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我对 R 很陌生，所以我的问题可能非常基本。我有以下数据框

                FM_1      SBM_1       FM_2        BP_1       BP_2       SBM_2
K00121 -0.1839897960 -0.8656314 -0.8411707 -0.69968109 -0.8031558 -0.70689896
K08660 -0.5250720652 -0.1513665 -0.2865290 -0.01167864 -0.4330590 -0.52919490
K07408 -0.3784026846 -0.1521273  0.1021097 -0.40613804 -0.4201983 -0.27915511
K13524 -0.4049012076 -0.8533916 -0.4431474 -0.15884372 -0.5256129 -0.54496893
K00600 -0.0009098706  0.2313674 -0.1080085 -0.07682120 -0.1740538  0.09553883
K00286 -0.2710184537 -0.2543416  0.1453829 -0.11907861  0.3392705 -0.19903857

我想创建一个具有相同行的新数据框，但作为列，具有相同前缀的列的平均值（即“FM”、“SBM”、“BP”）。我正在尝试函数 aggregate() 但我在使用“by”参数时遇到了麻烦。我做错了。有人可以给我一个提示吗？非常感谢。

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3 回答 3

3

这是另一种选择

> prefix <- unique(unlist(strsplit(names(df), "\\_[0-9]")))
> sapply(prefix, function(i) rowMeans(df[, grepl(i, names(df))]))
                FM        SBM         BP
K00121 -0.51258025 -0.7862652 -0.7514184
K08660 -0.40580053 -0.3402807 -0.2223688
K07408 -0.13814649 -0.2156412 -0.4131682
K13524 -0.42402430 -0.6991803 -0.3422283
K00600 -0.05445919  0.1634531 -0.1254375
K00286 -0.06281778 -0.2266901  0.1100959
于 2013-11-12T17:04:25.493 回答
3

df1 是你的数据框

   vars<-c("FM","SBM","BP")     
    sapply(vars,function(x)apply(df1[,grep(x,names(df1))],1,mean))
                        FM        SBM         BP
        K00121 -0.51258025 -0.7862652 -0.7514184
        K08660 -0.40580053 -0.3402807 -0.2223688
        K07408 -0.13814649 -0.2156412 -0.4131682
        K13524 -0.42402430 -0.6991803 -0.3422283
        K00600 -0.05445919  0.1634531 -0.1254375
        K00286 -0.06281778 -0.2266901  0.1100959
于 2013-11-12T17:06:07.543 回答
2

您可以使用meltand dcastfrom "reshape2"。假设您data.frame的名称为“mydf”，请尝试以下操作：

library(reshape2)

## melt your data. Since you have rownames, wrap in `as.matrix`
##   to get the rownames as a variable in the long data.frame
dfL <- melt(as.matrix(mydf))

## Your "Var2" column should be split to give us access to the "variable" 
##    and "time" values. (Only the "variable" part is required here.)
dfL <- cbind(dfL, colsplit(as.character(dfL$Var2), "_", c("var", "time")))

## The new data now look like this:
head(dfL)
#     Var1 Var2         value var time
# 1 K00121 FM_1 -0.1839897960  FM    1
# 2 K08660 FM_1 -0.5250720652  FM    1
# 3 K07408 FM_1 -0.3784026846  FM    1
# 4 K13524 FM_1 -0.4049012076  FM    1
# 5 K00600 FM_1 -0.0009098706  FM    1
# 6 K00286 FM_1 -0.2710184537  FM    1

## From here, it's easy to aggregate with `dcast`
dcast(dfL, Var1 ~ var, value.var="value", fun.aggregate=mean)
#     Var1         BP          FM        SBM
# 1 K00121 -0.7514184 -0.51258025 -0.7862652
# 2 K00286  0.1100959 -0.06281778 -0.2266901
# 3 K00600 -0.1254375 -0.05445919  0.1634531
# 4 K07408 -0.4131682 -0.13814649 -0.2156412
# 5 K08660 -0.2223688 -0.40580053 -0.3402807
# 6 K13524 -0.3422283 -0.42402430 -0.6991803

从“长”形式开始，您也可以使用aggregate(try aggregate(value ~ Var1 + var, dfL, mean))，但结果本身将是长形式。

于 2013-11-12T16:59:27.387 回答