2

我有这个data.frame。我需要做的是使用具有相同类型数据的所有其他列的方式创建 2 个额外的列。

这是dput(head())文件的

         structure(list(COLE_CODIGO_COLEGIO = c(182L, 3046L, 3327L, 3418L, 
3459L, 3525L), Utilidad12.1 = c(5.67960793875611e+84, 6.70878394856429e+83, 
3.10783852265787e+84, 1.80000434506295e+84, 7.92256239238555e+84, 
2.6032583906869e+83), genero.x = c(0.581395348837209, 0.525423728813559, 
0.636363636363636, 0.55, 0.488636363636364, 0.63768115942029), 
    Utilidad11.1 = c(1.15336934758534e+73, 3.77916671655328e+72, 
    4.7319512062371e+73, 7.97038534283783e+72, 1.32237182934735e+73, 
    6.55595718179632e+72), genero.y = c(0.527472527472527, 0.616666666666667, 
    0.526315789473684, 0.560975609756098, 0.450704225352113, 
    0.742268041237113), Utilidad10.1 = c(5.09613856124168e+72, 
    5.25727948275145e+71, 1.4350276514895e+72, 7.04968791434072e+72, 
    5.97923875650689e+72, 4.30547735977066e+72), genero.x = c(0.566371681415929, 
    0.315789473684211, 0.571428571428571, 0.578947368421053, 
    0.353658536585366, 0.743243243243243), Utilidad07.1 = c(1.401355333064e+71, 
    1.35282220680438e+71, 9.87359187347488e+71, 1.89236591312778e+72, 
    1.18195124980311e+73, 5.99279404969151e+70), genero.y = c(0.534883720930233, 
    0.522727272727273, 0.5, 0.560975609756098, 0.411764705882353, 
    0.714285714285714), Utilidad06.1 = c(4.47939696971958e+72, 
    2.91946645871643e+72, 3.16785158272574e+72, 1.16889985482301e+74, 
    5.83958109398712e+74, 5.37640536403147e+71), genero = c(0.5, 
    0.6, 0.357142857142857, 0.5, 0.422680412371134, 0.695652173913043
    )), .Names = c("COLE_CODIGO_COLEGIO", "Utilidad12.1", "genero.x", 
"Utilidad11.1", "genero.y", "Utilidad10.1", "genero.x", "Utilidad07.1", 
"genero.y", "Utilidad06.1", "genero"), row.names = c(NA, 6L), class = "data.frame")

我的意思是我想创建一个名称为“MeanUtilidad”的列，它是其中所有其他名称为“Utilidad”的列的平均值。还有一个名称为“Meangenero”的列，其中包含所有其他变量的平均值，其中名称为“genera”。

我希望我已经足够清楚地表达了我的问题。

我尝试使用以下代码Semestre1["UProm"]<-apply(Semestre1 [ ,2:dim(Semestre1)[2]],1,mean, na.rm=T)，但它考虑了每一列的平均值，我无法找到选择每个特定列的方法。

4

2 回答 2

4

你可以这样做：

dat$MeanUtilidad <- rowMeans(dat[grep("Utilidad", names(dat), value = TRUE)])
dat$MeanGenero   <- rowMeans(dat[grep("genero",   names(dat), value = TRUE)])
于 2013-05-09T23:23:18.070 回答
1

rowMeans将强制输入到矩阵。你可以使用Reduce(`+`, list)

例如

nUtilidad <- grep("Utilidad", names(dat), value = TRUE)
nGenero   <- grep("genero",   names(dat), value = TRUE)

dat$MeanUtilidad <- Reduce(`+`, dat[nUtilidad]) / length(nUtilidad)
dat$MeanGenero   <- Reduce(`+`, dat[nGenero]) / length(nGenero)

对这个小例子的 rowMeans 进行基准测试

library(microbenchmark)
microbenchmark(Reduce(`+`,dat[nUtilidad])/length(nUtilidad), rowMeans(dat[nUtilidad]))
## Unit: microseconds
##                                          expr     min       lq   median       uq     max neval
## Reduce(`+`, dat[nUtilidad])/length(nUtilidad)  75.561  77.6105  79.2025  80.4055 451.967   100
##                      rowMeans(dat[nUtilidad]) 178.477 179.9305 180.9955 182.6255 321.482   100
于 2013-05-09T23:36:57.257 回答