rdna-sequenceprotein-database

How to find percentage of occurrence of characters in an argument?

what should i do to calculate percentage of occurrence of characters in an argument if the data are

t<-c(UUU,UUC,UUA,UUG,CUU,CUC,CUA,CUG,AUU,AUC,AUA,AUG,GUU,GUC,GUA,GUG,UCU,UCC,UCA,UCG,CCU,CCC,CCA,CCG,ACU,ACC,ACA,ACG,GCU,GCC,GCA,GCG,UAU,UAC,UAA,UAG,CAU,CAC,CAA,CAG,AAU,AAC,AAA,AAG,GAU,GAC,GAA,GAG,UGU,UGC,UGA,UGG,CGU,CGC,CGA,CGG,AGU,AGC,AGA,AGG,GGU,GGC,GGA,GGG)

i want to make a function regarding this which may help me in future to calculate more problems in future.

suppose our argument would be-

(UUUUUCUUAUUGCUUCUCCUACUGAUUAUCAUAAUGGUUGUCGUAGUGUCUUCCUCAUCGCCUCCCCCACCGACUACCACAACGGCUGCCGCAGCGUAUUACUAAUAGCAUCACCAACAGAAUAACAAAAAGGAUGACGAAGAGUGUUGCUGAUGGCGUCGCCGACGGAGUAGCAGAAGAGGUGGCGGAGGG)

also, the reading frame would start right in the starting which separate in the number of 3(e.g-AUG,GUG) I got this code which is following but i want my answer in the form of list with two columns named count and percentage, please help me in modify this code to give percentage in required manner.

    seqn <- c("UUA","AUC","GUA", "UUA", "GAU", "UUA") #your sequence
l_seq <- length(seqn) 
u_seq <- unique(seqn) 
seq_long <- "UUUAUGGGCG"
seqn <- unlist(str_extract_all(seq_long, pattern = "[AUGC]{3}"))

colSums(sapply(u_seq, function(s) str_count(string = seqn,pattern = s)))/l_seq

help me in correcting this code i want my argument continuous like UGCUGCUAUGAAUGAUG

Solution

  • This might work for you:

    require(stringr)
bases <- c("U","A","G","C")
sapply(bases, function(b) str_count(string = c("UUA","AUC","GUA"),pattern = b))

     U A G C
[1,] 2 1 0 0
[2,] 1 1 0 1
[3,] 1 1 1 0

    EDIT: basic genetics

    EDIT2: as per your comment this might work

    seqn <- c("UUA","AUC","GUA", "UUA", "GAU", "UUA") #your sequence
l_seq <- length(seqn) #length of sequence
u_seq <- unique(seqn) #unique codons

# This calculates the fractions of the unique codons in your sequence
colSums(sapply(u_seq, function(s) str_count(string = seqn,pattern = s)))/l_seq

      UUA       AUC       GUA       GAU 
0.5000000 0.1666667 0.1666667 0.1666667

    EDIT3: As per your second question you can split your string in 3 letter codons like so:

    seq_long <- "UUUAUGGGCG"
seqn <- unlist(str_extract_all(seq_long, pattern = "[AUGC]{3}"))

    and run the code from EDIT2. If your sequence is not a multiple of 3 you will lose the last letters. You can solve this with some padding.