在R中按行排列数据
在R中按行排列数据
提问于 2017-04-13 10:00:17
我有OMIM基因列表(大约15,000个基因)与相应的疾病如下：
SLC6A8,CRTR,CCDS1 Cerebral creatine deficiency syndrome 1, 300352 (3)
BCAP31,BAP31,DXS1357E,DDCH Deafness, dystonia, and cerebral hypomyelination
ABCD1,ALD,AMN Adrenoleukodystrophy, 300100 (3), X-linked recessive
PLXNB3,PLXN6 NA 对于某些疾病，我们有多个与疾病相关的基因名称。我想把它组织起来，所以我每一行只有一个基因名和相关的疾病：
SLC6A8 Cerebral creatine deficiency syndrome 1, 300352 (3)
CRTR Cerebral creatine deficiency syndrome 1, 300352 (3)
CCDS1 Cerebral creatine deficiency syndrome 1, 300352 (3)这能在R中完成吗？
回答 1
Stack Overflow用户
回答已采纳
发布于 2017-04-13 10:20:28
不完全确定您有什么样的数据结构。下面是一个快速的解决方案，希望对您正在寻找的内容有所帮助：
splitFn <- function(x) expand.grid(df[x,"a"] %>% as.character %>% strsplit(., ",") %>% unlist, df[x, "b"])
ldply(1:nrow(df), splitFn)
Var1 Var2
1 SLC6A8 Cerebral creatine deficiency syndrome 1, 300352(3)
2 CRTR Cerebral creatine deficiency syndrome 1, 300352(3)
3 CCDS1 Cerebral creatine deficiency syndrome 1, 300352(3)
4 BCAP31 Deafness, dystonia, and cerebral hypomyelination
5 BAP31 Deafness, dystonia, and cerebral hypomyelination
6 DXS1357E Deafness, dystonia, and cerebral hypomyelination
7 DDCH Deafness, dystonia, and cerebral hypomyelination
8 ABCD1 Adrenoleukodystrophy, 300100(3), X-linked recessive
9 ALD Adrenoleukodystrophy, 300100(3), X-linked recessive
10 AMN Adrenoleukodystrophy, 300100(3), X-linked recessive
11 PLXNB3 <NA>
12 PLXN6 <NA>我用过的data.frame
df <- structure(list(a = structure(c(4L, 2L, 1L, 3L), .Label = c("ABCD1,ALD,AMN",
"BCAP31,BAP31,DXS1357E,DDCH", "PLXNB3,PLXN6", "SLC6A8,CRTR,CCDS1"
), class = "factor"), b = structure(c(1L, 3L, 2L, NA), .Label = c(" Cerebral
creatine deficiency syndrome 1, 300352(3)",
"Adrenoleukodystrophy, 300100(3), X-linked recessive", "Deafness, dystonia, and cerebral hypomyelination"
), class = "factor")), .Names = c("a", "b"), row.names = c(NA,
-4L), class = "data.frame")```页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/43389170
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