如何通过对某些列进行分组来重塑数据框架

Question

假设我有一个包含5列的R数据框，如下所示

time MeanVar1 SdVar1 MedianVar1 MeanVar2 SdVar2
1 -0.8453978 -1.636985 -0.6239832 -0.4366982 -1.7037374
2 -0.3000778 -1.034199  0.3292459 -0.6606399 -0.1525361

是否有一个简洁的方法来使dataFrame如下：

Var time Mean/Median SD
1 1 -0.8453978 -1.636985
1 2 -0.3000778 -1.034199
1 1 -0.6239832 N/A
1 2  0.3292459 N/A 
2 1 -0.4366982 -1.7037374
2 2 -0.6606399 -0.1525361

要么

Var time Mean/Median SD
MeanVar1 1 -0.8453978 -1.636985
MeanVar1 2 -0.3000778 -1.034199
MeanVar1 1 -0.6239832 N/A
MeanVar1 2  0.3292459 N/A 
MeanVar2 1 -0.4366982 -1.7037374
MeanVar2 2 -0.6606399 -0.1525361

我的总体意图是在同一图中绘制变量1的均值，SD与变量1，变量1的中值和平均值，变量1的SD。 因此，我觉得如果我以这种格式修改数据，我可以立即绘制它而不是分别绘制每一行。

由于我对Reshape和融化的知识有限，我无法做到这一点。

编辑：添加更多信息

样本输入（给定3行，共有100行）：

Label   trainingSize    Accuracy_Mean   Accuracy_SD Accuracy_SE Precision_Mean  Recall_Mean F1  Accuracy_Median PriorClass0_Mean    PriorClass0_SD  PriorClass0_SE  ProbabilityEstimate_0given0_Mean    ProbabilityEstimate_0given0_SD  ProbabilityEstimate_0given0_SE  ProbabilityEstimate_0given1_Mean    ProbabilityEstimate_0given1_SD  ProbabilityEstimate_0given1_SE

0perc_0repeat   0.4 0.5506  0.0531  0.0038  0.6374  0.2336  0.3419  0.5372  0.5278  0.0254  0.0018  0.6433  0.0028  0.0 0.4169  0.003   0.0
0perc_0repeat   0.4 0.5456  0.0482  0.0034  0.6465  0.2142  0.3218  0.5333  0.5304  0.0248  0.0018  0.6414  0.0028  0.0 0.4193  0.0027  0.0
0perc_0repeat   0.4 0.5574  0.0555  0.0039  0.6604  0.2197  0.3297  0.5404  0.529   0.0233  0.0016  0.6436  0.003   0.0 0.4163  0.0029  0.0

我试图策划

1) the iteration number(1:100) in X Axis and the points of 5 columns (Accuracy_Mean, Accuracy_Median, PriorClass0_Mean, ProbabilityEstimate_0given0_Mean, ProbabilityEstimate_0given1_Mean in the Y AXIS. 2) distribution (density obtained by 100 points) of 5 columns with error bars (either SD or SE) in a single plot using ggplot.

我有4列Precision_Mean，Recall_Mean，F1，Accuracy_Median不遵循均值，sd，se模式！

编辑1：1）

dput（droplevels（head（data，3）））结构（list（标签=结构（c（1L，1L，1L），。Label =“0perc_0repeat”，class =“factor”），trainingSize = c（0.4,0.4 ，0.4），Accuracy_Mean = c（0.5506,0.5456,0.5574），Accuracy_SD = c（0.0531,0.0482,0.0555），Accuracy_SE = c（0.0038,0.0034,0.0039），Precision_Mean = c（0.6374,0.6646,0.6604），Recall_Mean = c（0.2336,0.2142,0.2197），F1 = c（0.3419,0.3218,0.3297），Accuracy_Median = c（0.5372,0.5333,0.5404），PriorClass0_Mean = c（0.5278,0.5304,0.529），PriorClass0_SD = c（0.0254,0.0248， 0.0233），PriorClass0_SE = c（0.0018,0.0018,0.0016），ProbabilityEstimate_0given0_Mean = c（0.6433,0.6414,0.6646），ProbabilityEstimate_0given0_SD = c（0.0028,0.0028,0.003），ProbabilityEstimate_0given0_SE = c（0,0,0），ProbabilityEstimate_0given1_Mean = c （0.4169,0.4193,0.4163），ProbabilityEstimate_0given1_SD = c（0.003,0.0027,0.0029），ProbabilityEstimate_0given1_SE = c（0,0,0））,. Name = c（“label”，“trainingSize”，“Accuracy_Mean”，“Accuracy_SD “，”Accuracy_SE“，”Prec ision_Mean”， “Recall_Mean”， “F1”， “Accuracy_Median”， “PriorClass0_Mean”， “PriorClass0_SD”， “PriorClass0_SE”， “ProbabilityEstimate_0given0_Mean”， “ProbabilityEstimate_0given0_SD”， “ProbabilityEstimate_0given0_SE”， “ProbabilityEstimate_0given1_Mean”， “ProbabilityEstimate_0given1_SD”， “ProbabilityEstimate_0given1_SE” ），row.names = c（NA，3L），class =“data.frame”）

2）预期输出类似于：

Vars  Label  trainingSize  Mean  SD  SE

Vars：Mean，PriorClass0，ProbabilityEstimate_0given0，ProbabilityEstimate_0given1; （中位数，精度，召回，F1不是必需的，或者它们可以适合上表，SD，SE为N / A或0）。

Answer 1

merged.stack从我的“splitstackshape”包处理这在一定程度上，但它会从你的“SdVar”回收列中的值（所以我没有得到的NA值在您需要的输出显示）。

尽管如此，它可能是解决问题的开始，所以这是方法：

library(splitstackshape)
merged.stack(mydf, var.stubs = c("MeanVar|MedianVar", "SdVar"), sep = "var.stubs")
#    time .time_1 MeanVar|MedianVar      SdVar
# 1:    1       1        -0.8453978 -1.6369850
# 2:    1       1        -0.6239832 -1.6369850
# 3:    1       2        -0.4366982 -1.7037374
# 4:    2       1        -0.3000778 -1.0341990
# 5:    2       1         0.3292459 -1.0341990
# 6:    2       2        -0.6606399 -0.1525361

---

如果你真的想要那些NA值，也许这样就可以了：

merged.stack(
  mydf, var.stubs = c("MeanVar|MedianVar", "SdVar"), 
  sep = "var.stubs")[, SdVar := ifelse(
    duplicated(SdVar), NA, SdVar), by = time][]
#    time .time_1 MeanVar|MedianVar      SdVar
# 1:    1       1        -0.8453978 -1.6369850
# 2:    1       1        -0.6239832         NA
# 3:    1       2        -0.4366982 -1.7037374
# 4:    2       1        -0.3000778 -1.0341990
# 5:    2       1         0.3292459         NA
# 6:    2       2        -0.6606399 -0.1525361