I am struggling with R loess function in R.
I have a dataframe on which I would locally weighted polynomial regression
For each ‘Gene’ is associated a Count (log10 transformed) which gives information regarding the gene expression. For each Gene is associated an ‘Integrity’ measurement (span 0-100) which tells you the quality of the ‘Count’ measurement for each gene. As a general principle, higher is the ‘Integrity’, more reliable is the ‘Count’ for the specific Gene.
Below is reported a chunk of the dataframe
Sample dataframe:
Gene
Integrity
Count
ENSG00000198786.2
96.6937
3.55279
ENSG00000210194.1
96.68682
1.39794
ENSG00000212907.2
94.81709
2.396199
ENSG00000198886.2
93.87207
3.61595
ENSG00000198727.2
89.08319
3.238548
ENSG00000198804.2
88.82048
3.78326
I would like to use loess to predict the ‘true’ value of genes with low ‘Integrity’ values (since less reliable).
I) Should I pre-process my dataframe in order to correctly apply loess ? From a pletora of examples I observed sinusoidal distributions of points (A), while my dataset seem distributed in a ‘rollercoaster’-like fashion (B).
II) How should I run loess?
I cannot understand how to run loess with the correct syntax to differentially weighted observations:
-1 loess( Count ~ Integrity, weight=None)
-2 loess( Count ~ 1:nrow(dataframe), weight=Integrity)
I performed several tests. Fig. C-D used loess (stats), Fig. E-F run weightedloess (limma). I used 2 different packages since, from the loess docs it is clear that prior weights are set based on x distance between points. weightedloess function allow the user to give priors in order to perform regression.
Below is reported the basic sintax adopted to perform regression and to generate images.
C) loess(Count ~ Integrity),degree=2,span=0.1)
D) loess(Count ~ 1:nrow(df)),weigths=’Integrity’,degree=2,span=0.1)
E) weightedLowess(x=1:nrow(df), y=Count, weigths=’Integrity’, span=0.1)
F) weightedLowess(x=1:nrow(df), y=order(Count), weigths=’Integrity’, span=0.1)
Please find enclosed images cited in the question.
Sample Images
Related
I am using the R package mclust to separate data into clusters. For this, I am using a uni-dimensional that allows for variable variances of the normal distributions underlying the clustering (the "V" model in the package).
The function looks like this: Mclust(dataToCluster, G=possibleClusters, modelNames=c("V")). To define the number of clusters possible, I use an array possibleClusters, e. g. 1:4 to allow for one to four clusters.
As a result of the clustering, after automatic model selection by Mclust using the BIC, I get a result with parameters of a normal distribution. For a model with three clusters, it might look like this:
# output shortened and commented for better readibility
> result$parameters
# proportion of data points per cluster ("lambda")
$pro
[1] 0.3459566 0.3877521 0.2662913
# mean of normal distribution per cluster ("mu")
$mean
1 2 3
110.3197 204.0477 265.0929
# variances per cluster ("sigma sq")
$variance$sigmasq
[1] 342.5032 128.4648 254.9257
However, I do have some knowledge about what these parameters are supposed to look like a priori. For example, I might know that:
sigmasq must be between 100 and 1000 units
the mean value for adjacent clusters must be at least 40 units apart
if there are three clusters, the mean value of the third cluster must be at least 215 units
Here is a graphical example for possible results of the clustering (the x axis corresponds to the units of mean and sigma unsquared):
Taking into account the constraints given above, example plots A1 (according to rules 1 and 2) and B1 (according to rules 2 and 3) can't be correct. Instead, the results should look more like A2 and B2, which were produced using slightly different data. Note that, taking into account these constraints, the “best” number of clusters might change (A1 vs. A2).
I would like to know how to include this kind of a priori information when using the Mclust function. The function does have a parameter prior, which might allow for this but I wasn't able to figure out how this could work. How could I bring the constraints into the function?
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 have a few data sets consisting of frequencies for i distinct alleles/SNPs of some populations. Additionally I recorded some factors that are suspicious for having changed the frequencies of these alleles within the populations in the past due to their selectional effect. It is assumed that the selection impact can be described in the form of a simple linear regression for every selection factor.
Now I'd like to estimate how the allele frequencies are expected to be under identical selectional forces (thus, I set selection=1). These new allele frequencies a'_i are derived as
a'_i = a_i - function[a_i|selection=1]
with the current frequency a_i of the allele i of a population and function[a_i|selection=1] as the estimated allele frequency under the absence of selectional forces.
However, there are some constraints for the whole process:
The minimal values of a'_i allowed is 0.
The sum of all allele frequencies a'_i has to be 1.
Usually I'd solve this problem by applying multiple linear regressions. But then the constraints are not fulfilled ...
Any idea how to approach this analysis with constraints (maybe using linear equation/regression systems or structural equation modelling)?
Here is an example data set containing allele frequencies for the ABO major allele groups (p, q, r) as well as the selection variables (x, y, z).
Although this example file only contains 3 alleles and 3 influential variables, all my data sets contain up to ~1050 alleles/SNPs and always 8 selection variables that may have (but don't have to) an impact on the allele frequencies ...
Many thanks in advance for ideas, code snippets and hints!
I have the following code, which basically try to predict the Species from iris data using randomForest. What I'm really intersed in is to find what are the best features (variable) that explain the species classification. I found the package randomForestExplainer is the best
to serve the purpose.
library(randomForest)
library(randomForestExplainer)
forest <- randomForest::randomForest(Species ~ ., data = iris, localImp = TRUE)
importance_frame <- randomForestExplainer::measure_importance(forest)
randomForestExplainer::plot_multi_way_importance(importance_frame, size_measure = "no_of_nodes")
The result of the code produce this plot:
Based on the plot, the key factor to explain why Petal.Length and Petal.Width is the best factor are these (the explanation is based on the vignette):
mean_min_depth – mean minimal depth calculated in one of three ways specified by the parameter mean_sample,
times_a_root – total number of trees in which Xj is used for splitting the root node (i.e., the whole sample is divided into two based on the value of Xj),
no_of_nodes – total number of nodes that use Xj for splitting (it is usually equal to no_of_trees if trees are shallow),
It's not entirely clear to me why the high times_a_root and no_of_nodes is better? And low mean_min_depth is better?
What are the intuitive explanation for that?
The vignette information doesn't help.
You would like a statistical model or measure to be a balance between "power" and "parsimony". The randomForest is designed internally to do penalization as its statistical strategy for achieving parsimony. Furthermore the number of variables selected in any given sample will be less than the the total number of predictors. This allows model building when hte number of predictors exceeds the number of cases (rows) in the dataset. Early splitting or classification rules can be applied relatively easily, but subsequent splits become increasingly difficult to meet criteria of validity. "Power" is the ability to correctly classify items that were not in the subsample, for which a proxy, the so-called OOB or "out-of-bag" items is used. The randomForest strategy is to do this many times to build up a representative set of rules that classify items under the assumptions that the out-of-bag samples will be a fair representation of the "universe" from which the whole dataset arose.
The times_a_root would fall into the category of measuring the "relative power" of a variable compared to its "competitors". The times_a_root statistic measures the number of times a variable is "at the top" of a decision tree, i.e., how likely it is to be chosen first in the process of selecting split criteria. The no_of_node measures the number of times the variable is chosen at all as a splitting criterion among all of the subsampled.
From:
?randomForest # to find the names of the object leaves
forest$ntree
[1] 500
... we can see get a denominator for assessing the meaning of the roughly 200 values in the y-axis of the plot. About 2/5ths of the sample regressions had Petal.Length in the top split criterion, while another 2/5ths had Petal.Width as the top variable selected as the most important variable. About 75 of 500 had Sepal.Length while only about 8 or 9 had Sepal.Width (... it's a log scale.) In the case of the iris dataset, the subsamples would have ignored at least one of the variables in each subsample, so the maximum possible value of times_a_root would have been less than 500. Scores of 200 are pretty good in this situation and we can see that both of these variables have a comparable explanatory ability.
The no_of_nodes statistic totals up the total number of trees that had that variable in any of its nodes, remembering that the number of nodes would be constrained by the penalization rules.
I would like to use R to perform hierarchical clustering with two groups of variables describing the same samples. One group is microarray gene expression data (for specific genes) that have been normalized and batch effect corrected. The other group also has some quantitative clinical parameters that describe the same samples. However, these clinical variables have not been normalized or subjected to any kind of transformation(i.e. raw continuous values).
For example, one variable of these could have range of values from 2 to 35, whereas another from 0.1 to 0.9, etc.
Thus, as my ultimate goal in to implement hierarchical clustering and use both groups simultaneously (merged in a matrix/dataframe), in order to inspect which of these clinical variables cluster with specific genes, etc:
1) Is an initial transformation in the group of the clinical variables necessary before merging with the genes and perform the clustering ? For example: log2 transformation, which has also been done to part of my gene expression data !!
2) Or, a row scaling (that is the total features in the input data) would take into account this discrepancy ?
3) For a similar analysis/approach, like constructing a correlation plot of the above total variables, would a simple scaling be sufficient?
Without having seen your gene expression data, I can only provide you some general suggestions based on your description, in the context of the 3 questions you asked:
1) You should definitely check the distribution of each group. In R, you may use one or more of the following function to visualize the distribution:
hist(expression_data) ##histogram
plot(density(expression_data)) ##density plot; alternative to histogram
qqnorm(expression_data); qqline(expression_data) #QQ plot
Since my understanding is that one of your expression data group is log2 transformed, that particular group should have a normal distribution (i.e. a bell curve shape in the histogram and a straight line in the QQ plot). Whether to transform the group that has not yet been transformed will depend on what you want to do with the data. For instance, if you want to use a t-test to compare the two groups, then you definitely need a transformation, as there is a normality assumption associated with a t-test. With regard to hierarchical clustering, if you decide to use both groups in a single clustering analysis, then why would you ever keep one transformed and the other not?
2) Scaling by features is a reasonable approach. Here is a clustering lecture from a Utah State Univ. stats course, with an example. scale=TRUE is an option for you if you decide to use heatmap function in R.
3) I don't think there is a definitive answer to your third question. It has to depend on how many available features you have and what analyses you will be doing downstream. Similar to question 1, I would argue that simple scaling may be sufficient for visualizing your data by hierarchical clustering. However, do keep in mind that, say you decide to perform a linear model (which is very common with microarray gene expression data), you might want to consider more sophisticated data scaling.