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## A. BASIC *Frequency() EXAMPLES
## ---------------------------------------------------------------------
data(yeastSEQCHR1)
yeast1 <- DNAString(yeastSEQCHR1)
dinucleotideFrequency(yeast1)
trinucleotideFrequency(yeast1)
oligonucleotideFrequency(yeast1, 4)
## Get the counts of tetranucleotides overlapping by one nucleotide:
oligonucleotideFrequency(yeast1, 4, step=3)
## Get the counts of adjacent tetranucleotides, starting from the first
## nucleotide:
oligonucleotideFrequency(yeast1, 4, step=4)
## Subset the sequence to change the starting nucleotide (here we start
## counting from third nucleotide):
yeast2 <- subseq(yeast1, start=3)
oligonucleotideFrequency(yeast2, 4, step=4)
## Get the less and most represented 6-mers:
f6 <- oligonucleotideFrequency(yeast1, 6)
f6[f6 == min(f6)]
f6[f6 == max(f6)]
## Get the result as an array:
tri <- trinucleotideFrequency(yeast1, as.array=TRUE)
tri["A", "A", "C"] # == trinucleotideFrequency(yeast1)["AAC"]
tri["T", , ] # frequencies of trinucleotides starting with a "T"
## With input made of multiple sequences:
library(drosophila2probe)
probes <- DNAStringSet(drosophila2probe)
dfmat <- dinucleotideFrequency(probes) # a big matrix
dinucleotideFrequency(probes, simplify.as="collapsed")
dinucleotideFrequency(probes, simplify.as="collapsed", as.matrix=TRUE)
## ---------------------------------------------------------------------
## B. OBSERVED DINUCLEOTIDE FREQUENCY VERSUS EXPECTED DINUCLEOTIDE
## FREQUENCY
## ---------------------------------------------------------------------
## The expected frequency of dinucleotide "ab" based on the frequencies
## of its individual letters "a" and "b" is:
## exp_Fab = Fa * Fb / N if the 2 letters are different (e.g. CG)
## exp_Faa = Fa * (Fa-1) / N if the 2 letters are the same (e.g. TT)
## where Fa and Fb are the frequencies of "a" and "b" (respectively) and
## N the length of the sequence.
## Here is a simple function that implements the above formula for a
## DNAString object 'x'. The expected frequencies are returned in a 4x4
## matrix where the rownames and colnames correspond to the 1st and 2nd
## base in the dinucleotide:
expectedDinucleotideFrequency <- function(x)
{
# Individual base frequencies.
bf <- alphabetFrequency(x, baseOnly=TRUE)[DNA_BASES]
(as.matrix(bf) %*% t(bf) - diag(bf)) / length(x)
}
## On Celegans chrI:
library(BSgenome.Celegans.UCSC.ce2)
chrI <- Celegans$chrI
obs_df <- dinucleotideFrequency(chrI, as.matrix=TRUE)
obs_df # CG has the lowest frequency
exp_df <- expectedDinucleotideFrequency(chrI)
## A sanity check:
stopifnot(as.integer(sum(exp_df)) == sum(obs_df))
## Ratio of observed frequency to expected frequency:
obs_df / exp_df # TA has the lowest ratio, not CG!
## ---------------------------------------------------------------------
## C. nucleotideFrequencyAt()
## ---------------------------------------------------------------------
nucleotideFrequencyAt(probes, 13)
nucleotideFrequencyAt(probes, c(13, 20))
nucleotideFrequencyAt(probes, c(13, 20), as.array=FALSE)
## nucleotideFrequencyAt() can be used to answer questions like: "how
## many probes in the drosophila2 chip have T, G, T, A at position
## 2, 4, 13 and 20, respectively?"
nucleotideFrequencyAt(probes, c(2, 4, 13, 20))["T", "G", "T", "A"]
## or "what's the probability to have an A at position 25 if there is
## one at position 13?"
nf <- nucleotideFrequencyAt(probes, c(13, 25))
sum(nf["A", "A"]) / sum(nf["A", ])
## Probabilities to have other bases at position 25 if there is an A
## at position 13:
sum(nf["A", "C"]) / sum(nf["A", ]) # C
sum(nf["A", "G"]) / sum(nf["A", ]) # G
sum(nf["A", "T"]) / sum(nf["A", ]) # T
## See ?hasLetterAt for another way to get those results.
## ---------------------------------------------------------------------
## D. oligonucleotideTransitions()
## ---------------------------------------------------------------------
## Get nucleotide transition matrices for yeast1
oligonucleotideTransitions(yeast1)
oligonucleotideTransitions(yeast1, 2, as.prob=TRUE)
## ---------------------------------------------------------------------
## E. ADVANCED *Frequency() EXAMPLES
## ---------------------------------------------------------------------
## Note that when dropping the dimensions of the 'tri' array, elements
## in the resulting vector are ordered as if they were obtained with
## 'fast.moving.side="left"':
triL <- trinucleotideFrequency(yeast1, fast.moving.side="left")
all(as.vector(tri) == triL) # TRUE
## Convert the trinucleotide frequency into the amino acid frequency
## based on translation:
tri1 <- trinucleotideFrequency(yeast1)
names(tri1) <- GENETIC_CODE[names(tri1)]
sapply(split(tri1, names(tri1)), sum) # 12512 occurrences of the stop codon
## When the returned vector is very long (e.g. width >= 10), using
## 'with.labels=FALSE' can improve performance significantly.
## Here for example, the observed speed up is between 25x and 500x:
f12 <- oligonucleotideFrequency(yeast1, 12, with.labels=FALSE) # very fast!
## With the use of 'step', trinucleotideFrequency() is a very fast way to
## calculate the codon usage table in an ORF (or a set of ORFs).
## Taking the same example as in '?codons':
file <- system.file("extdata", "someORF.fa", package="Biostrings")
my_ORFs <- readDNAStringSet(file)
## Strip flanking 1000 nucleotides around each ORF and remove first
## sequence as it contains an intron:
my_ORFs <- DNAStringSet(my_ORFs, start=1001, end=-1001)[-1]
## Codon usage for each ORF:
codon_usage <- trinucleotideFrequency(my_ORFs, step=3)
## Codon usage across all ORFs:
global_codon_usage <- trinucleotideFrequency(my_ORFs, step=3,
simplify.as="collapsed")
stopifnot(all(colSums(codon_usage) == global_codon_usage)) # sanity check
## Some related functions:
dict1 <- mkAllStrings(LETTERS[1:3], 4)
dict2 <- mkAllStrings(LETTERS[1:3], 4, fast.moving.side="left")
stopifnot(identical(reverse(dict1), dict2))
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