根据零之间的数据添加新列 [英] Add a new column, based on data in between zeroes

问题描述

我每秒收集一次功率数据（功率）（样本）。因此，我的data.frame的结构如下：

I have power data (Power) collected every second (Sample). My data.frame is therefore structured as follows:

Test <- data.frame(Sample = c(1:20), 
                   Power = c(0,0,0,0,0,50,67,100,92,0,0,0,36,89,36,0,0,0,89,90))

功率输入的数量取决于人类骑自行车努力并偶尔休息的力量。因此，电源不会以有序的方式出现。由于没有标记来指示何时开始和停止工作，因此我想包括这个细节。当功率> 0且可以基于数据组一起评估每个工作的开始/停止时，可以描述工作。

The number of power entries is dependent upon a human performing an effort on a bike and resting sporadically. Therefore, power does not appear in an ordered fashion. As there are no markers to indicate when an effort starts and stops, I want to include this detail. An effort can be characterised when power > 0 and the start/ stop of each effort can be assessed based on data group together.

我现在希望包括一个新列（标记），用于查找分组在一起并用零分隔的功率数据。例如，我的预期输出将是：

I now wish to include a new column (Marker) that looks for power data grouped together and separated by zeroes. For example, my anticipated output would be:

Test$Marker <- c("Rest","Rest","Rest","Rest","Rest","Effort 1","Effort 1","Effort 1","Effort 1",
                 "Rest","Rest","Rest","Effort 2","Effort 2","Effort 2","Rest","Rest","Rest",
                 "Effort 3","Effort 3")

不幸的是，我的原始数据的长度大于3000行，因此手动执行此操作将很繁琐！

Unfortunately my raw data is > 3000 rows long, so to do this manually would be tedious! How do I please go about doing this in R?

推荐答案

以R为基数的选项：

indx1 = with(rle(Test$Power>0),rep(values,lengths))
indx2 = with(rle(Test$Power>0),rep(cumsum(values),lengths))
Test$Effort[indx1] = paste0("Effort",indx2[indx1])
Test$Effort[!indx1]="Rest"

输出：

   Sample Power  Effort
1       1     0    Rest
2       2     0    Rest
3       3     0    Rest
4       4     0    Rest
5       5     0    Rest
6       6    50 Effort1
7       7    67 Effort1
8       8   100 Effort1
9       9    92 Effort1
10     10     0    Rest
11     11     0    Rest
12     12     0    Rest
13     13    36 Effort2
14     14    89 Effort2
15     15    36 Effort2
16     16     0    Rest
17     17     0    Rest
18     18     0    Rest
19     19    89 Effort3
20     20    90 Effort3

3000行约0.0038秒;）希望这会有所帮助！

About 0.0038 seconds for 3,000 rows ;) Hope this helps!

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