对于data.frame中的每一行，获取R中前n个值的标准偏差 [英] For each row in a data.frame, get the standard deviation of the previous n values in R

问题描述

我正在努力获取前n个值的标准偏差。或者我最近5天。

I am struggling to get the standard deviation of the previous n values. Or in my case the last 5 days.

我以下面的代码为例：

df<- data.frame(date = seq(as.Date("2019-12-01"), as.Date("2020-03-31"), by="days"),
                TRM= runif(122, min=3500, max=4100))
> df
          date      TRM
1   2019-12-01 3540.028
2   2019-12-02 3673.536
3   2019-12-03 3827.182
4   2019-12-04 3824.791
5   2019-12-05 3906.753
6   2019-12-06 3528.100
7   2019-12-07 3650.191
# ... with more rows

然后我使用 mutate 添加一些我需要的信息，将显示最后一行：

Then I use mutate to add some information that I need, I will show you the last rows:

df<-mutate(df, diferencia = TRM - lag(TRM, 1),
           VAR=diferencia/lag(TRM, 1))
>df
          date      TRM  diferencia          VAR
118 2020-03-27 3779.479 -262.366328 -0.064912515
119 2020-03-28 3773.771   -5.708207 -0.001510316
120 2020-03-29 4097.078  323.307069  0.085672159 
121 2020-03-30 3752.619 -344.459061 -0.084074332 
122 2020-03-31 3707.442  -45.176979 -0.012038788 

所以我需要的是以下内容：

So what I need is the following:

1. 创建具有 sd 的列


2. 每行的 sd 必须仅包含 VAR列的最后5天。


3. 如果所有这些都可以通过 dply 完成，那就太好了。 （不是必需的）

1. Create a column that have the sd for the column "VAR".
2. That the sd for each row must contain only the last 5 days of the column "VAR".
3. If all this could be done with dply, would be great. (Not necessary)

例如，对于第122行，结果为：

For example, for the row 122 the result would be this:

 > sd(df[118:122,4])
[1] 0.06630885

那又怎样我要得到的是 df 的所有行的此值，我以5天为例，但是我想修改范围：

So what I what to get is this value for all the rows of my df, I used 5 days as an example but I would like to modify the range:

          date      TRM  diferencia          VAR  diff5days
118 2020-03-27 3779.479 -262.366328 -0.064912515 0.05801765
119 2020-03-28 3773.771   -5.708207 -0.001510316 0.04799908
120 2020-03-29 4097.078  323.307069  0.085672159 0.06207932
121 2020-03-30 3752.619 -344.459061 -0.084074332 0.07522609
122 2020-03-31 3707.442  -45.176979 -0.012038788 0.06630885

谢谢！

推荐答案

这里是使用Base R的解决方案：

Here is a solution using Base R:

df<- data.frame(date = seq(as.Date("2019-12-01"), as.Date("2020-03-31"), by="days"),
                TRM= runif(122, min=3500, max=4100))
df$stDev <- NA

for(i in 5:nrow(df)) df$stDev[i] <- sd(df$TRM[(i - 4):i])

...以及输出：

> head(df,n = 10)
         date      TRM rownum     stDev
1  2019-12-01 3553.666      1        NA
2  2019-12-02 4054.015      2        NA
3  2019-12-03 3976.555      3        NA
4  2019-12-04 3825.628      4        NA
5  2019-12-05 4036.383      5 208.01581
6  2019-12-06 3787.414      6 122.38142
7  2019-12-07 3886.663      7 103.45743
8  2019-12-08 3930.801      8  97.10099
9  2019-12-09 3626.911      9 155.10571
10 2019-12-10 3781.731     10 117.29726
>

我们可以验证前三行的结果，如下所示：

We can verify the results for the first three rows as follows:

> # verify first three results
> sd(df$TRM[1:5])
[1] 208.0158
> sd(df$TRM[2:6])
[1] 122.3814
> sd(df$TRM[3:7])
[1] 103.4574
>

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