Theory, R functions & Examples
Section: Ordination analysis
In this example, we will apply constrained ordination (tb-RDA) on Vltava river valley dataset. We will ask how much variance in species composition can be explained by two variables, soil pH and soil depth. Both are important factors for plant growth, and moreover, in the study area, they are somewhat correlated (shallower soils have lower pH since the prevailing geological substrate is acid).
First, upload the Vltava river valley data:
vltava.spe <- read.delim ('https://raw.githubusercontent.com/zdealveindy/anadat-r/master/data/vltava-spe.txt', row.names = 1) vltava.env <- read.delim ('https://raw.githubusercontent.com/zdealveindy/anadat-r/master/data/vltava-env.txt') spe <- vltava.spe # rename variables to make them shorter env <- vltava.env[, c('pH', 'SOILDPT')] # select only two explanatory variables
vegan and calculate tb-RDA based on Hellinger pre-transformed species composition data. Note that since the original data represent estimates of percentage cover, it is better to log transform these values first before Hellinger transformation is done (using function
log1p, which calculates log (x+1) to avoid log (0)):
library (vegan) spe.log <- log1p (spe) # species data are in percentage scale which is strongly rightskewed, better to transform them spe.hell <- decostand (spe.log, 'hell') # we are planning to do tb-RDA, this is Hellinger pre-transformation tbRDA <- rda (spe.hell ~ pH + SOILDPT, data = env) # calculate tb-RDA with two explanatory variables tbRDA
The result printed by
rda function is the following:
Call: rda(formula = spe.hell ~ pH + SOILDPT, data = env) Inertia Proportion Rank Total 0.70476 1.00000 Constrained 0.06250 0.08869 2 Unconstrained 0.64226 0.91131 94 Inertia is variance Eigenvalues for constrained axes: RDA1 RDA2 0.04023 0.02227 Eigenvalues for unconstrained axes: PC1 PC2 PC3 PC4 PC5 PC6 PC7 PC8 0.07321 0.04857 0.04074 0.03144 0.02604 0.02152 0.01917 0.01715 (Showed only 8 of all 94 unconstrained eigenvalues)
and, the same with comments:
The two variables explain 8.9% of the variance (the row
Constrained and column
Proportion in the table above, can be calculated also as the sum of eigenvalues for the constrained axes divided by total variance (inertia): (0.04023+0.02227) /0.70476=0.08869. The first constrained axis (RDA1) explains 0.04023/0.70476=5.7% of the variance, while the second (RDA2) explains 0.02227/0.70476=3.2%. Note that the first unconstrained axis (PC1) represents 0.07321/0.70476=10.4% of the total variance, which is more than the variance explained by both explanatory variables together; the first two unconstrained explain (0.07321+0.04857)/0.70476=17.3%. This means that the dataset may be structured by some strong environmental variable(s) different from pH and soil depth (we will check this below).
The relationship between the variation represented by individual (constrained and unconstrained) ordination axes can be displayed using the barplot of percentage variance explained by individual axes (ie their eigenvalue divided by total inertia):
constrained_eig <- tbRDA$CCA$eig/tbRDA$tot.chi*100 unconstrained_eig <- tbRDA$CA$eig/tbRDA$tot.chi*100 expl_var <- c(constrained_eig, unconstrained_eig) barplot (expl_var[1:20], col = c(rep ('red', length (constrained_eig)), rep ('black', length (unconstrained_eig))), las = 2, ylab = '% variation')
(note that all information about the eigenvalues and total inertia is in the object calculated by
vegan's ordination function (
rda in this case, stored in the list
tbRDA), you just need to search a bit inside to find it - consider using the function
str to check the structure of tbRDA first).
Let's see the ordination diagram:
By the way, what may be those environmental variables associated with unconstrained axes? The
vltava.env dataset contains a number of other measured variables which we may fit as supplementary to the first and second unconstrained axis to see which of them is most related to which of them. But here we will do an alternative thing: we will use mean Ellenberg indicator values (mEIV) calculated for each plot based on the species composition and tabulated Ellenberg species indicator values (ecological optima of species along several main environmental gradients). This approach will illustrate the situation as if in the field we measured only soil pH and depth (which are rather easy to measure), and we use these indirect estimates to get an idea about which other factors may be important.
ordiplot (tbRDA, choices = c(3,4), type = 'n') points (tbRDA, choices = c(3,4), display = 'sites', pch = as.character (vltava.env$GROUP), col = vltava.env$GROUP) ef <- envfit (tbRDA, vltava.env[,23:28], choices = c(3,4), permutations = 0) plot (ef)
***VECTORS PC1 PC2 r2 LIGHT -0.93135 -0.36411 0.6282 TEMP -0.97246 -0.23305 0.2352 CONT -0.86643 -0.49929 0.0885 MOIST 0.44495 -0.89556 0.4706 REACT 0.95614 -0.29291 0.1166 NUTR 0.94383 -0.33044 0.4519
The highest R2 of regression with the first two axes have light and moisture, with light associated mostly with the first unconstrained axis and the moisture mostly with the second. It seems that these two ecological factors, not related to soil pH and soil depth, are important for the studied vegetation, but were not measured; light passing through the canopy of the forest has a strong effect on the species composition of the herb understory (herbs makes most of the species in this analysis since the temperate forest is rather poor for woody species), and moisture also (flooded alluvial forests at the bottom parts of the valley, veg. type 2, have very different species composition from dry open forests on the upper parts of the valley slopes).
Note one more thing: when applying the function
envfit on mean Ellenberg indicator values, I did not test for the significance (I set the argument
permutation = 0). This has a meaning: both mean Ellenberg indicator values and sample scores on ordination axes are calculated from the same matrix of species composition, and to directly test their relationship would be wrong (they are not independent, and we get high probability to get significant result even if the species Ellenberg indicator values are randomly generated). Check the section Analysis of species attributes for detail explanation on how to solve this .
This example is using Difference between cookies, pastries and pizzas dataset, which I found on reddit.com, posted by author everest4ever. As the post on reddit.com goes, the author attended the Christmas party at his office with “Christmass cookie competition”, which sparked a “huge debate about what are eligible entries for the cookie competition (e.g. are mini-pizzas cookies?)”. The author decided to approach the discussion rigorously and did the following: “I scraped 1931 recipes from the Food Network that contain the keywords cookies (my group of interest), pastry, or pizza (two control groups). Next, I extracted the ingredient list and pooled similar ingredients together (e.g. salt, seasalt, Kosher salt), coming up with a total of 133 unique ingredients. I ended up with a 1931×133 matrix, where each row is one recipe, and each column is whether this recipe contains a certain ingredient (0 or 1)”. The author did PCA analysis on the data accompanied by some clustering and predictions, just to prove that “NO IAN AND JOSEPH YOUR FUCKING EGG TARTS AREN'T COOKIES, NO MATTER HOW GOOD THEY WERE!!”. I think we can also use this dataset for a simple constrained ordination exercise. First import the data:
recipes.ingr <- read.delim ('https://raw.githubusercontent.com/zdealveindy/anadat-r/master/data/cookie_dataset_everest4ever_composition.txt', row.names = 1) recipes.type <- read.delim ('https://raw.githubusercontent.com/zdealveindy/anadat-r/master/data/cookie_dataset_everest4ever_type.txt', row.names = 1)
Data represent a matrix of presence/absence of different ingredients in individual recipes (each row of
recipes.ingr matrix is a recipe, each column one ingredients). To know which recipe is classified how, we need a variable
type_of_food in the data frame
To get familiar with data, let's first calculate DCA:
library (vegan) DCA <- decorana (recipes.ingr) DCA
Call: decorana(veg = recipes.ingr) Detrended correspondence analysis with 26 segments. Rescaling of axes with 4 iterations. DCA1 DCA2 DCA3 DCA4 Eigenvalues 0.5633 0.2598 0.2224 0.2185 Decorana values 0.5918 0.2622 0.2352 0.2138 Axis lengths 6.2276 4.1302 5.6396 3.5449
The output shows that the length of the first axis is 6.2 S.D. units, so unimodal ordination methods is advisable. The ordination diagram which displays recipes of cookies, pastries and pizzas by different symbols and colours, is more informative:
type_num <- as.numeric (recipes.type$type_of_food) ordiplot (DCA, type = 'n') points (DCA, display = 'sites', col = type_num, pch = type_num) legend ('topright', col = 1:3, pch = 1:3, legend = levels (recipes.type$type_of_food))
type_num contains numerical values 1, 2 and 3 in place of Cookies, Pastries and Pizzas from the original
type_of_food variable in
recipes.type data frame, so as we can use these values as colors and symbols in ordination diagram).
It seems that pizzas are somewhat different from the rest (although part of pastries is close), while pastries and cookies form a cloud with big overlap. Let's try to ask the following question: can the classification of a recipe into cookies/pastries/pizzas (done largely subjectively by authors of that recipes based on their opinion how each category item should look like) explain the difference in “ingredients composition” of individual recipes? This is task for constrained ordination. Since the first DCA axis is long, we use CCA for it, with recipes dependent variable and assignment into the type as explanatory. Note that explanatory variable is categorical with three levels (
type <- recipes.type$type_of_food CCA <- cca (recipes.ingr ~ type) CCA
Call: cca(formula = recipes.ingr ~ type) Inertia Proportion Rank Total 14.28961 1.00000 Constrained 0.60311 0.04221 2 Unconstrained 13.68650 0.95779 132 Inertia is scaled Chi-square Eigenvalues for constrained axes: CCA1 CCA2 0.4649 0.1382 Eigenvalues for unconstrained axes: CA1 CA2 CA3 CA4 CA5 CA6 CA7 CA8 0.30447 0.28409 0.26249 0.25278 0.23942 0.21521 0.21069 0.20674 (Showed only 8 of all 132 unconstrained eigenvalues)
(note that I saved the column
type_of_food from the data frame
recipes.type into a variable
type, not really because I want to simplify the
cca (it would need to be
CCA <- cca (recipes.ingr ~ type_of_food, data = recipes.type), but that is still fine), but because this will make it simple to display individual factor levels onto ordination diagram (the levels are displayed as the Name_of_variableName_of_category, which would be too long with the original names).
We got two constrained axes (explanatory variable is qualitative with three factor levels -> number of contrained axes = number of levels - 1), the first exlaining more than 3 time more than the second (eigCCA1 = 0.4649, eigCCA2 = 0.1382, which means that the first axis represents 0.4649/14.28961 = 3.3 % of variance (eigenvalue/total inertia), while the second 0.1382/14.28961 = 1.0%. Ordination diagram shows that the first axis is mostly separating pizzas (right) and cookies+pastries (left), while the second axis is mostly separating cookies (up) from pastries (bottom):
ordiplot (CCA, display = c('si', 'cn'), type = 'n') points (CCA, display = 'si', col = type_num, pch = type_num) text (CCA, display = 'cn', col = 'navy', cex = 1.5) legend ('topright', col = 1:3, pch = 1:3, legend = levels (recipes.type$type_of_food))
Few more things. First, we should ask whether the CCA ordination is significant and whether it is worth to interpret it:
Permutation test for cca under reduced model Permutation: free Number of permutations: 999 Model: cca(formula = recipes.ingr ~ type) Df ChiSquare F Pr(>F) Model 2 0.6031 42.479 0.001 *** Residual 1928 13.6865 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Indeed, it is. And how about individual axes, are they both significant? We will use the argument
by = “axis” in the function
anova - see this explanation what it means:
Permutation test for cca under reduced model Forward tests for axes Permutation: free Number of permutations: 999 Model: cca(formula = recipes.ingr ~ type) Df ChiSquare F Pr(>F) CCA1 1 0.4649 65.493 0.001 *** CCA2 1 0.1382 19.465 0.001 *** Residual 1928 13.6865 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Both axes are significant, which means that even the distinction between cookies and pastries along the second axis is important. So I would say, without further testing, that not only pizza is (quite obviously) different from cookies, but also much more ambiguous category pastries are different regarding ingredients they use. Btw, let's see these ingredients (display species in CCA ordination diagram):
ordiplot (CCA, display = c('sp', 'cn'), type = 'n') orditorp (CCA, display = 'sp', priority = colSums (recipes.ingr))
Note that I did not display all 133 ingredients (“species”); otherwise the diagram gets too cluttered. I used the low-level graphical function
orditorp which is adding only some labels and draws others as symbols. It has argument
priority (the species with the highest priority will be more likely plotted as text, with lower as symbols, if there is not enough space); the priority here is the overall frequency of ingredence in the dataset (
colSums applied on the
recipes.ingr data frame). The diagram shows clear triangle, with each corner representing one type of food. There is a gradient of ingredients connecting pizzas with pastries and pastries with cookies, but almost no ingredients connecting pizzas and cookies (except
pine nuts, which can perhaps make it in both pizzas and cookies - but I have no idea). Some ingredients are shared among all three (obviously water and seems that also honey), some are only for that type of food (basil for pizza, puff pastry for pastries and cookies(??) and ice cream for cookies. Please, see the diagram and guess which item in your opinion should be where (carrots in pastry? not sure...).