我有一个像变量k的数据帧。列all_possible_names包含更多ILMN代码的标识符。 现在,我想在列all_possible_names中搜索数据帧标识符中可用的标识符。
z <- matrix(c(0,0,1,1,0,0,1,1,0,0,0,0,1,0,1,1,0,1,1,1,1,0,0,0,"RND1 | AB849382 | uc001aeu.1","WDR | AB361738 | uc001aif.1", "PLAC8 | AB271612 | uc001amd.1","TYBSA | AB859482","GRA | AB758392 | uc001aph.1","TAF | AB142353"), nrow=6,
dimnames=list(c("ILMN_1651838","ILMN_1652371","ILMN_1652464","ILMN_1652952","ILMN_1653026","ILMN_1653103"),c("A","B","C","D","all_possible_names")))
k<-as.data.frame(z)
search<-c("AB361738","RND1", "LIS")
identifier <- as.data.frame(search)
结果必须如下:
search Names
1 AB361738 WDR | AB361738 | uc001aif.1
2 RND1 RND1 | AB849382 | uc001aeu.1
3 LIS NA
创建此数据框后,可以创建最终输出。列名必须只包含以uc0开头的命名。
最终结果将是:
search Names
1 AB361738 uc001aif.1
2 RND1 uc001aeu.1
3 LIS NA
任何人都可以帮我吗?
非常感谢, Lisanne
答案 0 :(得分:1)
可能不是最好的方式,而是一种方式:
firstStep<-lapply(srch, grep, k$all_possible_names, fixed=TRUE, value=TRUE)
res<-lapply(firstStep, function(subres){
prts<-unlist(strsplit(subres, " | ", fixed=TRUE))
prts[which(substr(prts, 1, 3)=="uc0")]
})
这会将结果作为列表返回,因为您可能无法确定每个搜索字符串只有一个结果。