重新调整因子以便于在R中的DESeq2模型中用作嵌套因子

Question

我使用DESeq2包装适合GLM,并且具有个体(RatID)嵌套在治疗(饮食)中的情况.该软件包的作者建议,每个饮食中的个体从1:N重新调整(其中N是特定饮食中的RatID数),而不是其原始ID /因子水平(DESeq2小插图,第35页).

数据看起来像这样(实际上有更多的列和行,但为简单起见省略):

     Diet Extraction RatID
199 HAMSP          8    65
74   HAMS          9   108
308  HAMS         18   100
41  HAMSA          3    83
88  HAMSP         12    11
221 HAMSP         14    66
200 HAMSA          8    57
155 HAMSB          1   105
245 HAMSB         19    50
254  HAMS         21    90
182 HAMSB          4     4
283 HAMSA         23    59
180 HAMSP          4    22
71  HAMSP          9   112
212  HAMS         12    63
220 HAMSP         14    54
56   HAMS          7    81
274 HAMSP          1    11
114  HAMS         17   102
143 HAMSP         22    93

这是dput()结构的输出:

data = structure(list(Diet = structure(c(4L, 1L, 1L, 2L, 4L, 4L, 2L, 
        3L, 3L, 1L, 3L, 2L, 4L, 4L, 1L, 4L, 1L, 4L, 1L, 4L), .Label = c("HAMS", 
        "HAMSA", "HAMSB", "HAMSP", "LAMS"), class = "factor"), Extraction = c(8L, 
        9L, 18L, 3L, 12L, 14L, 8L, 1L, 19L, 21L, 4L, 23L, 4L, 9L, 12L, 
        14L, 7L, 1L, 17L, 22L), RatID = structure(c(61L, 7L, 3L, 76L, 
        9L, 62L, 52L, 6L, 46L, 81L, 37L, 54L, 20L, 12L, 59L, 50L, 74L, 
        9L, 4L, 84L), .Label = c("1", "10", "100", "102", "103", "105", 
        "108", "109", "11", "110", "111", "112", "113", "13", "14", "16", 
        "17", "18", "20", "22", "23", "24", "25", "26", "27", "28", "29", 
        "3", "30", "31", "32", "34", "35", "36", "37", "39", "4", "40", 
        "42", "43", "45", "46", "48", "49", "5", "50", "51", "52", "53", 
        "54", "55", "57", "58", "59", "6", "60", "61", "62", "63", "64", 
        "65", "66", "67", "68", "69", "70", "71", "73", "77", "78", "79", 
        "8", "80", "81", "82", "83", "85", "86", "88", "89", "90", "91", 
        "92", "93", "94", "95", "96", "98", "99"), class = "factor")), .Names = c("Diet", 
        "Extraction", "RatID"), row.names = c(199L, 74L, 308L, 41L, 88L, 
        221L, 200L, 155L, 245L, 254L, 182L, 283L, 180L, 71L, 212L, 220L, 
        56L, 274L, 114L, 143L), class = "data.frame")

有人可以指定一种优雅的方法来为Diet中的RatID生成新的因子水平,作为上述data.frame的附加列.这可以通过data.table的roll函数完成吗？

期望的输出(手动完成):

    Diet Extraction RatID newCol
1  HAMSP          8    65      1
2   HAMS          9   108      1
3   HAMS         18   100      2
4  HAMSA          3    83      1
5  HAMSP         12    11      2
6  HAMSP         14    66      3
7  HAMSA          8    57      2
8  HAMSB          1   105      1
9  HAMSB         19    50      2
10  HAMS         21    90      3
11 HAMSB          4     4      3
12 HAMSA         23    59      3
13 HAMSP          4    22      4
14 HAMSP          9   112      5
15  HAMS         12    63      4
16 HAMSP         14    54      6
17  HAMS          7    81      5
18 HAMSP          1    11      2
19  HAMS         17   102      6
20 HAMSP         22    93      7

注意:每次治疗中没有相同数量的大鼠.我也喜欢不重新排序数据中的行的解决方案(如果可能的话).

编辑:对于RatIDs没有"自然"顺序,只要饮食中有1:1的映射,就可以了.

Answer 1

您可以将“RatID”转换为“因子”并将其强制回“数字”

 library(data.table)#v1.9.4+
 setDT(data)[, newCol:=as.numeric(factor(RatID, 
                       levels=unique(RatID))), Diet]
 #      Diet Extraction RatID newCol
 # 1: HAMSP          8    65      1
 # 2:  HAMS          9   108      1
 # 3:  HAMS         18   100      2
 # 4: HAMSA          3    83      1
 # 5: HAMSP         12    11      2
 # 6: HAMSP         14    66      3
 # 7: HAMSA          8    57      2
 # 8: HAMSB          1   105      1
 # 9: HAMSB         19    50      2
 #10:  HAMS         21    90      3
 #11: HAMSB          4     4      3
 #12: HAMSA         23    59      3
 #13: HAMSP          4    22      4
 #14: HAMSP          9   112      5
 #15:  HAMS         12    63      4
 #16: HAMSP         14    54      6
 #17:  HAMS          7    81      5
 #18: HAMSP          1    11      2
 #19:  HAMS         17   102      6
 #20: HAMSP         22    93      7

或者使用match

 setDT(data)[, newCol:=match(RatID, unique(RatID)), Diet]

或类似的选项base R

data$newCol <- with(data, ave(as.numeric(levels(RatID))[RatID],
       Diet, FUN=function(x) match(x, unique(x))))