是否有一种简单的方法来评估范围并检查整数是否在该范围内?
除了这篇文章Check to see if a value is within a range in R?之外,我没有找到其他相关内容。
实施例
range <- cut(rep(1,5),4) # Create intervals
range.test <- range[2]
# Now I want to check whether integer 1L is within the range.test (Of course it is)
Code comes here.
我尝试使用findInterval
并将range.test
转换为向量,或使用seq
,inrange
或其他功能但失败了。
由于所有分析都基于data.table
,并且这部分分析构成了整个实践的一部分,其中输出首选为一个data.table,因此我将标记data.table
设置为确保一致性。
修改
data.table
。
dt <- data.table(structure(list(Time = c("2016-01-04 09:05:06", "2016-01-04 09:20:00","2016-01-04 09:30:00", "2016-01-04 09:30:01", "2016-01-04 09:30:02","2016-01-04 09:30:05", "2016-01-04 09:30:06", "2016-01-04 09:31:35","2016-01-04 09:31:38", "2016-01-04 09:32:33"), Price = c(105,104.1, 104.1, 103.9, 104.1, 104, 104.1, 104.1, 104.1, 104), Volume = c(9500L,23500L, 18500L, 12500L, 16118L, 13000L, 2500L, 300L, 500L, 500L), Flag = c(1L, 0L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 0L), Ticker = c("0001","0001", "0001", "0001", "0001", "0001", "0001", "0001", "0001","0001")), .Names = c("Time", "Price", "Volume", "Flag", "Ticker"), class = c("data.table", "data.frame"), row.names = c(NA, -10L)))
Time Price Volume Flag Ticker 1: 2016-01-04 09:05:06 105.0 9500 1 0001 2: 2016-01-04 09:20:00 104.1 23500 0 0001 3: 2016-01-04 09:30:00 104.1 18500 1 0001 4: 2016-01-04 09:30:01 103.9 12500 0 0001 5: 2016-01-04 09:30:02 104.1 16118 1 0001 6: 2016-01-04 09:30:05 104.0 13000 0 0001 7: 2016-01-04 09:30:06 104.1 2500 1 0001 8: 2016-01-04 09:30:07 104.1 1500 1 0001 9: 2016-01-04 09:30:08 104.3 500 1 0001 10: 2016-01-04 09:30:10 104.0 1000 0 0001 11: 2016-01-04 09:30:11 103.9 1000 0 0001 12: 2016-01-04 09:30:15 104.0 3500 1 0001 13: 2016-01-04 09:30:17 104.3 2000 1 0001 14: 2016-01-04 09:30:19 104.3 1500 1 0001 15: 2016-01-04 09:30:20 104.4 500 1 0001 16: 2016-01-04 09:30:21 104.4 1500 1 0001 17: 2016-01-04 09:30:22 104.4 1000 1 0001 18: 2016-01-04 09:30:24 104.4 1500 1 0001 19: 2016-01-04 09:30:25 104.0 2000 0 0001 20: 2016-01-04 09:30:27 104.1 3500 1 0001 21: 2016-01-04 09:30:35 104.0 500 0 0001 22: 2016-01-04 09:31:14 104.1 5000 1 0001 23: 2016-01-04 09:31:15 104.1 500 1 0001 24: 2016-01-04 09:31:18 104.1 2500 1 0001 25: 2016-01-04 09:31:25 104.1 3000 1 0001 26: 2016-01-04 09:31:29 104.0 2000 0 0001 27: 2016-01-04 09:31:30 104.1 500 1 0001 28: 2016-01-04 09:31:35 104.1 300 1 0001 29: 2016-01-04 09:31:38 104.1 500 1 0001 30: 2016-01-04 09:32:33 104.0 500 0 0001
# First get the distribution of the Volume
distribution <- dt[Flag == 1, sum(Volume), by = cut(Price, 5)][, percentage := list(V1/sum(V1))]
# Get the max range bin
Max_range <- distribution[which.max(percentage), cut]
# Get the Closing price
Closing_price <- dt[.N, Price]
# Check whether the closing price is in the Max_range
Code comes here[?????]
所以问题就出现了:对于具体的Ticker
,如何检查收盘价是否在特定范围内?只需要True
或False
。如果closing_price
位于Max_range
范围内,则相应的Signal
将为True
,否则为False
。
编辑2
添加了所需的输出
所需的输出
Ticker Signal 1: 0001 False
所以我想创建一个函数来检查Signal是True
还是False
,然后在data.table中更新。
非常感谢!
答案 0 :(得分:1)
如果有一个超出给定范围的值,我是否理解你想要找到每个股票代码(001,002等)?
如果出现这个问题,您可以使用dplyr中的group_by函数和逻辑表达式:
group_by(dt,Ticker) %>%
summarise(Signal=any(with(.,Price>max_price & Price<min_price)))
答案 1 :(得分:0)
range.test
对象是levels(range.test)
的因子变量:
levels(range.test)
[1] "(0.999,0.9995]" "(0.9995,1]" "(1,1.0005]" "(1.0005,1.001]"
当你把它传递给findInterval作为第二个参数时,它被强制转换为数值2,所以这就是结果:
> findInterval(1,2)
[1] 0
应该发生的是因为1小于2.如果你真的想要一个数值序列,范围从0.999到1.001,有5个值,你可以使用seq:
> seq( 0.999, 1.001, length=5)
[1] 0.9990 0.9995 1.0000 1.0005 1.0010
然后,您可以测试该矢量的数量为1.000的区间:
> findInterval( 1, seq( 0.999, 1.001, length=5) )
[1] 3