,-1.1144,-1.2170,-1.2715,-1.1537,-1.5077,-1.1305,-1.3396,-1.2107,-1.5458,-1.1482,-1.1224,
-1.3690,-1.2058,-1.1685,-1.3400,-1.5033,-1.2152,-1.3805,-1.1439,-1.5183,-1.4288,-1.1252,-1.2330,-1.2511,
-1.5429,-1.3333,-1.1851,-1.1367,-1.3952,-1.1240,-1.2113,-1.1632,-1.1965,-0.9917,-0.7416,-0.7729,-1.1279,
-0.9323,-0.9372,-0.7013,-1.1746,-0.9191,-0.9356,-0.7873,-1.1957,-0.9838,-0.5825,-1.0738,-0.9302,-0.7713,
-0.9407,-0.7774,-0.8160,-0.9861,-1.0440,-0.9896,-0.6478,-0.8865,-1.0601,-1.0640,-0.9898,-0.5989,-0.7375,
-0.7689,-0.9799,-0.9147,-1.1048,-0.9735,-0.8591,-0.7913,-1.0085,-0.7231,-0.9688,-0.9272,-0.9395,-0.9494,
-0.7859,-1.0817,-0.7262,-0.9915,-0.9329,-1.0953,-1.0425,-1.0806,-1.0132,-0.8514,-1.0785,-1.1109,-0.8542,
-1.0849,-0.9665,-0.5940,-0.6145,-0.7830,-0.9601,-0.8996,-0.7717,-0.7447,-1.0406,-1.0067,-0.5710,-0.9839,
-1.0594,-0.7069,-1.1202,-0.9705,-1.0100,-0.6377,-1.0632,-0.9450,-0.9163,-0.7865,-1.0090,-1.1005,-1.0049,
-0.8042,-1.0781,-0.6829,-0.5962,-1.0759,-0.7918,-0.9732,-0.7353,-0.5615,-1.2002,-0.9295,-0.9944,-1.1570,
-0.9524,-0.9257,-0.9360,-1.1328,-0.7661),
"cluster"=c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,
2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2))
# How do you run DBSCAN against the points within each cluster?
I first thought I'd try to use the group_by function in dplyr but DBSCAN requires a data matrix input and group_by doesn't work for matrices.
matrix <- as.matrix(df[, -1])
set.seed(1234)
db = matrix %>%
group_by(cluster) %>%
dbscan(matrix, 0.4, 4)
#Error in UseMethod("group_by_") :
# no applicable method for 'group_by_' applied to an object of class "c('matrix', 'double', 'numeric')"
I've also tried using by() but get duplicate results for each cluster grouping, which isn't right:
by(data = df, INDICES = df$cluster, FUN = function(x) {
out <- dbscan(as.matrix(df[, c(2:3)]),eps=.0215,minPts=4)
})
#df$cluster: 1
#DBSCAN clustering for 200 objects.
#Parameters: eps = 0.0215, minPts = 4
#The clustering contains 10 cluster(s) and 138 noise points.
#
# 0 1 2 3 4 5 6 7 8 9 10
#138 11 12 4 5 8 2 4 8 4 4
#
#Available fields: cluster, eps, minPts
#--------------------------------------------------------------------------
#df$cluster: 2
#DBSCAN clustering for 200 objects.
#Parameters: eps = 0.0215, minPts = 4
#The clustering contains 10 cluster(s) and 138 noise points.
#
# 0 1 2 3 4 5 6 7 8 9 10
#138 11 12 4 5 8 2 4 8 4 4
#
#Available fields: cluster, eps, minPts
Can anyone point me in the right direction?
To be clear, dbscan::dbscan works fine on data.frame objects. You do not need to convert to matrix. It returns an object that includes a vector with the same dimension as the number of records in your input. The issue is that dplyr exposes variables to other functions as individual vectors, rather than as data.frame or matrix objects. You are free to do something like: df %>% group_by(cluster) %>% mutate( dbscan_cluster = dbscan::dbscan( data.frame(X, Y), eps = 0.0215, minPts = 4 )[["cluster"]] ) dplyr is not necessary, by also works, you just need to supply a generic function rather than one that directly references the source object directly. Your data must already be ordered by cluster. df$dbscan_cluster <- unlist( by( df, INDICES = df$cluster, function(x) dbscan::dbscan(x[,c(2,3)], eps = 0.0215, minPts = 4)[["cluster"]] ) ) However, you can still get garbage results if you don't have a good way to pick your epsilon. You might consider using dbscan::optics instead.
Fixing an error in R- "Incorrect number of dimensions" in the Dunn Test
I am trying to use the Dunn test for a comparison but I am getting an error: "Error in Psort[1, i] : incorrect number of dimensions" the data I am trying to use is this sort of idea (but sample size is bigger): Frequency Height 1 10 2 11 1 9 1 8 2 15 1 9 2 11 2 13 the code I used was dunnTest(Height ~ Frequency, data=Data, method="bh") is my problem that my frequency is only split into two groups? cause for another factor my frequency had three groups and it worked fine. If this is the problem, is there another test I can do that will perform a similar/the same function? Thanks!
The Dunn test is equivalent to the Wilcox test (wilcox.test) if you adjust values of input parameters (disable the exact calculation of p value, disable the continuity correction, more here). For your data, one obtains: > wilcox.test(df$Frequency, df$Height, correct = FALSE, exact = FALSE) Wilcoxon rank sum test data: df$Frequency and df$Height W = 0, p-value = 0.0006346 alternative hypothesis: true location shift is not equal to 0 I think you are using the dunnTest function from the FSA package. This function fails for two groups. Data df <- read.table(text="Frequency Height 1 10 2 11 1 9 1 8 2 15 1 9 2 11 2 13", header=TRUE)
Creation prediction function for kmean in R
I want create predict function which predicts for which cluster, observation belong
data(iris)
mydata=iris
m=mydata[1:4]
train=head(m,100)
xNew=head(m,10)
rownames(train)<-1:nrow(train)
norm_eucl=function(train)
train/apply(train,1,function(x)sum(x^2)^.5)
m_norm=norm_eucl(train)
result=kmeans(m_norm,3,30)
predict.kmean <- function(cluster, newdata)
{
simMat <- m_norm(rbind(cluster, newdata),
sel=(1:nrow(newdata)) + nrow(cluster))[1:nrow(cluster), ]
unname(apply(simMat, 2, which.max))
}
## assign new data samples to exemplars
predict.kmean(m_norm, x[result$cluster, ], xNew)
After i get the error
Error in predict.kmean(m_norm, x[result$cluster, ], xNew) :
unused argument (xNew)
i understand that i am making something wrong function, cause I'm just learning to do it, but I can't understand where exactly.
indeed i want adopt similar function of apcluster ( i had seen similar topic, but for apcluster)
predict.apcluster <- function(s, exemplars, newdata)
{
simMat <- s(rbind(exemplars, newdata),
sel=(1:nrow(newdata)) + nrow(exemplars))[1:nrow(exemplars), ]
unname(apply(simMat, 2, which.max))
}
## assign new data samples to exemplars
predict.apcluster(negDistMat(r=2), x[apres#exemplars, ], xNew)
how to do it?
Rather than trying to replicate something, let's come up with our own function. For a given vector x, we want to assign a cluster using some prior k-means output. Given how k-means algorithm works, what we want is to find which cluster's center is closest to x. That can be done as predict.kmeans <- function(x, newdata) apply(newdata, 1, function(r) which.min(colSums((t(x$centers) - r)^2))) That is, we go over newdata row by row and compute the corresponding row's distance to each of the centers and find the minimal one. Then, e.g., head(predict(result, train / sqrt(rowSums(train^2))), 3) # 1 2 3 # 2 2 2 all.equal(predict(result, train / sqrt(rowSums(train^2))), result$cluster) # [1] TRUE which confirms that our predicting function assigned all the same clusters to the training observations. Then also predict(result, xNew / sqrt(rowSums(xNew^2))) # 1 2 3 4 5 6 7 8 9 10 # 2 2 2 2 2 2 2 2 2 2 Notice also that I'm calling simply predict rather than predict.kmeans. That is because result is of class kmeans and a right method is automatically chosen. Also notice how I normalize the data in a vectorized manner, without using apply.
R multiclass/multinomial classification ROC using multiclass.roc (Package ‘pROC’)
I am having difficulties understanding how the multiclass.roc parameters should look like. Here a snapshot of my data: > head(testing.logist$cut.rank) [1] 3 3 3 3 1 3 Levels: 1 2 3 > head(mnm.predict.test.probs) 1 2 3 9 1.013755e-04 3.713862e-02 0.96276001 10 1.904435e-11 3.153587e-02 0.96846413 12 6.445101e-23 1.119782e-11 1.00000000 13 1.238355e-04 2.882145e-02 0.97105472 22 9.027254e-01 7.259787e-07 0.09727389 26 1.365667e-01 4.034372e-01 0.45999610 > I tried calling multiclass.roc with: multiclass.roc( response=testing.logist$cut.rank, predictor=mnm.predict.test.probs, formula=response~predictor ) but naturally I get an error: Error in roc.default(response, predictor, levels = X, percent = percent, : Predictor must be numeric or ordered. When it's a binary classification problem I know that 'predictor' should contain probabilities (one per observation). However, in my case, I have 3 classes, so my predictor is a list of rows that each have 3 columns (or a sublist of 3 values) correspond to the probability for each class. Does anyone know how should my 'predictor' should look like rather than what it's currently look like ?
The pROC package is not really designed to handle this case where you get multiple predictions (as probabilities for each class). Typically you would assess your P(class = 1) multiclass.roc( response=testing.logist$cut.rank, predictor=mnm.predict.test.probs[,1]) And then do it again with P(class = 2) and P(class = 3). Or better, determine the most likely class: predicted.class <- apply(mnm.predict.test.probs, 1, which.max) multiclass.roc( response=testing.logist$cut.rank, predictor=predicted.class) Consider multiclass.roc as a toy that can sometimes be helpful but most likely won't really fit your needs.
Generating random graph in r
I would like to generate a grandom graph in R using any of the packages. The desired output would be a two column matrix with the first column listing agents and the second column their connections of the following form: 1 3 1 4 1 6 1 7 2 2 2 5 3 9 3 11 3 32 3 43 3 2 4 5 I would like to be able to specify the average degree and minimum and maximum number of contacts. What is the easiest way of doing this?
Since you don't specify the need for anything other than just a graph we ca do this very simply:
actor <- sample(1:4, 10, replace=TRUE)
receiver <- sample(3:43, 10, replace=TRUE)
graph <- cbind(actor,receiver)
if you want something more specific have a look at igraph for instance
library(igraph)
graph <- erdos.renyi.game(21, 0.3, type=c("gnp", "gnm"),
directed = FALSE, loops = FALSE)
# here the 0.3 is the probability of ties and 21 is the number of nodes
# this is a one mode network
or using package bipartite which focuses specifically on two mode networks:
library(bipartite)
web <- genweb(N1 = 5, N2 = 10, dens = 2)
web2edges(web,return=TRUE)
# here N1 is the number of nodes in set 1 and N2 the number of nodes in set 2
# and dens the average number of ties per node
There are many things to take into account, for instance if you want to constrain the degree distribution, probablity of ties between agents etc.