Question

假设我有4个矩阵：

matrix<-list(

MA0275.1 = structure(c(0, 76, 0, 24, 0, 100, 0, 0, 0, 0, 
                       100, 0, 0, 0, 100, 0, 72, 11, 16, 0, 53, 0, 0, 47), .Dim = c(4L, 
                                                                                    6L), .Dimnames = list(c("A", "C", "G", "T"), NULL), id = "MA0275.1", accession = "ASG1"), 
MA0276.1 = structure(c(0, 220, 8, 35, 0, 291, 0, 3, 61, 21, 
                       133, 10, 58, 54, 101, 12, 130, 0, 54, 0, 0, 11, 8, 147, 33, 
                       150, 8, 35, 80, 0, 92, 26, 0, 8, 249, 19, 0, 0, 256, 18), .Dim = c(4L, 
                                                                                          10L), .Dimnames = list(c("A", "C", "G", "T"), NULL), id = "MA0276.1", accession = "ASH1"), 
MA0277.1 = structure(c(63, 13, 13, 13, 100, 0, 0, 0, 100, 
                       0, 0, 0, 88, 13, 0, 0, 75, 0, 25, 0, 0, 0, 100, 0, 78, 16, 
                       3, 3, 81, 6, 6, 6, 63, 13, 13, 13), .Dim = c(4L, 9L), .Dimnames = list(
                         c("A", "C", "G", "T"), NULL), id = "MA0277.1", accession = "AZF1"), 
MA0278.1 = structure(c(64, 217, 425, 292, 104, 552, 150, 
                       192, 484, 111, 114, 288, 78, 401, 186, 333, 455, 51, 370, 
                       122, 248, 34, 670, 46, 98, 724, 143, 33, 52, 918, 7, 20, 
                       348, 346, 280, 24, 12, 3, 977, 6, 966, 5, 23, 4, 26, 6, 962, 
                       4, 9, 10, 4, 975, 47, 930, 7, 15, 892, 42, 16, 49, 487, 123, 
                       320, 68, 288, 140, 317, 254, 373, 110, 81, 434, 178, 367, 
                       184, 268, 402, 140, 341, 114, 435, 229, 241, 94), .Dim = c(4L, 
                                                                                  21L), .Dimnames = list(c("A", "C", "G", "T"), NULL), id = "MA0278.1", accession = "BAS1"))

这些矩阵用于评分序列（亲和力）。我的功能使我能够给出1个亲和度分数，因此只使用第一个矩阵。

但如果在data.frame（）中按矩阵定义得分，那就太好了。我尝试用for循环来做这个。

sequence<-"GCCTTTCCTTCTCTTCTCCGCGTGTGGAGGGAGCCAGCGCTTAGGCCGGAGCGAGCCTGGGGGCCGCCCGCCGTGAAGACATCGCGGGGACCGATTCACC"
for (i in matrix) {
    score<-affinity(i,sequence)
}

这给了我1个矩阵的数值。所以for循环不能正常工作。我希望它能给我所有矩阵的所有亲和力分数。

亲和力的功能：

affinity<-function (pwm, seq, Rmax = NULL, lambda = 0.7, pseudo.count = 1, 
    gc.content = 0.5, slide = FALSE) 
{
    if (is.null(seq) || is.na(seq) || mode(seq) != "character") {
        stop("sequence must be a character string of length >= ncol(pwm)")
    }
    gap.pos = sapply(1:nchar(seq), function(i) substr(seq, i, 
        i) == "-")
    seq = gsub("-", "", seq)
    if (nchar(seq) < ncol(pwm)) {
        stop("sequence must be a character string of length >= ncol(pwm)")
    }
    Rmax = ifelse(is.null(Rmax), exp(0.584 * ncol(pwm) - 5.66), 
        Rmax)
    pwm = pwm + pseudo.count
    at.content = 1 - gc.content
    pwm = apply(pwm, 2, function(p) {
        maxAT = max(p[c(1, 4)])
        maxCG = max(p[c(2, 3)])
        if (maxAT > maxCG) {
            transformed = c(log(maxAT/p[1])/lambda, log((maxAT/at.content) * 
                (gc.content/p[2]))/lambda, log((maxAT/at.content) * 
                (gc.content/p[3]))/lambda, log(maxAT/p[4])/lambda)
        }
        else {
            transformed = c(log((maxCG/gc.content) * (at.content/p[1]))/lambda, 
                log(maxCG/p[2])/lambda, log(maxCG/p[3])/lambda, 
                log((maxCG/gc.content) * (at.content/p[4]))/lambda)
        }
        if (maxAT == maxCG) {
            transformed = log(maxAT/p)/lambda
        }
        return(transformed)
    })
    if (slide) {
        z = .C("R_affinity", as.double(pwm), ncol(pwm), as.character(seq), 
            as.integer(nchar(seq)), as.double(Rmax), as.double(lambda), 
            double(length = nchar(seq) - ncol(pwm) + 1), PACKAGE = "tRap")
        res = z[[7]]
        gapped = numeric(length = nchar(seq) - ncol(pwm) + 1)
        gapped[!gap.pos] = res
        res = gapped
    }
    else {
        z = .C("R_affinity_sum", as.double(pwm), ncol(pwm), as.character(seq), 
            as.integer(nchar(seq)), as.double(Rmax), as.double(lambda), 
            double(length = 1), PACKAGE = "tRap")
        res = z[[7]]
    }
    return(res)
}

Answer 1

尝试lapply

lapply(matrix, affinity, sequence)

PS：调用您的数据matrix

是非常糟糕的主意

借助矩阵对数据进行评分

1 个答案: