I am trying to create a survival plot in R for deaths from exposure to a fungal disease over a number of weeks. I have the death week(continuous), whether they were alive (TRUE/FALSE), as well as categorical variables for diet (high/low) and sex(male/female). I have run a coxph model:
surv1 <- coxph(Surv(week_died,alive) ~ exposed + diet + sex,
data=surv)
I would like to plot a survival curve, with individual lines for exposed males on high and low diets, and the same for females on high and low diets (resulting in 4 individual survival curves on the same plot). if I use this then I only get a single curve.
plot(survfit(surv1), ylim=c(), xlab="Weeks")
I have also tried to use the ggsurv function created by Edwin Thoen (http://www.r-statistics.com/2013/07/creating-good-looking-survival-curves-the-ggsurv-function/) but keep getting an error for "invalid line type". I have tried to work out what could be causing this and think it would be this last ifelse statement - but I am not sure.
pl <- if(strata == 1) {ggsurv.s(s, CI , plot.cens, surv.col ,
cens.col, lty.est, lty.ci,
cens.shape, back.white, xlab,
ylab, main)
} else {ggsurv.m(s, CI, plot.cens, surv.col ,
cens.col, lty.est, lty.ci,
cens.shape, back.white, xlab,
ylab, main)}
Does anyone have any idea on what is causing this error/how to fix it or if I completely trying to do the wrong thing to plot these curves.
Many thanks!
survfit on a coxph model without any other specifications gives the survival curve for a case whose covariate predictors are the average of the population that the model was created with. From the help for survfit.coxph
Serious thought has been given to removing the default value for newdata, which is to use a single "psuedo" subject with covariate values equal to the means of the data set, since the resulting curve(s) almost never make sense. ... Two particularly egregious examples are factor variables and interactions. Suppose one were studying interspecies transmission of a virus, and the data set has a factor variable with levels ("pig", "chicken") and about equal numbers of observations for each. The “mean” covariate level will be 1/2 – is this a flying pig? ... Users are strongly advised to use the newdata argument.
So after you have computed surv1,
sf <- survfit(surv1,
newdata = expand.grid(diet = unique(surv$diet),
sex = unique(surv$sex)))
plot(sf)
sf should also work as an argument to ggsurv, though I've not tested it.
Related
How to run Latent Class Growth Modelling (LCGM) with a multinomial response variable in R (using the flexmix package)?
And how to stratify each class by a binary/categorical dependent variable?
The idea is to let gender shape the growth curve by cluster (cf. Mikolai and Lyons-Amos (2017, p. 194/3) where the stratification is done by education. They used Mplus)
I think I might have come close with the following syntax:
lcgm_formula <- as.formula(rel_stat~age + I(age^2) + gender + gender:age)
lcgm <- flexmix::stepFlexmix(.~ .| id,
data=d,
k=nr_of_classes, # would be 1:12 in real analysis
nrep=1, # would be 50 in real analysis to avoid local maxima
control = list(iter.max = 500, minprior = 0),
model = flexmix::FLXMRmultinom(lcgm_formula,varFix=T,fixed = ~0))
,which is close to what Wardenaar (2020,p. 10) suggests in his methodological paper for a continuous outcome:
stepFlexmix(.~ .|ID, k = 1:4,nrep = 50, model = FLXMRglmfix(y~ time, varFix=TRUE), data = mydata, control = list(iter.max = 500, minprior = 0))
The only difference is that the FLXMRmultinom probably does not support varFix and fixed parameters, altough adding them do produce different results. The binomial equivalent for FLXMRmultinom in flexmix might be FLXMRglm (with family="binomial") as opposed FLXMRglmfix so I suspect that the restrictions of the LCGM (eg. fixed slope & intercept per class) are not specified they way it should.
The results are otherwise sensible, but model fails to put men and women with similar trajectories in the same classes (below are the fitted probabilities for each relationship status in each class by gender):
We should have the following matches by cluster and gender...
1<->1
2<->2
3<->3
...but instead we have
1<->3
2<->1
3<->2
That is, if for example men in class one and women in class three would be forced in the same group, the created group would be more similar than the current first row of the plot grid.
Here is the full MVE to reproduce the code.
Got similar results with another dataset with diffent number of classes and up to 50 iterations/class. Have tried two alternative ways to predict the probabilities, with identical results. I conclude that the problem is most likely in the model specification (stepflexmix(...,model=FLXMRmultinom(...) or this is some sort of label switch issue.
If the model would be specified correctly and the issue is that similar trajectories for men/women end up in different classes, is there a way to fix that? By for example restricting the parameters?
Any assistance will be highly appreciated.
This seems to be a an identifiability issue apparently common in mixture modelling. In other words the labels are switched so that while there might not be a problem with the modelling as such, men and women end up in different groups and that will have to be dealt with one way or another
In the the new linked code, I have swapped the order manually and calculated the predictions with by hand.
Will be happy to hear, should someone has an alternative approach to deal with the label swithcing issue (like restricting parameters or switching labels algorithmically). Also curious if the model could/should be specified in some other way.
A few remarks:
I believe that this is indeed performing a LCGM as we do not specify random effects for the slopes or intercepts. Therefore I assume that intercepts and slopes are fixed within classes for both sexes. That would mean that the model performs LCGM as intended. By the same token, it seems that running GMM with random intercept, slope or both is not possible.
Since we are calculating the predictions by hand, we need to be able to separate parameters between the sexes. Therefore I also added an interaction term gender x age^2. The calculations seems to slow down somewhat, but the estimates are similar to the original. It also makes conceptually sense to include the interaction for age^2 if we have it for age already.
varFix=T,fixed = ~0 seem to be reduntant: specifying them do not change anything. The subsampling procedure (of my real data) was unaffected by the set.seed() command for some reason.
The new model specification becomes:
lcgm_formula <- as.formula(rel_stat~ age + I(age^2) +gender + age:gender + I(age^2):gender)
lcgm <- flexmix::flexmix(.~ .| id,
data=d,
k=nr_of_classes, # would be 1:12 in real analysis
#nrep=1, # would be 50 in real analysis to avoid local maxima (and we would use the stepFlexmix function instead)
control = list(iter.max = 500, minprior = 0),
model = flexmix::FLXMRmultinom(lcgm_formula))
And the plots:
I am using the useful gratia package by Gavin Simpson to extract the difference in two smooths for two different levels of a factor variable. The smooths are generated by the wonderful mgcv package. For example
library(mgcv)
library(gratia)
m1 <- gam(outcome ~ s(dep_var, by = fact_var) + fact_var, data = my.data)
diff1 <- difference_smooths(m1, smooth = "s(dep_var)")
draw(diff1)
This give me a graph of the difference between the two smooths for each level of the "by" variable in the gam() call. The graph has a shaded 95% credible interval (CI) for the difference.
Statistical significance, or areas of statistical significance at the 0.05 level, is assessed by whether or where the y = 0 line crosses the CI, where the y axis represents the difference between the smooths.
Here is an example from Gavin's site where the "by" factor variable had 3 levels.
The differences are clearly statistically significant (at 0.05) over nearly all of the graphs.
Here is another example I have generated using a "by" variable with 2 levels.
The difference in my example is clearly not statistically significant anywhere.
In the mgcv package, an approximate p value is outputted for a smooth fit that tests the null hypothesis that the coefficients are all = 0, based on a chi square test.
My question is, can anyone suggest a way of calculating a p value that similarly assesses the difference between the two smooths instead of solely relying on graphical evidence?
The output from difference_smooths() is a data frame with differences between the smooth functions at 100 points in the range of the smoothed variable, the standard error for the difference and the upper and lower limits of the CI.
Here is a link to the release of gratia 0.4 that explains the difference_smooths() function
enter link description here
but gratia is now at version 0.6
enter link description here
Thanks in advance for taking the time to consider this.
Don
One way of getting a p value for the interaction between the by factor variables is to manipulate the difference_smooths() function by activating the ci_level option. Default is 0.95. The ci_level can be manipulated to find a level where the y = 0 is no longer within the CI bands. If for example this occurred when ci_level = my_level, the p value for testing the hypothesis that the difference is zero everywhere would be 1 - my_level.
This is not totally satisfactory. For example, it would take a little manual experimentation and it may be difficult to discern accurately when zero drops out of the CI. Although, a function could be written to search the accompanying data frame that is outputted with difference_smooths() as the ci_level is varied. This is not totally satisfactory either because the detection of a non-zero CI would be dependent on the 100 points chosen by difference_smooths() to assess the difference between the two curves. Then again, the standard errors are approximate for a GAM using mgcv, so that shouldn't be too much of a problem.
Here is a graph where the zero first drops out of the CI.
Zero dropped out at ci_level = 0.88 and was still in the interval at ci_level = 0.89. So an approxiamte p value would be 1 - 0.88 = 0.12.
Can anyone think of a better way?
Reply to Gavin Simpson's comments Feb 19
Thanks very much Gavin for taking the time to make your comments.
I am not sure if using the criterion, >= 0 (for negative diffs), is a good way to go. Because of the draws from the posterior, there is likely to be many diffs that meet this criterion. I am interpreting your criterion as sample the posterior distribution and count how many differences meet the criterion, calculate the percentage and that is the p value. Correct me if I have misunderstood. Using this approach, I consistently got p values at around 0.45 - 0.5 for different gam models, even when it was clear the difference in the smooths should be statistically significant, at least at p = 0.05, because the confidence band around the smooth did not contain zero at a number of points.
Instead, I was thinking perhaps it would be better to compare the means of the posterior distribution of each of the diffs. For example
# get coefficients for the by smooths
coeff.level1 <- coef(gam.model1)[31:38]
coeff.level0 <- coef(gam.model1)[23:30]
# these indices are specific to my multi-variable gam.model1
# in my case 8 coefficients per smooth
# get posterior coefficients variances for the by smooths' coefficients
vp_level1 <- gam.model1$Vp[31:38, 31:38]
vp_level0 <- gam.model1$Vp[23:30, 23:30]
#run the simulation to get the distribution of each
#difference coefficient using the joint variance
library(MASS)
no.draws = 1000
sim <- mvrnorm(n = no.draws, (coeff.level1 - coeff.level0),
(vp_level1 + vp_level0))
# sim is a no.draws X no. of coefficients (8 in my case) matrix
# put the results into a data.frame.
y.group <- data.frame(y = as.vector(sim),
group = c(rep(1,no.draws), rep(2,no.draws),
rep(3,no.draws), rep(4,no.draws),
rep(5,no.draws), rep(6,no.draws),
rep(7,no.draws), rep(8,no.draws)) )
# y has the differences sampled from their posterior distributions.
# group is just a grouping name for the 8 sets of differences,
# (one set for each difference in coefficients)
# compare means with a linear regression
lm.test <- lm(y ~ as.factor(group), data = y.group)
summary(lm.test)
# The p value for the F statistic tells you how
# compatible the data are with the null hypothesis that
# all the group means are equal to each other.
# Same F statistic and p value from
anova(lm.test)
One could argue that if all coefficients are not equal to each other then they all can't be equal to zero but that isn't what we want here.
The basis of the smooth tests of fit given by summary(mgcv::gam.model1)
is a joint test of all coefficients == 0. This would be from a type of likelihood ratio test where model fit with and without a term are compared.
I would appreciate some ideas how to do this with the difference between two smooths.
Now that I got this far, I had a rethink of your original suggestion of using the criterion, >= 0 (for negative diffs). I reinterpreted this as meaning for each simulated coefficient difference distribution (in my case 8), count when this occurs and make a table where each row (my case, 8) is for one of these distributions with two columns holding this count and (number of simulation draws minus count), Then on this table run a chi square test. When I did this, I got a very low p value when I believe I shouldn't have as 0 was well within the smooth difference CI across almost all the levels of the exposure. Maybe I am still misunderstanding your suggestion.
Follow up thought Feb 24
In a follow up thought, we could create a variable that represents the interaction between the by factor and continuous variable
library(dplyr)
my.dat <- my.dat %>% mutate(interact.var =
ifelse(factor.2levels == "yes", 1, 0)*cont.var)
Here I am assuming that factor.2levels has the levels ("no", "yes"), and "no" is the reference level. The ifelse function creates a dummy variable which is multiplied by the continuous variable to generate the interactive variable.
Then we place this interactive variable in the GAM and get the usual statistical test for fit, that is, testing all the coefficients == 0.
#GavinSimpson actually posted a method of how to get the difference between two smooths and assess its statistical significance here in 2017. Thanks to Matteo Fasiolo for pointing me in that direction.
In that approach, the by variable is converted to an ordered categorical variable which causes mgcv::gam to produce difference smooths in comparison to the reference level. Statistical significance for the difference smooths is then tested in the usual way with the summary command for the gam model.
However, and correct me if I have misunderstood, the ordered factor approach causes the smooth for the main effect to now be the smooth for the reference level of the ordered factor.
The approach I suggested, see the main post under the heading, Follow up thought Feb 24, where the interaction variable is created, gives an almost identical result for the p value for the difference smooth but does not change the smooth for the main effect. It also does not change the intercept and the linear term for the by categorical variable which also both changed with the ordered variable approach.
I am using multivariate GAM models to learn more about fog trends in multiple regions. Fog is determined by visibility going below a certain threshold (< 400 meters). Our GAM model is used to determine the response of visibility to a range of meteorological variables.
However, my challenge right now is that I'd really like the y-axis to be the actual visibility observations rather than the centered smoothed. It is interesting to see how visibility is impacted by the covariates relative to the mean visibility in that location, but it's difficult to compare this for multiple locations where the mean visibility is different (and thus the 0 point in which visibility is enhanced or diminished has little comparable meaning).
In order to compare the results of multiple locations, I'm trying to make the y-axis actual visibility observations, and then I'll put a line at the visibility threshold we're interested in looking at (400 m)
to evaluate what the predictor variables values are like below that threshold (eg what temperatures are associated with visibility below 400 m).
I'm still a beginner when it comes to GAMs and R in general, but I've figured out a few helpful pieces so far.
Helpful things so far:
Attempt 1. how to extract gam fit for each variable in model
Extracting data used to make a smooth plot in mgcv
Attempt 2. how to use predict function to reconstruct a univariable model
http://zevross.com/blog/2014/09/15/recreate-the-gam-partial-regression-smooth-plots-from-r-package-mgcv-with-a-little-style/
Attempt 3. how to get some semblance of a y-axis that looks like visibility observations using "fitted" -- though I don't think this is
the correct approach since I'm not taking the intercept into account
http://gsp.humboldt.edu/OLM/R/05_03_GAM.html
simulated data
install.packages("mgcv") #for gam package
require(mgcv)
install.packages("pspline")
require(pspline)
#simulated GAM data for example
dataSet <- gamSim(eg=1,n=400,dist="normal",scale=2)
visibility <- dataSet[[1]]
temperature <- dataSet[[2]]
dewpoint <- dataSet[[3]]
windspeed <- dataSet[[4]]
#Univariable GAM model
gamobj <- gam(visibility ~ s(dewpoint))
plot(gamobj, scale=0, page=1, shade = TRUE, all.terms=TRUE, cex.axis=1.5, cex.lab=1.5, main="Univariable Model: Dew Point")
summary(gamobj)
AIC(gamobj)
abline(h=0)
Univariable Model of Dew Point
https://imgur.com/1uzP34F
ATTEMPT 2 -- predict function with univariable model, but didn't change y-axis
#dummy var that spans length of original covariate
maxDP <-max(dewpoint)
minDP <-min(dewpoint)
DPtrial.seq <-seq(minDP,maxDP,length=3071)
DPtrial.seq <-data.frame(dewpoint=DPtrial.seq)
#predict only the DP term
preds <- predict(gamobj, type="terms", newdata=DPtrial.seq, se.fit=TRUE)
#determine confidence intervals
DPplot <-DPtrial.seq$dewpoint
fit <-preds$fit
fit.up95 <-fit-1.96*preds$se.fit
fit.low95 <-fit+1.96*preds$se.fit
#plot
plot(DPplot, fit, lwd=3,
main="Reconstructed Dew Point Covariate Plot")
#plot confident intervals
polygon(c(DPplot, rev(DPplot)),
c(fit.low95,rev(fit.up95)), col="grey",
border=NA)
lines(DPplot, fit, lwd=2)
rug(dewpoint)
Reconstructed Dew Point Covariate Plot
https://imgur.com/VS8QEcp
ATTEMPT 3 -- changed y-axis using "fitted" but without taking intercept into account
plot(dewpoint,fitted(gamobj), main="Fitted Response of Y (Visibility) Plotted Against Dew Point")
abline(h=mean(visibility))
rug(dewpoint)
Fitted Response of Y Plotted Against Dew Point https://imgur.com/RO0q6Vw
Ultimately, I want a horizontal line where I can investigate the predictor variable relative to 400 meters, rather than just the mean of the response variable. This way, it will be comparable across multiple sites where the mean visibility is different. Most importantly, it needs to be for multiple covariates!
Gavin Simpson has explained the method in a couple of posts but unfortunately, I really don't understand how I would hold the mean of the other covariates constant as I use the predict function:
Changing the Y axis of default plot.gam graphs
Any deeper explanation into the method for doing this would be super helpful!!!
I'm not sure how helpful this will be as your Q is a little more open ended than we'd typically like on SO, but, here goes.
Firstly, I think it would help to think about modelling the response variable, which I assume is currently visibility. This is going to be a continuous variable, bounded at 0 (perhaps the data never reach zero?) which suggests modelling the data as conditionally distributed either
gamma (family = Gamma(link = 'log')) for visibility that never takes a value of zero.
Tweedie (family = tw()) for data that do have zeroes.
An alternative approach would be to model the occurrence of fog; if this is defined as an event <400m visibility then you could turn all your observations into 0/1 values for being a fog event or otherwise. Then you'd model the data as conditionally distributed Bernoulli, using family = binomial().
Having decided on a modelling approach, we need to model the response. This should be done using a multiple regression type of approach, with a GAM including multiple predictors. This way you get to estimate the effect of each potential predictor variable on the response while controlling for the effects of the other predictors. If you just do this using a single predictor at a time, say dewpoint, that variable could well "explain" variation in the data that might be due to another predictor, windspeed say, and you wouldn't know it.
Furthermore, there may well be interactions between predictors that you'll want to control for if they exist, which can only be done in
Then, to finally get to the crux of your problem, having fitted the multi-predictor model to "explain" visibility, you will need to predict from the model for sets of likely conditions. To look at how the visibility varies with dewpoint in a model where other predictor variables have effects, you need to fix the other variables at some reasonable values; one option is to set them to their mean (or modal value in the case of any factor predictor variables), or some other value indicative of typically values for that variable. You'll have to use your domain knowledge for this.
If you have interactions in the model, then you'll need to vary the two variables in the interaction, whilst holding all other variable fixed at some values.
Let's assume you don't have interactions and are interested in dewpoint but the model also includes windspeed. The mean windspeed for the values used to fit the model can be found from the cmX component of the fitted model. Of you could just calculate this from the observed windpseed values or set it to some known number you want to use. Denote the fitted by m, and the data frame with your data in it by df, then we can create new data to predict at over the range of dewpoint, whilst holding windspeed fixed.
mn.windspd <- m$cmX['windspeed']
## or
mn.windspd <- with(df, mean(windspeed))
## or set it some some value
mn.windspd <- 10 # say
Then you can do
preddata <- with(df,
expand.grid(dewpoint = seq(min(dewpoint),
max(dewpoint),
length = 300),
windspeed = mn.windspd))
Then you use this to predict from the fitted model:
pred <- predict(m, newdata = preddata, type = "link", se.fit = TRUE)
pred <- as.data.frame(pred)
Now we want to put these predictions back on to the response scale, and we want a confidence interval so we have to create that first before back transforming:
ilink <- family(m)$linkinv
pred <- transform(pred,
Fitted = ilink(fit),
Upper = ilink(fit + (2 * se.fit)),
Lower = ilink(fit - (2 * se.fit)),
dewpoint = preddata = dewpoint)
Now you can visualised the effect of dewpoint on the response whilst keeping windspeed fixed.
In your case, you will have to extend this to keeping temperature constant also, but that is done in the same way
mn.windspd <- m$cmX['windspeed']
mn.temp <- m$cmX['temperature']
preddata <- with(df,
expand.grid(dewpoint = seq(min(dewpoint),
max(dewpoint),
length = 300),
windspeed = mn.windspd,
temperature = mn.temp))
and then follow the steps above to do the prediction.
For one or two variables varying I have a function data_slice() in my gratia package which will do the above expand.grid() stuff for you so you don't have to specify the mean values of the other covariates:
preddata <- data_slice(m, 'dewpoint', n = 300)
technically this finds the value in the data closest to the median value (for the covariates not varying). If you want means, then do
fixdf <- data.frame(windspeed = mn.windspd, temperature = mn.temp)
preddata <- data_slice(m, 'dewpoint', data = fixdf, n = 300)
If you have an interaction, say between dewpoint and windspeed then you need to vary two variables. This is pretty easy again with expand.grid():
mn.temp <- m$cmX['temperature']
preddata <- with(df,
expand.grid(dewpoint = seq(min(dewpoint),
max(dewpoint),
length = 100),
windspeed = seq(min(windspeed),
max(windspeed),
length = 300),
temperature = mn.temp))
This will create a 100 x 100 grid of values of the covariates to predict at, whilst holding temperature constant.
For data_slice() you'd need to do:
fixdf <- data.frame(temperature = mn.temp)
preddata <- data_slice(m, 'dewpoint', 'windpseed',
data = fixdf, n = 300)
And extending this on to more covariates you want to vary, is also easy following this pattern with expand.grid(); I have yet to implement more than 2 variables varying in data_slice.
I have been unable to resolve an error when using abline(). I keep getting the warning message: In abline(model): only using the first two of 7 regression coefficients. I've been searching and seen many instances of others with this error but their examples are for multiple linear functions. I'm new to R and below is a simple example I'm using to work with. Thanks for any help!
year = c('2010','2011','2012','2013','2014','2015','2016')
population = c(25244310,25646389,26071655,26473525,26944751,27429639,27862596)
Texas=data.frame(year,population)
plot(population~year,data=Texas)
model = lm(population~year,data=Texas)
abline(model)
You probably want something like the following where we make sure that year is interpreted as a numeric variable in your model:
plot(population ~ year, data = Texas)
model <- lm(population ~ as.numeric(as.character(year)), data = Texas)
abline(model)
This makes lm to estimate an intercept (corresponding to a year 0) and slope (the mean increase in population each year), which is correctly interpreted by abline as can also be seen on the plot.
The reason for the warning is that year becomes a factor with 7 levels and so your lm call estimate the mean value for the refence year 2010 (the intercept) and 6 contrasts to the other years. Hence you get many coefficients and abline only uses the first two incorrectly.
Edit: With that said, you probably want change the way year is stored to a numeric. Then your code works, and plot also makes a proper scatter plot as regression line.
Texas$year <- as.numeric(as.character(Texas$year))
plot(population ~ year, data = Texas, pch = 16)
model <- lm(population ~ year, data = Texas)
abline(model)
Note that the as.character is needed in general, but it works in lm without it by coincidence (because the years are consecutive)
I have an issue with Random Forest with the Importance / varImPlot function, I hope someone could help me with?
I tried to code versions but I am confused about the (different) results:
1.)
rffit = randomForest(price~.,data=train,mtry=x,ntree=500)
rfvalpred = predict(rffit,newdata=test)
varImpPlot(rffit)
importance(rffit)
Shows the plot and the data of “importance”, however only “IncNodePurity”. And the data is different the plot and the data, I tried with "Scale" but did not work.
2.)
rf.analyzed_data = randomForest(price~.,data=train,mtry=x,ntree=500,importance=TRUE)
yhat.rf = predict(rf.analyzed_data,newdata=test)
varImpPlot(rf.analyzed_data)
importance(rf.analyzed_data)
In that case it does not produce any plot anymore and the importance data is showing “%IncMSE” and “IncNodePurity” data but the “IncNodePurity” data is different to first code?
Questions:
1.) Any idea why data is different for “IncNodePurity”?
2.) Any idea why no “%IncMSE” is shown in the first version?
3.) Why no plot is shown in the second version?
Many thanks!!
Ed
1) IncNodePurity is derived from the loss function, and you get that measure for free just by training the model. On the downside it is a more unstable estimate as results may vary from each model run. It is also more biased as it favors variables with many levels. I guess your found the differences are due to randomness.
2) VI, %IncMSE takes a little extra time to compute and is therefore optional. Roughly all values in data set needs to be shuffled and every OOB sample needs to be predicted once for every tree times for every variable. As the package randomForest is designed, you have to compute VI during training. importance must be set to TRUE. varImpPlot cannot plot it as it has not been computed.
3) Not sure. In this code example I see both plots at least.
library(randomForest)
#data
X = data.frame(replicate(6,rnorm(1000)))
y = with(X, X1^2 + sin(X2*pi) + X3*X4)
train = data.frame(y=y,X=X)
#training
rf1=randomForest(y~.,data=train,importance=F)
rf2=randomForest(y~.,data=train, importance=T)
#plotting importnace
varImpPlot(rf1) #plot only with IncNodePurity
varImpPlot(rf2) #bi-plot also with %IncMSE