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我正在尝试使用多个函数聚合数据。它运行良好，除了输出不是它所说的那样。

df<-data.frame(PLTID=rep(paste(letters), 5), R=rnorm(130, 150, 40), G=rnorm(130, 100, 50), B=rnorm(130, 200, 25))

agg.data<-aggregate(data=df, . ~ PLTID, FUN = function(x) c(mean=mean(x, na.rm=TRUE), sd=sd(x, na.rm=TRUE), n=length(x)))

> head(agg.data)
  PLTID    R.mean      R.sd       R.n    G.mean      G.sd       G.n     B.mean       B.sd        B.n
1     a 144.29202  28.49934   5.00000  87.85852  76.28156   5.00000 192.230731  29.349837   5.000000
2     b 148.41993  31.37764   5.00000  84.14367  21.59658   5.00000 224.862334  15.769459   5.000000
3     c 158.89111  17.75320   5.00000 114.40942  53.81999   5.00000 205.931132   5.790658   5.000000
4     d 153.70118  34.68649   5.00000 137.97890  54.79668   5.00000 209.587936  21.877864   5.000000
5     e 154.28002  27.33376   5.00000 124.87306  71.09103   5.00000 225.774236  17.720090   5.000000
6     f 144.08148  37.30555   5.00000  57.15275  44.68034   5.00000 204.709050  18.207047   5.000000

这就是我想要的……除非我以后想访问这些列，但它们实际上并不存在。

> names(agg.data)

[1] "PLTID" "R"     "G"     "B"



> str(agg.data)
'data.frame':   26 obs. of  4 variables:
 $ PLTID: Factor w/ 26 levels "a","b","c","d",..: 1 2 3 4 5 6 7 8 9 10 ...
 $ R    : num [1:26, 1:3] 144 148 159 154 154 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : NULL
  .. ..$ : chr  "mean" "sd" "n"
 $ G    : num [1:26, 1:3] 87.9 84.1 114.4 138 124.9 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : NULL
  .. ..$ : chr  "mean" "sd" "n"
 $ B    : num [1:26, 1:3] 192 225 206 210 226 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : NULL
  .. ..$ : chr  "mean" "sd" "n"

> sessionInfo()
R version 3.0.2 (2013-09-25)
Platform: x86_64-apple-darwin10.8.0 (64-bit)

locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8

attached base packages:
[1] grid      stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] data.table_1.8.10 plyr_1.8          gridExtra_0.9.1   rgl_0.93.963      ggplot2_0.9.3.1  

loaded via a namespace (and not attached):
 [1] colorspace_1.2-4   dichromat_2.0-0    digest_0.6.3       gtable_0.1.2       labeling_0.2       MASS_7.3-29        munsell_0.4.2      proto_0.3-10       RColorBrewer_1.0-5
[10] reshape2_1.2.2     scales_0.2.3       stringr_0.6.2      tools_3.0.2
4

1 回答 1

4

通常，当我遇到这个问题时，我会使用 ado.call(data.frame, ...)来“展平”输出。请参见下面的示例：

agg.data.2 <- do.call(data.frame, agg.data)
head(agg.data.2)
#   PLTID   R.mean     R.sd R.n    G.mean      G.sd G.n   B.mean     B.sd B.n
# 1     a 154.5559 33.43817   5 137.17159  26.21461   5 196.7478 28.59095   5
# 2     b 156.8467 32.62164   5 131.35181  48.94969   5 206.3617 17.97154   5
# 3     c 134.1172 21.83323   5  50.35166  71.57083   5 176.2187 38.82059   5
# 4     d 143.5015 43.31296   5 104.60027 103.66002   5 210.0669 18.40418   5
# 5     e 141.8565 27.02871   5  76.82525  47.76355   5 194.3454 30.87080   5
# 6     f 143.3337 12.14870   5 105.95271  42.22400   5 189.3063 12.11930   5
str(agg.data.2)
# 'data.frame': 26 obs. of  10 variables:
# $ PLTID : Factor w/ 26 levels "a","b","c","d",..: 1 2 3 4 5 6 7 8 9 10 ...
# $ R.mean: num  155 157 134 144 142 ...
# $ R.sd  : num  33.4 32.6 21.8 43.3 27 ...
# $ R.n   : num  5 5 5 5 5 5 5 5 5 5 ...
# $ G.mean: num  137.2 131.4 50.4 104.6 76.8 ...
# $ G.sd  : num  26.2 48.9 71.6 103.7 47.8 ...
# $ G.n   : num  5 5 5 5 5 5 5 5 5 5 ...
# $ B.mean: num  197 206 176 210 194 ...
# $ B.sd  : num  28.6 18 38.8 18.4 30.9 ...
# $ B.n   : num  5 5 5 5 5 5 5 5 5 5 ...
于 2013-10-30T10:29:27.303 回答