我有一个超过100,000行的data.table(A)。共有3列。
chrom start end
1: chr1 6484847 6484896
2: chr1 6484896 6484945
3: chr1 6484945 6484994
4: chr1 6484994 6485043
5: chr1 6485043 6485092
---
183569: chrX 106893605 106893654
183570: chrX 106893654 106893703
183571: chrX 106893703 106893752
183572: chrX 106893752 106893801
183573: chrX 106893801 106894256
我想生成一个名为“gene”的新列,它为来自另一个data.table的每个行提供一个标签,该行有~90行(B)。见下文:
chrom start end gene
1: chr1 6484847 6521004 ESPN
2: chr1 41249683 41306124 KCNQ4
3: chr1 55464616 55474465 BSND
42: chrX 82763268 82764775 POU3F4
43: chrX 100600643 100603957 TIMM8A
44: chrX 106871653 106894256 PRPS1
如果data.table A中的行起始值在data.table B的行开始值和结束值内,则需要A中的行相应地标记正确的基因。
例如,生成的完整数据。表A将是
chrom start end gene
1: chr1 6484847 6484896 ESPN
2: chr1 6484896 6484945 ESPN
3: chr1 6484945 6484994 ESPN
4: chr1 6484994 6485043 ESPN
5: chr1 6485043 6485092 ESPN
---
183569: chrX 106893605 106893654 TIMM8A
183570: chrX 106893654 106893703 TIMM8A
183571: chrX 106893703 106893752 TIMM8A
183572: chrX 106893752 106893801 TIMM8A
183573: chrX 106893801 106894256 TIMM8A
我尝试过一些嵌套循环来做这件事,但这似乎需要花费太长时间。我认为必须有一种方法来使用data.table包,但我似乎无法弄明白。
非常感谢任何和所有建议。
答案 0 :(得分:4)
虽然在基础R中可以做到这一点(或者可能使用data.table
),但我强烈建议使用GenomicRanges
;它是一个非常强大而灵活的R / Bioconductor库,专为这类任务而设计。
以下是使用GenomicRanges::findOverlaps
的示例:
# Sample data
df1 <- read.table(text =
"chrom start end
chr1 6484847 6484896
chr1 6484896 6484945
chr1 6484945 6484994
chr1 6484994 6485043
chr1 6485043 6485092", sep = "", header = T, stringsAsFactors = F);
df2 <- read.table(text =
"chrom start end gene
chr1 6484847 6521004 ESPN
chr1 41249683 41306124 KCNQ4
chr1 55464616 55474465 BSND
chrX 82763268 82764775 POU3F4
chrX 100600643 100603957 TIMM8A
chrX 106871653 106894256 PRPS1", sep = "", header = TRUE, stringsAsFactors = F);
# Convert to GRanges objects
gr1 <- with(df1, GRanges(chrom, IRanges(start = start, end = end)));
gr2 <- with(df2, GRanges(chrom, IRanges(start = start, end = end), gene = gene));
# Find features from gr1 that overlap with gr2
m <- findOverlaps(gr1, gr2);
# Add gene annotation as metadata to gr1
mcols(gr1)$gene[queryHits(m)] <- mcols(gr2)$gene[subjectHits(m)];
gr1;
#GRanges object with 5 ranges and 1 metadata column:
# seqnames ranges strand | gene
# <Rle> <IRanges> <Rle> | <character>
# [1] chr1 [6484847, 6484896] * | ESPN
# [2] chr1 [6484896, 6484945] * | ESPN
# [3] chr1 [6484945, 6484994] * | ESPN
# [4] chr1 [6484994, 6485043] * | ESPN
# [5] chr1 [6485043, 6485092] * | ESPN
# -------
# seqinfo: 1 sequence from an unspecified genome; no seqlengths
答案 1 :(得分:1)
除了GRanges
/IRanges
solution by Maurits Evers之外,还有另一种data.table
方法,使用非equi join 和更新加入。
A[B, on = .(chrom, start >= start, start <= end), gene := i.gene][]
chrom start end gene 1: chr1 6484847 6484896 ESPN 2: chr1 6484896 6484945 ESPN 3: chr1 6484945 6484994 ESPN 4: chr1 6484994 6485043 ESPN 5: chr1 6485043 6485092 ESPN 6: chrX 106893605 106893654 PRPS1 7: chrX 106893654 106893703 PRPS1 8: chrX 106893703 106893752 PRPS1 9: chrX 106893752 106893801 PRPS1 10: chrX 106893801 106894256 PRPS1
根据OP,A
和B
已经是data.table
个对象。因此,这种方法避免了对GRanges
个对象的强制。
library(data.table)
A <- fread("rn chrom start end
1: chr1 6484847 6484896
2: chr1 6484896 6484945
3: chr1 6484945 6484994
4: chr1 6484994 6485043
5: chr1 6485043 6485092
183569: chrX 106893605 106893654
183570: chrX 106893654 106893703
183571: chrX 106893703 106893752
183572: chrX 106893752 106893801
183573: chrX 106893801 106894256", drop = 1L)
B <- fread("rn chrom start end gene
1: chr1 6484847 6521004 ESPN
2: chr1 41249683 41306124 KCNQ4
3: chr1 55464616 55474465 BSND
42: chrX 82763268 82764775 POU3F4
43: chrX 100600643 100603957 TIMM8A
44: chrX 106871653 106894256 PRPS1", drop = 1L)