[英]R: mapply over objects of two lists and return a list of data frames
我有兩個GRange列表,我試圖將countOverlaps函數應用於列表的每個組合,並返回如下結果列表:
library(GenomicRanges)
gr1 <- GRanges(seqnames = c("chr1", "chr2"), ranges = IRanges(c(7,13), width = 3), strand = c("+", "-"))
gr2 <- GRanges(seqnames = c("chr1", "chr3"), ranges = IRanges(c(5,13), width = 3), strand = c("+", "-"))
grlA <- GRangesList("a" = gr1, "b" = gr2)
gr1 <- GRanges(seqnames = c("chr1", "chr2"), ranges = IRanges(c(1,13), width = 3), strand = c("+", "-"))
gr2 <- GRanges(seqnames = c("chr1", "chr3"), ranges = IRanges(c(3,13), width = 3), strand = c("+", "-"))
grlB <- GRangesList("c" = gr1, "d" = gr2)
我想在grlA中獲得對象“ a”和對象“ b”的列表,其中包含針對grlB的每個值的函數結果:
(列出$ a,$ b和c,d的數據框)
$ c
b
$ d
b
這樣可以獲取列表的所有組合:
comb_apply <- function(f,..., MoreArgs=list()){
exp <- unname(as.list(expand.grid(...,stringsAsFactors = FALSE)))
do.call(mapply, c(list(FUN=f, SIMPLIFY=FALSE, MoreArgs=MoreArgs), exp))
}
# This function is thanks to Michael Lawrence's help posted in the bioconductor package
t= comb_apply(function(i, j) countOverlaps(grlA[[i]], grlB[[j]]), seq_along(grlA), seq_along(grlB))
names(t)=apply(expand.grid(names(grlA), names(grlB)), 1, paste, collapse="_")
但是然后要獲得我想要的內容(數據幀列表),我需要使用grep命令來選擇屬於grlB的數據幀並將它們保存在單獨的列表中,但這確實很慢。
new=list()
for (i in names(grlB)) {
df = as.data.frame(t[grep(i,names(t))])
new[[length(new)+1]] <- df
}
有沒有grep的另一種方法我可以做到這一點? 謝謝!
此數據不應采用列表結構,因為它具有可預測且一致的結構。 我將其放入數據框,然后將其成形為大致與所需格式相同的格式。
library(dplyr)
library(tidyr)
t %>%
as.data.frame %>%
mutate(ID = 1:n()) %>%
gather(variable, value, -ID) %>%
separate(variable, c("A", "B")) %>%
spread(ID, value) %>%
group_by(B) %>%
do(result = my_function(.) )
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