Question

我在人类的22条染色体中分布着不同数量的基因，每条染色体的位置从0个碱基对开始，我试图找到一种方法将基因区间均匀分布在一个圆圈中，保持每个基因之间的相对位置和每个基因的长度，但重建新的位置，使基因可以在每条染色体上均匀分布，并在每条染色体之间留下一个空间。这是数据的一个例子（完整的数据集包括所有染色体）：

df = structure(list(Chr = structure(c(1L, 1L, 1L, 2L, 3L, 4L, 5L, 
6L, 7L, 8L, 9L, 10L, 11L, 12L, 12L, 13L, 17L, 20L, 22L, 22L), .Label = c("chr1", 
"chr2", "chr3", "chr4", "chr5", "chr6", "chr7", "chr8", "chr9", 
"chr10", "chr11", "chr12", "chr13", "chr14", "chr15", "chr16", 
"chr17", "chr18", "chr19", "chr20", "chr21", "chr22"), class = "factor"), 
    start = c(19068972, 25996369, 235879265, 46650500, 57732485, 
    44224566, 127510071, 33694865, 2297266, 105108497, 35252252, 
    64633822, 125738394, 416309, 93636009, 50070191, 72389245, 
    36432660, 19608500, 31498612), stop = c(20068972L, 26996369L, 
    236879265L, 47650500L, 58732485L, 45224566L, 128510071L, 
    34694865L, 3297266L, 106108497L, 36267753L, 65633822L, 126754018L, 
    1416309L, 94636009L, 51070191L, 73389245L, 37432660L, 20608500L, 
    32498612L), Gene = c("KIAA0090", "ZNF593", "GPR137B", "MCFD2", 
    "ABHD6", "GUF1", "FBN2", "HMGA1", "GNA12", "LRP12", "GBA2", 
    "NRBF2", "ST3GAL4", "WNK1", "SOCS2", "DLEU2", "FADS6", "BPI", 
    "TRMT2A", "PISD")), .Names = c("Chr", "start", "stop", "Gene"
), class = "data.frame", row.names = c(1L, 2L, 3L, 4L, 
5L, 6L,7L, 8L, 9L, 10L, 11L, 12L, 13L, 
14L, 15L, 16L, 17L, 18L, 19L, 20L))

我想要实现的是从每个染色体的0开始，重新分配基因间隔，使每个基因之间的空间相等（以及下一个基因之前的某个空间）：

out = structure(list(Chr = structure(c(1L, 1L, 1L, 2L, 3L, 4L, 5L, 
6L, 7L, 8L, 9L, 10L, 11L, 12L, 12L, 13L, 17L, 20L, 22L, 22L), .Label = c("chr1", 
"chr2", "chr3", "chr4", "chr5", "chr6", "chr7", "chr8", "chr9", 
"chr10", "chr11", "chr12", "chr13", "chr14", "chr15", "chr16", 
"chr17", "chr18", "chr19", "chr20", "chr21", "chr22"), class = "factor"), 
    start = c(2000000, 4000000, 6000000, 2000000, 2000000, 
    2000000, 2000000, 2000000, 2000000, 2000000, 2000000, 
    2000000,2000000, 2000000, 4000000, 2000000, 2000000, 
    2000000, 2000000, 2000000), stop = c(3000000L, 5000000L, 
    7000000L, 3000000L, 3000000L, 3000000L, 3000000L, 
    3000000L, 3000000L, 3000000L, 3000000L, 3000000L, 3000000L, 
    3000000L, 5000000L, 3000000L, 3000000L, 3000000L, 3000000L, 
    3000000L), Gene = c("KIAA0090", "ZNF593", "GPR137B", "MCFD2", 
    "ABHD6", "GUF1", "FBN2", "HMGA1", "GNA12", "LRP12", "GBA2", 
    "NRBF2", "ST3GAL4", "WNK1", "SOCS2", "DLEU2", "FADS6", "BPI", 
    "TRMT2A", "PISD")), .Names = c("Chr", "start", "stop", "Gene"
), class = "data.frame", row.names = c(1L, 2L, 3L, 4L, 
5L, 6L,7L, 8L, 9L, 10L, 11L, 12L, 13L, 
14L, 15L, 16L, 17L, 18L, 19L, 20L))

对于这个数据子集，可以说它们之间有1000000个碱基对，但是我遇到的困难是决定如何选择这个值。我是否将周长分成我所有染色体上的基因数量，并尝试从中找到正确的区间？谢谢你的任何建议！

-fra

Answer 1

好的，这是一个部分答案，可能会给你一个想法，让你完成整个事情。请注意，我的结果显示chr1上的重叠，但它可能是我的数学。我会让你跟踪它，因为我不确定这是你需要的解决方案，我无法测试绘图方面。

# Focus on chr1, 20 as they have multiple genes
df2 <- df[c(1:3, 19:20),]

Norm <- function(chrom) { # run on one chromosome at a time
    start <- chrom$start
    stop <- chrom$stop
    totLength <- max(stop) - min(start) 
    # simple normalization & offset
    newSt <- start/totLength
    newSt <- newSt - min(newSt)
    newEnd <- stop/totLength
    newEnd <- newEnd - min(newSt)
    totMax <- max(newSt, newEnd)
    newSt <- newSt/totMax
    newEnd <- newEnd/totMax
    return(data.frame(start = newSt, stop = newEnd))
}

normAll <- function(df) {
    #chromLvls <- levels(df$Chr) # this would work except your example data is truncated
    # and some levels are  missing
    chromLvls <- unique(as.character(df$Chr))
    noChrom <- length(chromLvls)
    drop <- 1:nrow(df)
    for (i in 1:noChrom) {
        df2 <- subset(df, df$Chr == chromLvls[i])
        df2[,c(2,3)] <- Norm(df2)
        df <- rbind(df, df2)
        }
    df <- df[-drop,] # row no's are mangled, may not matter
 }

res <- normAll(df2)

addGapBtwGenes <- function(df, gap = 0.05) {
    # gap is a fraction on [0...1]
    # this acts on a subset composed of just one chromosome
    for (i in 1:nrow(df)) {
        df$start[i] <- df$start[i] + (i-1)*gap
        df$stop[i] <- df$stop [i]+ (i)*gap
        # this denormalizes things but that probably doesn't matter
        }
    return(df)
    }

gapAllGenes <- function(df, gap = 0.05) {
    #chromLvls <- levels(df$Chr) # this would work except your example data is truncated
    # and some levels are  missing
    chromLvls <- unique(as.character(df$Chr))
    noChrom <- length(chromLvls)
    drop <- 1:nrow(df)
    for (i in 1:noChrom) {
        df2 <- subset(df, df$Chr == chromLvls[i])
        if (nrow(df2) == 1) { # no gap needed
           df <- rbind(df, df2)
           next
          }
        df2 <- addGapBtwGenes(df2, gap = gap)
        df <- rbind(df, df2)
        }
    df <- df[-drop,] # row no's are mangled, may not matter
 }

res2 <- gapAllGenes(res)

你可以编写一个名为addGapBtwChrom的函数来控制这个差距，除非绘图软件允许这样做。

上面给出了res：

      Chr      start       stop     Gene
6    chr1 0.00000000 0.08472237 KIAA0090
7    chr1 0.02924442 0.11396679   ZNF593
8    chr1 0.91527763 1.00000000  GPR137B
191 chr22 0.00000000 0.63413477   TRMT2A
201 chr22 0.36586523 1.00000000     PISD

和res2：

       Chr      start      stop     Gene
61    chr1 0.00000000 0.1347224 KIAA0090
71    chr1 0.07924442 0.2139668   ZNF593
81    chr1 1.01527763 1.1500000  GPR137B
1911 chr22 0.00000000 0.6841348   TRMT2A
2011 chr22 0.41586523 1.1000000     PISD

也许这接近你想要的。重新阅读你的一些评论，我看到我把所有的chromsomes都做了大致相同的长度，但是，我再等一下，看看你对它的看法。

如何在R中的圆圈间均匀分布基因间隔？

1 个答案: