In this class session we are going to plot the degree distribution of the undirected human
protein-protein interaction network (PPI), without using igraph. We'll obtain the interaction data from the Pathway Commons SIF file (in the shared/ folder) and we'll
manually compute the degree of each vertex (protein) in the network. We'll then compute
the count N(k) of vertices that have a given vertex degree k, for all k values.
Finally, we'll plot the degree distribution and discuss whether it is consistent with the
results obtained in the Jeong et al. article for the yeast PPI.
We're going to use ggplot2 in this notebook
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library(ggplot2)
Step 1: load in the SIF file as the data frame sif_data (with column names species1, interaction_type, and species2), using the function read.table
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Let's take a look at the top of the data frame, using the head function:
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head(sif_data)
Step 2: restrict the interactions to undirected protein-protein interactions. We will do this by subsetting the data frame to only rows for which the interaction_type column contains either in-complex-with or interacts-with. We'll also subset the data frame to include only the species1 and species2 columns. There are two ways to do it, (i) using standard indexing and (ii) using the subset function.
Here is how you would do it using standard indexing:
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interac_ppi <- sif_data[sif_data$interaction_type %in% c("in-complex-with",
"interacts-with"), c(1,3)]
Now you can try it using the subset function; this code is arguably more "readable":
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Use the head function to show the first few rows of the data frame, and use the nrow function to see how many rows it is:
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Step 3: for each interaction, reorder species1 and species2 (if necessary) so that
species1 < species2 (in terms of lexicographic order); we are doing this because there are many cases where a pair of proteins "X" and "Y" appear in the database as both [X,Y] and [Y,X], and we want those edges to be recognized as the same so that they are not separately counted. We'll do this using subset twice (once for species1 < species2 and once for species1 >= species2) and then stacking the two data frames that were returned from the two calls to subset, using the rbind function. Save it as a data frame interac_ppi_ordered.
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The number of rows in interac_ppi_ordered should be the same as for interac_ppi; let's verify that.
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nrow(interac_ppi_ordered)
Step 4: get only the unique interaction pairs of proteins (ignoring the interaction type), using the unique function; save as the object interac_ppi_unique.
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How many rows do we have now? Let's print it out
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nrow(interac_ppi_unique)
Step 5: compute the degree of each vertex. We'll do this by making a single character vector that is the concatenation of the species1 and species2 columns of interac_ppi_unique, and then we'll use the table function to compute the number of times each protein name occurs (you could also do the same thing by using the sort and rle functions, but table is easier to remember). Save the results as vertex_degrees.
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Let's have a look at vertex_degrees
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head(vertex_degrees)
Step 6: Calculate the histogram of N(k) vs. k, "suggesting" 30 bin "breaks" (i.e., bin edges) to R; hist will actually pick a different number of bins based on its own histogramming algorithm, so we'll find out exactly how many bin breaks that hist picked and then we'll call it nbreaks_actual, and we'll work with that.
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nbreaks <- 30
hist_res <- hist(vertex_degrees, plot=FALSE, nclass=nbreaks)
hist_counts <- hist_res$counts
hist_breaks <- hist_res$breaks
nbreaks_actual <- length(hist_breaks)
kvals <- (hist_breaks[1:(nbreaks_actual-1)] + hist_breaks[-1])/2
Step 7: Plot N(k) vs. k, on log-log scale. Restrict the horizontal axis to only values up to (but not including) the first zero value of N(k).
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indend <- min(which(hist_counts == 0))
indend <- if (length(indend) == 0) { length(kvals) } else { indend - 1 }
plot(kvals[1:indend],
hist_counts[1:indend],
type="p",
main="",
log="xy",
xlab=expression(k),
ylab=expression(N[k]))
Let's also plot the data using ggplot2 instead of R base graphics. We'll use the functions ggplot, geom_point
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print(ggplot(data=data.frame(k=kvals[1:indend], Nk=hist_counts[1:indend]), aes(x=k, y=Nk)) +
geom_point() +
theme_classic(base_size=24) +
scale_x_log10() +
scale_y_log10() +
ylab(expression(N[k])))
Step 8: Do a linear fit to the log10(N(k)) vs. log10(k) data. Restrict your analysis to the k bin values for which the log(Nk) vs. log(k) relationship is linear-looking, which is the first four points on the plot. You'll use lm to fit the linear model (make sure to log10 transform the data using the log10 function), and summary to print the results from the linear model.
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# put the data from the linear range, into a data frame
degdist_df <- data.frame(log10Nk=log10(hist_counts[1:4]),
log10k=log10(kvals[1:4]))
# linear regression
summary(lm(log10Nk ~ log10k,
data=degdist_df))
What is the slope? Can you find it in the summary output? (Hint: it's under the "Coefficients" part).
Now let's compute the slope for the degree distribution Fig. 1b in the Jeong et al. article, for the yeast PPI. The change in ordinate over the linear range is about -6.5 in units of natural logarithm. The change in abscissa over the linear range is approximately log(45)-log(2), so we can compute the Jeong et al. slope thus:
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round(-6.5/(log(45)-log(2)), digits=2)
How close is the human PPI slope to the Jeong. et al. slope?