How to get CI 95% for coefficients of linear model using simpleboot package - r
I'm trying to predict a linear model (basic linear regressions with 4 predictors) with the procedure lm(). This works all fine.
What I want to do now is bootstrapping the model. After a quick research on Google I found out about the package simpleboot which seemed to be quite easy to understand.
I can easily bootstrap the lm.object using something like this:
boot_mod <- lm.boot(mod,R=100,rows=TRUE)
and afterwards print the object boot_mod.
I can also access the list in which the coefficients for each bootstrap sample are among other metrics such as RSS, R² and so on.
Can anyone tell me how I can save all coefficients from the boot list in a list or dataframe?
The result would look like this at best:
boot_coef
sample coef 1 coef 2 coef 3...
1 1,1 1,4 ...
2 1,2 1,5 ...
3 1,3 1,6 ...
library(tidyverse)
library(simpleboot)
### Some Dummy-Data in a dataframe
a <- c(3,4,5,6,7,9,13,12)
b <- c(5,9,14,22,12,5,12,18)
c <- c(7,2,8,7,12,5,3,1)
df <- as_data_frame(list(x1=a,x2=b,y=c))
### Linear model
mod <- lm(y~x1+x2,data=df)
### Bootstrap
boot_mod <- lm.boot(mod,R=10,rows = TRUE)
You can also use the function sample of the same package simpleboot:
given the output from either lm.boot or loess.boot, you can specify what kind of information you want to extract:
samples(object, name = c("fitted", "coef", "rsquare", "rss"))
It outputs either a vector or matrix depending on the entity extracted.
Source:
https://rdrr.io/cran/simpleboot/man/samples.html
Here is a tidyverse option to save all coefficients from the boot.list:
library(tidyverse)
as.data.frame(boot_mod$boot.list) %>%
select(ends_with("coef")) %>% # select coefficients
t(.) %>% as.data.frame(.) %>% # model per row
rownames_to_column("Sample") %>% # set sample column
mutate(Sample = parse_number(Sample))
# output
Sample (Intercept) x1 x2
1 1 5.562417 -0.2806786 0.12219191
2 2 8.261905 -0.8333333 0.54761905
3 3 9.406171 -0.5863124 0.07783740
4 4 8.996784 -0.6040479 0.06737891
5 5 10.908036 -0.7249561 -0.03091908
6 6 8.914262 -0.5094340 0.05549390
7 7 7.947724 -0.2501127 -0.08607481
8 8 6.255539 -0.2033771 0.07463971
9 9 5.676581 -0.2668020 0.08236743
10 10 10.118126 -0.4955047 0.01233728
Related
Batch distribution fitting using Tidyverse and fitdistrplus
I have a dataset that is as follows (10,000+ Rows): P_ID SNUM RNUM X ID_233 10 2 40.31 ID_233 10 3 23.21 ID_234 12 5 11.00 ID_234 12 6 0.31 ID_234 13 1 0.00 ID_235 10 2 66.23 From this dataset, I want to fit each distinct P_ID to a Gamma distribution (ignoring the testing of how well the sampled data fits the distribution) Using the fitdistrplus package, I can achieve this by extracting the X for an individual P_ID into a vector and then run it through fw <- fitdist(data,"gamma") and then extract the shape and rate descriptive variables out, but this is all very manual. I would like to find a method using tidyverse to go from the data frame above to: P_ID Distrib G_Shape G_Rate ID_233 Gamma 1.21557116 0.09206639 ID_234 Gamma 3.23234542 0.34566432 ID_235 Gamma 2.34555553 0.92344521 How would i achieve this with Tidyverse and Pipes and not doing a succession of for loops?
You could apply fitdist for every individual using group_by and extract shape and rate values out of each model. library(dplyr) library(purrr) library(fitdistrplus) data %>% group_by(P_ID) %>% summarise(model = list(fitdist(X, "gamma"))) %>% mutate(G_Shape = map_dbl(model, pluck, 'estimate', 'shape'), G_rate = map_dbl(model, pluck, 'estimate', 'rate')) -> result result
Adjusted survival curve based on weigthed cox regression
I'm trying to make an adjusted survival curve based on a weighted cox regression performed on a case cohort data set in R, but unfortunately, I can't make it work. I was therefore hoping that some of you may be able to figure it out why it isn't working.
In order to illustrate the problem, I have used (and adjusted a bit) the example from the "Package 'survival'" document, which means im working with:
data("nwtco")
subcoh <- nwtco$in.subcohort
selccoh <- with(nwtco, rel==1|subcoh==1)
ccoh.data <- nwtco[selccoh,]
ccoh.data$subcohort <- subcoh[selccoh]
ccoh.data$age <- ccoh.data$age/12 # Age in years
fit.ccSP <- cch(Surv(edrel, rel) ~ stage + histol + age,
data =ccoh.data,subcoh = ~subcohort, id=~seqno, cohort.size=4028, method="LinYing")
The data set is looking like this:
seqno instit histol stage study rel edrel age in.subcohort subcohort
4 4 2 1 4 3 0 6200 2.333333 TRUE TRUE
7 7 1 1 4 3 1 324 3.750000 FALSE FALSE
11 11 1 2 2 3 0 5570 2.000000 TRUE TRUE
14 14 1 1 2 3 0 5942 1.583333 TRUE TRUE
17 17 1 1 2 3 1 960 7.166667 FALSE FALSE
22 22 1 1 2 3 1 93 2.666667 FALSE FALSE
Then, I'm trying to illustrate the effect of stage in an adjusted survival curve, using the ggadjustedcurves-function from the survminer package:
library(suvminer)
ggadjustedcurves(fit.ccSP, variable = ccoh.data$stage, data = ccoh.data)
#Error in survexp(as.formula(paste("~", variable)), data = ndata, ratetable = fit) :
# Invalid rate table
But unfortunately, this is not working. Can anyone figure out why? And can this somehow be fixed or done in another way?
Essentially, I'm looking for a way to graphically illustrate the effect of a continuous variable in a weighted cox regression performed on a case cohort data set, so I would, generally, also be interested in hearing if there are other alternatives than the adjusted survival curves?
Two reasons it is throwing errors. The ggadjcurves function is not being given a coxph.object, which it's halp page indicated was the designed first object. The specification of the variable argument is incorrect. The correct method of specifying a column is with a length-1 character vector that matches one of the names in the formula. You gave it a vector whose value was a vector of length 1154. This code succeeds: fit.ccSP <- coxph(Surv(edrel, rel) ~ stage + histol + age, data =ccoh.data) ggadjustedcurves(fit.ccSP, variable = 'stage', data = ccoh.data) It might not answer your desires, but it does answer the "why-error" part of your question. You might want to review the methods used by Therneau, Cynthia S Crowson, and Elizabeth J Atkinson in their paper on adjusted curves: https://cran.r-project.org/web/packages/survival/vignettes/adjcurve.pdf
Clustering with Mclust results in an empty cluster
I am trying to cluster my empirical data using Mclust. When using the following, very simple code: library(reshape2) library(mclust) data <- read.csv(file.choose(), header=TRUE, check.names = FALSE) data_melt <- melt(data, value.name = "value", na.rm=TRUE) fit <- Mclust(data$value, modelNames="E", G = 1:7) summary(fit, parameters = TRUE) R gives me the following result: ---------------------------------------------------- Gaussian finite mixture model fitted by EM algorithm ---------------------------------------------------- Mclust E (univariate, equal variance) model with 4 components: log-likelihood n df BIC ICL -20504.71 3258 8 -41074.13 -44326.69 Clustering table: 1 2 3 4 0 2271 896 91 Mixing probabilities: 1 2 3 4 0.2807685 0.4342499 0.2544305 0.0305511 Means: 1 2 3 4 1381.391 1381.715 1574.335 1851.667 Variances: 1 2 3 4 7466.189 7466.189 7466.189 7466.189 Edit: Here my data for download https://www.file-upload.net/download-14320392/example.csv.html I do not readily understand why Mclust gives me an empty cluster (0), especially with nearly identical mean values to the second cluster. This only appears when specifically looking for an univariate, equal variance model. Using for example modelNames="V" or leaving it default, does not produce this problem. This thread: Cluster contains no observations has a similary problem, but if I understand correctly, this appeared to be due to randomly generated data? I am somewhat clueless as to where my problem is or if I am missing anything obvious. Any help is appreciated!
As you noted the mean of cluster 1 and 2 are extremely similar, and it so happens that there's quite a lot of data there (see spike on histogram):
set.seed(111)
data <- read.csv("example.csv", header=TRUE, check.names = FALSE)
fit <- Mclust(data$value, modelNames="E", G = 1:7)
hist(data$value,br=50)
abline(v=fit$parameters$mean,
col=c("#FF000080","#0000FF80","#BEBEBE80","#BEBEBE80"),lty=8)
Briefly, mclust or gmm are probabilistic models, which estimates the mean / variance of clusters and also the probabilities of each point belonging to each cluster. This is unlike k-means provides a hard assignment. So the likelihood of the model is the sum of the probabilities of each data point belonging to each cluster, you can check it out also in mclust's publication
In this model, the means of cluster 1 and cluster 2 are near but their expected proportions are different:
fit$parameters$pro
[1] 0.28565736 0.42933294 0.25445342 0.03055627
This means if you have a data point that is around the means of 1 or 2, it will be consistently assigned to cluster 2, for example let's try to predict data points from 1350 to 1400:
head(predict(fit,1350:1400)$z)
1 2 3 4
[1,] 0.3947392 0.5923461 0.01291472 2.161694e-09
[2,] 0.3945941 0.5921579 0.01324800 2.301397e-09
[3,] 0.3944456 0.5919646 0.01358975 2.450108e-09
[4,] 0.3942937 0.5917661 0.01394020 2.608404e-09
[5,] 0.3941382 0.5915623 0.01429955 2.776902e-09
[6,] 0.3939790 0.5913529 0.01466803 2.956257e-09
The $classification is obtained by taking the column with the maximum probability. So, same example, everything is assigned to 2:
head(predict(fit,1350:1400)$classification)
[1] 2 2 2 2 2 2
To answer your question, no you did not do anything wrong, it's a fallback at least with this implementation of GMM. I would say it's a bit of overfitting, but you can basically take only the clusters that have a membership.
If you use model="V", i see the solution is equally problematic:
fitv <- Mclust(Data$value, modelNames="V", G = 1:7)
plot(fitv,what="classification")
Using scikit learn GMM I don't see a similar issue.. So if you need to use a gaussian mixture with spherical means, consider using a fuzzy kmeans:
library(ClusterR)
plot(NULL,xlim=range(data),ylim=c(0,4),ylab="cluster",yaxt="n",xlab="values")
points(data$value,fit_kmeans$clusters,pch=19,cex=0.1,col=factor(fit_kmeans$clusteraxis(2,1:3,as.character(1:3))
If you don't need equal variance, you can use the GMM function in the ClusterR package too.
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$cluster, function(x) dbscan::dbscan(x[,c(2,3)], eps = 0.0215, minPts = 4)[["cluster"]] ) ) However, you can still get garbage results if you don't have a good way to pick your epsilon. You might consider using dbscan::optics instead.
predict.lm after regression with missing data in Y
I don't understand how to generate predicted values from a linear regression using the predict.lm command when some value of the dependent variable Y are missing, even though no independent X observation is missing. Algebraically, this isn't a problem, but I don't know an efficient method to do it in R. Take for example this fake dataframe and regression model. I attempt to assign predictions in the source dataframe but am unable to do so because of one missing Y value: I get an error. # Create a fake dataframe x <- c(1,2,3,4,5,6,7,8,9,10) y <- c(100,200,300,400,NA,600,700,800,900,100) df <- as.data.frame(cbind(x,y)) # Regress X and Y model<-lm(y~x+1) summary(model) # Attempt to generate predictions in source dataframe but am unable to. df$y_ip<-predict.lm(testy) Error in `$<-.data.frame`(`*tmp*`, y_ip, value = c(221............. replacement has 9 rows, data has 10 I got around this problem by generating the predictions using algebra, df$y<-B0+ B1*df$x, or generating the predictions by calling the coefficients of the model df$y<-((summary(model)$coefficients[1, 1]) + (summary(model)$coefficients[2, 1]*(df$x)) ; however, I am now working with a big data model with hundreds of coefficients, and these methods are no longer practical. I'd like to know how to do it using the predict function. Thank you in advance for your assistance!
There is built-in functionality for this in R (but not necessarily obvious): it's the na.action argument/?na.exclude function. With this option set, predict() (and similar downstream processing functions) will automatically fill in NA values in the relevant spots. Set up data: df <- data.frame(x=1:10,y=100*(1:10)) df$y[5] <- NA Fit model: default na.action is na.omit, which simply removes non-complete cases. mod1 <- lm(y~x+1,data=df) predict(mod1) ## 1 2 3 4 6 7 8 9 10 ## 100 200 300 400 600 700 800 900 1000 na.exclude removes non-complete cases before fitting, but then restores them (filled with NA) in predicted vectors: mod2 <- update(mod1,na.action=na.exclude) predict(mod2) ## 1 2 3 4 5 6 7 8 9 10 ## 100 200 300 400 NA 600 700 800 900 1000
Actually, you are not using correctly the predict.lm function. Either way you have to input the model itself as its first argument, hereby model, with or without the new data. Without the new data, it will only predict on the training data, thus excluding your NA row and you need this workaround to fit the initial data.frame: df$y_ip[!is.na(df$y)] <- predict.lm(model) Or explicitly specifying some new data. Since the new x has one more row than the training x it will fill the missing row with a new prediction: df$y_ip <- predict.lm(model, newdata = df)