根据一列排列数据帧，消除不必要的响应 [英] arrange dataframe based on one column eliminating the unwanted responses
问题描述
我有此数据
日期信号
1 2009-01-13 09:55:00 4645.00 4838.931 5358.883 Buy2
2 2009-01-14 09:55:00 4767.50 4718.254 5336.703 Buy1
3 2009-01-15 09:55:00 4485.00 4653.316 5274.384 Buy2
4 2009-01-16 09:55:00 4580.00 4537.693 5141.435 Buy1
5 2009-01-19 09:55:00 4532.00 4548.088 4891.041 Buy2
6 2009-01-27 09:55:00 4190.00 4183.503 4548.497 Buy1
7 2009-01-30 09:55:00 4436.00 4155.236 4377.907 Sell1
8 2009-02-02 09:55:00 4217.00 4152.626 4390.802 Sell2
9 2009-02-09 09:55:00 4469.00 4203.437 4376.277 Sell1
10 2009-02-12 09:55:00 4469.90 4220.845 4503.798 Sell2
11 2009-02-13 09:55:00 4553.00 4261.980 4529.777 Sell1
12 2009-02 -16 09:55:00 4347.20 4319.656 4564.387 Sell2
13 2009-02-17 09:55:00 4161.05 4371.474 4548.912 Buy2
14 2009-02-27 09:55:00 3875.55 3862.085 4101.929 Buy1
15 2009-03-02 09:55:00 3636.00 3846.423 4036.020 Buy2
16 2009-03-12 09:55:00 3420.00 3372.665 3734.949 Buy1
17 2009-03-13 09 ：55：00 3656.00 3372.100 3605.357 Sell1
18 2009-03-17 09:55:00 3650.00 3360.421 3663.322 Sell2
19 2009-03-18 09:55:00 3721.00 3363.735 3682.293 Sell1
20 2009-03-20 09:55:00 3687.00 3440.651 3784.778 Sell2
，并且必须将其安排在这种形式
2 2009-01-14 09:55:00 4767.50 4718.254 5336.703 Buy1
7 2009-01- 30 09:55:00 4436.00 4155.236 4377.907 Sell1
8 2009-02-02 09:55:00 4217.00 4152.626 4390.802 Sell2
13 2009-02-17 09:55:00 4161.05 4371.474 4548.912 Buy2
14 2009-02-27 09:55:00 3875.55 3862.085 4101.929 Buy1
17 2009-03-13 09:55:00 3656.00 3372.100 3605.357 Sell1
18 2009-03-17 09:55： 00 3650.00 3360.421 3663.322 Sell2
因此数据按以下顺序排列Buy1 Sell1 Sell2 Buy2并消除中间观察值。
我已经尝试了几个dplyr：filter命令，但是没有一个给出期望的输出。
如果我很了解您的问题，以下代码可以解决。改编自此讨论。
这个想法是将序列定义为模式：
模式<-c（ Buy1， Sell1， Sell2， Buy2）
然后在您的列中找到该模式的位置：
library（zoo）
pos< -which（rollapply（data $ signal，4，same，pattern，fill = FALSE，align = left））
并提取模式位置之后的行：
行<-unlist（lapply（pos，函数（x，n）seq（x，x + n-1），4））
data_filtered<-data [rows]
Voilà。
编辑
<因为我误解了您的问题，所以这里有一个新的解决方案。
您想在列中检索序列 Buy1， Sell1， Sell2， Buy2，并消除不适合该序列的观察值。我没有看到简单的矢量化解决方案，因此这里有一个循环来解决这个问题。根据数据的大小，您可能想在RCPP中实现类似的算法或以某种方式对其进行矢量化。
序列<-c（ Buy1， Sell1， Sell2， Buy2）
保持<-逻辑（length（data（signal）））
s<-0
for（i in seq（1，length（data $ signal）））{
if（sequence [s +1] == data $ signal [i]）{
keep [i ]<-T
s<-（（s + 1）%% 4
}否则{
keep [i]<-F
}
}
data_filtered<-data [keep，]
告诉我这项工作更好。
如果有人有矢量解决方案，我会很好奇。
I have this data
date signal
1 2009-01-13 09:55:00 4645.00 4838.931 5358.883 Buy2
2 2009-01-14 09:55:00 4767.50 4718.254 5336.703 Buy1
3 2009-01-15 09:55:00 4485.00 4653.316 5274.384 Buy2
4 2009-01-16 09:55:00 4580.00 4537.693 5141.435 Buy1
5 2009-01-19 09:55:00 4532.00 4548.088 4891.041 Buy2
6 2009-01-27 09:55:00 4190.00 4183.503 4548.497 Buy1
7 2009-01-30 09:55:00 4436.00 4155.236 4377.907 Sell1
8 2009-02-02 09:55:00 4217.00 4152.626 4390.802 Sell2
9 2009-02-09 09:55:00 4469.00 4203.437 4376.277 Sell1
10 2009-02-12 09:55:00 4469.90 4220.845 4503.798 Sell2
11 2009-02-13 09:55:00 4553.00 4261.980 4529.777 Sell1
12 2009-02-16 09:55:00 4347.20 4319.656 4564.387 Sell2
13 2009-02-17 09:55:00 4161.05 4371.474 4548.912 Buy2
14 2009-02-27 09:55:00 3875.55 3862.085 4101.929 Buy1
15 2009-03-02 09:55:00 3636.00 3846.423 4036.020 Buy2
16 2009-03-12 09:55:00 3420.00 3372.665 3734.949 Buy1
17 2009-03-13 09:55:00 3656.00 3372.100 3605.357 Sell1
18 2009-03-17 09:55:00 3650.00 3360.421 3663.322 Sell2
19 2009-03-18 09:55:00 3721.00 3363.735 3682.293 Sell1
20 2009-03-20 09:55:00 3687.00 3440.651 3784.778 Sell2
and have to arrange it in this form
2 2009-01-14 09:55:00 4767.50 4718.254 5336.703 Buy1
7 2009-01-30 09:55:00 4436.00 4155.236 4377.907 Sell1
8 2009-02-02 09:55:00 4217.00 4152.626 4390.802 Sell2
13 2009-02-17 09:55:00 4161.05 4371.474 4548.912 Buy2
14 2009-02-27 09:55:00 3875.55 3862.085 4101.929 Buy1
17 2009-03-13 09:55:00 3656.00 3372.100 3605.357 Sell1
18 2009-03-17 09:55:00 3650.00 3360.421 3663.322 Sell2
So that data is arranged in order of Buy1 Sell1 Sell2 Buy2 and eliminating the middle observations. I have tried several dplyr:filter commands but none is giving the desired output.
If I have well understood your problem, the following code should solve it. It is adapted from this discussion.
The idea is to define your sequence as a pattern:
pattern <- c("Buy1", "Sell1", "Sell2", "Buy2")
Then find the position of this pattern in your column:
library(zoo)
pos <- which(rollapply(data$signal, 4, identical, pattern, fill = FALSE, align = "left"))
and extract the rows following the position of your patterns:
rows <- unlist(lapply(pos, function(x, n) seq(x, x+n-1), 4))
data_filtered <- data[rows,]
Voilà.
EDIT
Since I had misunderstood your problem, here is a new solution. You want to retrieve the sequence "Buy1", "Sell1", "Sell2", "Buy2" in your column, and eliminate the observations that do not fit in this sequence. I do not see a trivial vectorised solution, so here is a loop to solve that. Depending on the size of your data, you may want to implement a similar algorithm in RCPP or vectorise it in some ways.
sequence <- c("Buy1", "Sell1", "Sell2", "Buy2")
keep <- logical(length(data$signal))
s <- 0
for (i in seq(1, length(data$signal))){
if (sequence[s +1] == data$signal[i]){
keep[i] <- T
s <- (s + 1) %% 4
} else {
keep[i] <- F
}
}
data_filtered <- data[keep,]
Tell me if this work better. If anyone has a vectorised solution, I would be curious to see it.
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