i'm trying to do some things in pineScript.
using this condition: if close > ema 34 -> plot "Long".
now, the "Long" is displayed each time, the condition is true.
but i only want to see the plot the first time, when it's happening, and the other results should be ignored.
another condition is probably missing here, but what does it have to look like?
can someone help me please?your text
//#version=5
indicator("EMA_Long", overlay = true)
ema = ta.ema(close,34)
longCondition = close \> ema
label = "Long Triangle"
plotshape(longCondition, label, location=location.belowbar, color=color.yellow, style=shape.triangleup, text = "Long", textcolor = color.yellow)
plot(ema)`
You can use :
longCondition = close > ema and not(close[1] > ema[1])
This way, your long Condition will be true only when it is happening
Related
let's say i have the following list :
pop = [5,4,3,4,2,5,1,6,7,6,10,13,5,8,8]
I then put them into a DataFrame, and for each row, append the value of the first column to a list. I then plot and frame the list to produce the animation :
df = DataFrame(pop, :auto)
a = Animation()
li = []
for i in df.x1
append!(li,i)
plt = plot(li, ylim=(0,20), xlim=(0,length(df.x1)))
frame(a, plt)
end
gif(a)
The frame rate for Animation object is too fast, i would like to slow it down, how do i do that?
the easy answer is gif(a, fps = whatever i want the frame rate to be)
Here is my code right now:
f=function(Symbol, start, end, interval){
getSymbols(Symbols=Symbol, from=start, to= end)
Symbol=data.frame(Symbol)
a=length(Symbol$Symbol.Adjusted)
b=a/interval
c=ceiling(b)
origData=as.data.frame(matrix(`length<-`(Symbol$Symbol.Adjusted, c * interval), ncol = interval, byrow = TRUE))
return(origData)
}
f("SPY", "2012-01-01", "2013-12-31", 10)
Next I need to Get the adjusted close price and consider this price data only for following tasks. Split daily stock adjusted close price into N blocks as rows in a data frame. So that each block containing M days (columns) data, where M equals to the time interval value. It’s referred as origData in my code.
The function is supposed to return the data frame origData, but whenever I try running this it tells me that the Symbol data frame is empty. How do I need to change my function to get the data frame output?
#IRTFM's observations are correct. Incorporating those changes you can change your function to :
library(quantmod)
f = function(Symbol, start, end, interval){
getSymbols(Symbols=Symbol, from=start, to= end)
data= get(Symbol)
col = data[, paste0(Symbol, '.Adjusted')]
a=length(col)
b=a/interval
c=ceiling(b)
origData= as.data.frame(matrix(`length<-`(col, c * interval),
ncol = interval, byrow = TRUE))
return(origData)
}
f("SPY", "2012-01-01", "2013-12-31", 10)
I haven't figured out what the set of expressions inside the data.matrix call is supposed to do and you made no effort to explain your intent. However, your error occurs farther up the line. If you put in a debugging call to str(Symbol) you will see that Symbol will evaluate to "SPY" but that is just a character value and not an R object name. The object you wnat is named SPY and the way to retrieve an object's value when you can only have access to a character value is to use the R function get, So try adding this after the getSymbols call inside the function:
library(quantmod) # I'm assuming this was the package in use
...
Symbol=data.frame( get(Symbol) )
str(Symbol) # will print the result at your console
....
# then perhaps you can work on what you were trying inside the data.matrix call
You will also find that the name Symbol.Adjusted will not work (since R is not a macro language). You will need to do something like:
a=length( Symbol[[ paste0(Symbol, ".Adjusted")]] )
Oh wait. You overwrote the value for Symbol. That won't work. You need to use a different name for your dataframe. So why don't you edit your question to fix the errors I've identified so far and also describe what you are trying to do when you were using as.data.frame.
I have the following MATLAB code and I'm working to translating it to R:
nproc=40
T=3
lambda=4
tarr = zeros(1, nproc);
i = 1;
while (min(tarr(i,:))<= T)
tarr = [tarr; tarr(i, :)-log(rand(1, nproc))/lambda];
i = i+1;
end
tarr2=tarr';
X=min(tarr2);
stairs(X, 0:size(tarr, 1)-1);
It is the Poisson Process from the renewal processes perspective. I've done my best in R but something is wrong in my code:
nproc<-40
T<-3
lambda<-4
i<-1
tarr=array(0,nproc)
lst<-vector('list', 1)
while(min(tarr[i]<=T)){
tarr<-tarr[i]-log((runif(nproc))/lambda)
i=i+1
print(tarr)
}
tarr2=tarr^-1
X=min(tarr2)
plot(X, type="s")
The loop prints an aleatory number of arrays and only the last is saved by tarr after it.
The result has to look like...
Thank you in advance. All interesting and supportive comments will be rewarded.
Adding on to the previous comment, there are a few things which are happening in the matlab script that are not in the R:
[tarr; tarr(i, :)-log(rand(1, nproc))/lambda]; from my understanding, you are adding another row to your matrix and populating it with tarr(i, :)-log(rand(1, nproc))/lambda].
You will need to use a different method as Matlab and R handle this type of thing differently.
One glaring thing that stands out to me, is that you seem to be using R: tarr[i] and M: tarr(i, :) as equals where these are very different, as what I think you are trying to achieve is all the columns in a given row i so in R that would look like tarr[i, ]
Now the use of min is also different as R: min() will return the minimum of the matrix (just one number) and M: min() returns the minimum value of each column. So for this in R you can use the Rfast package Rfast::colMins.
The stairs part is something I am not familiar with much but something like ggplot2::qplot(..., geom = "step") may work.
Now I have tried to create something that works in R but am not sure really what the required output is. But nevertheless, hopefully some of the basics can help you get it done on your side. Below is a quick try to achieve something!
nproc <- 40
T0 <- 3
lambda <- 4
i <- 1
tarr <- matrix(rep(0, nproc), nrow = 1, ncol = nproc)
while(min(tarr[i, ]) <= T0){
# Major alteration, create a temporary row from previous row in tarr
temp <- matrix(tarr[i, ] - log((runif(nproc))/lambda), nrow = 1)
# Join temp row to tarr matrix
tarr <- rbind(tarr, temp)
i = i + 1
}
# I am not sure what was meant by tarr' in the matlab script I took it as inverse of tarr
# which in matlab is tarr.^(-1)??
tarr2 = tarr^(-1)
library(ggplot2)
library(Rfast)
min_for_each_col <- colMins(tarr2, value = TRUE)
qplot(seq_along(min_for_each_col), sort(min_for_each_col), geom="step")
As you can see I have sorted the min_for_each_col so that the plot is actually a stair plot and not some random stepwise plot. I think there is a problem since from the Matlab code 0:size(tarr2, 1)-1 gives the number of rows less 1 but I cant figure out why if grabbing colMins (and there are 40 columns) we would create around 20 steps. But I might be completely misunderstanding! Also I have change T to T0 since in R T exists as TRUE and is not good to overwrite!
Hope this helps!
I downloaded GNU Octave today to actually run the MatLab code. After looking at the code running, I made a few tweeks to the great answer by #Croote
nproc <- 40
T0 <- 3
lambda <- 4
i <- 1
tarr <- matrix(rep(0, nproc), nrow = 1, ncol = nproc)
while(min(tarr[i, ]) <= T0){
temp <- matrix(tarr[i, ] - log(runif(nproc))/lambda, nrow = 1) #fixed paren
tarr <- rbind(tarr, temp)
i = i + 1
}
tarr2 = t(tarr) #takes transpose
library(ggplot2)
library(Rfast)
min_for_each_col <- colMins(tarr2, value = TRUE)
qplot(seq_along(min_for_each_col), sort(min_for_each_col), geom="step")
Edit: Some extra plotting tweeks -- seems to be closer to the original
qplot(seq_along(min_for_each_col), c(1:length(min_for_each_col)), geom="step", ylab="", xlab="")
#or with ggplot2
df1 <- cbind(min_for_each_col, 1:length(min_for_each_col)) %>% as.data.frame
colnames(df1)[2] <- "index"
ggplot() +
geom_step(data = df1, mapping = aes(x = min_for_each_col, y = index), color = "blue") +
labs(x = "", y = "")
I'm not too familiar with renewal processes or matlab so bear with me if I misunderstood the intention of your code. That said, let's break down your R code step by step and see what is happening.
The first 4 lines assign numbers to variables.
The fifth line creates an array with 40 (nproc) zeros.
The sixth line (which doesnt seem to be used later) creates an empty vector with mode 'list'.
The seventh line starts a while loop. I suspect this line is supposed to say while the min value of tarr is less than or equal to T ...
or it's supposed to say while i is less than or equal to T ...
It actually takes the minimum of a single boolean value (tarr[i] <= T). Now this can work because TRUE and FALSE are treated like numbers. Namely:
TRUE == 1 # returns TRUE
FALSE == 0 # returns TRUE
TRUE == 0 # returns FALSE
FALSE == 1 # returns FALSE
However, since the value of tarr[i] depends on a random number (see line 8), this could lead to the same code running differently each time it is executed. This might explain why the code "prints an aleatory number of arrays ".
The eight line seems to overwrite the assignment of tarr with the computation on the right. Thus it takes the single value of tarr[i] and subtracts from it the natural log of runif(proc) divided by 4 (lambda) -- which gives 40 different values. These fourty different values from the last time through the loop are stored in tarr.
If you want to store all fourty values from each time through the loop, I'd suggest storing it in say a matrix or dataframe instead. If that's what you want to do, here's an example of storing it in a matrix:
for(i in 1:nrow(yourMatrix)){
//computations
yourMatrix[i,] <- rowCreatedByComputations
}
See this answer for more info about that. Also, since it's a set number of values per run, you could keep them in a vector and simply append to the vector each loop like this:
vector <- c(vector,newvector)
The ninth line increases i by one.
The tenth line prints tarr.
the eleveth line closes the loop statement.
Then after the loop tarr2 is assigned 1/tarr. Again this will be 40 values from the last time through the loop (line 8)
Then X is assigned the min value of tarr2.
This single value is plotted in the last line.
Also note that runif samples from the uniform distribution -- if you're looking for a Poisson distribution see: Poisson
Hope this helped! Let me know if there's more I can do to help.
I'm a bit of a r newbie, and have am a little stuck at the way forward to run a correlation on time-series data where the second vector is much longer and I want to run a rolling time window.
My data looks something like this :
set.seed(1)
# "Target sample" (this is always of known fixed length N, e.g. 20 )
target <- data.frame(Date=rep(seq(Sys.Date(),by="1 day",length=20)),Measurement=rnorm(2))
# "Potential Sample" (this is always much longer and of unknown length,e.g. 730 in this example)
potential <- data.frame(Date=rep(seq(Sys.Date()-1095,by="1 day",length=730)),Measurement=rnorm(2))
What I would like to do is take a rolling window of size N (i.e matching the size of target sample), incrementing the roll by one day at a time, and then print two columns for each window :
WindowStartDate and the result of cor(target,potentialWindow)
So in pseudo-code (using the generated example above) :
Start at Sys.Date()-1095, take window size N values
Print (or,probably better, put in to new data frame) Sys.Date()-1095 and result of cor(target,potentialWindow)
Roll forward +1 day to Sys.Date()-1094 , take window size N values
Print (or, probably better, put in to new data frame) Sys.Date()-1094 and result of cor(target,potentialWindow)
etc. etc.
N.B. The roll forward +1 day is obviously a variable that could be tweaked depending on desired overlap.
Here's a way we can do it. Note that in your original example you only specified rnorm(2), which worked because R can recycle arguments, but it's probably not what you wanted. We just need to initialize a few things, and then send it through a for loop.
It seems like we can just pull the date you want from the potential data set, but if you want to use the Sys.Date() - X formula, I've shown how to do that as well.
set.seed(1)
# "Target sample" (this is always of known fixed length N, e.g. 20 )
target <- data.frame(Date = rep(seq(Sys.Date(), by = "1 day", length = 20)),
Measurement = rnorm(20))
# "Potential Sample" (this is always much longer and of unknown length,e.g. 730 in this example)
potential <- data.frame(Date = rep(seq(Sys.Date() - 1095, by = "1 day", length = 730)),
Measurement = rnorm(730))
#initialize values
N <- 20
len_potential <- nrow(potential) - (N - 1)
time_start <- 1096
result.df <- data.frame(Day = potential[1,1],
Corr = numeric(len_potential),
Day2 = potential[1,1],
stringsAsFactors = FALSE
)
#use a for loop
for(i in 1:len_potential){
result.df[i,1] = as.Date(potential[i,1])
result.df[i,2] = cor(target[,2], potential[i:(i+N-1), 2])
result.df[i,3] = Sys.Date() - (time_start - i)
}
Also, as a note on posting questions to SO, sometimes it is helpful to provide desired output.
I have something like this within a function:
x <- as.POSIXct((substr((dataframe[z, ])$variable, 1, 8)), tz = "GMT",
format = "%H:%M:%S")
print(x)
if ( (x >= as.POSIXct("06:00:00", tz = "GMT", format = "%H:%M:%S")) &
(x < as.POSIXct("12:00:00", tz = "GMT", format = "%H:%M:%S")) ){
position <- "first"
}
but I get this output:
character(0)
Error in if ((as.numeric(departure) - as.numeric(arrival)) < 0) { : argument is of length zero
how can I fix this so my comparison works and it prints the correct thing?
some examples of the dataframe$variable column:
16:33:00
15:34:00
14:51:00
07:26:00
05:48:00
11:10:00
17:48:00
06:17:00
08:22:00
11:31:00
Welcome to Stack Overflow!
First, the reason you've gotten some down votes is most likely because you haven't given much in your question to go on. For one thing, you haven't shown us what
(dataframe[z, ])$variable
is, which makes it hard for us to formulate a complete answer. You seem to be trying to extract a single value from a dataframe, is that right? If so, I've never seen it done that way, try replacing the above with:
dataframe$variable[z]
My guess is what you're trying to achieve is a comparison of an entire column of the dataframe called "variable", since that's generally more useful...
Having said that, I often come up against issues with time data, and from what I've heard, my experiences are not uncommon. When I'm dealing with just times, as it appears you are here, I prefer the chron::times format over POSIXct (POSIX is a date-time format, so a date is always included, it also tries to correct for timezone changes, as well as daylight savings changes, which tends to get in my way more than help). If you've got your data in the format you've specified in your first as.POSIXct call, you won't even need to specify that in calling the times function instead.
x <- chron::times( dataframe$variable )
print(x)
position <- ifelse ( x >= chron::times( "06:00:00" ) &
x < chron::times( "12:00:00" ),
"first", "not first"
)
This will output a vector "position", with a result for all values taken from dataframe$variable. Does that achieve what you're hoping for?
From here, if you did want to extract the comparison result for the particular row "z" in dataframe, you can still do that with
position[z]
EDIT to add:
It might be worth checking for missing values in "variable". This should return TRUE:
sum( is.na( dataframe$variable ) ) == 0
Also check for any that aren't correctly formatted. Again, this should return TRUE:
sum( is.na( chron::times( dataframe$variable ) ) ) == 0
EDIT to add:
As per the comments, it looks like some values in your "variables" column aren't converting properly. You should be able to find them with
subset( dataframe, is.na( chron::times( variable ) ) )
That should let you see what's wrong. It may be a single cell, or it may be a number of them. You'll need to tidy up that data, which you can do in a few ways. You could go through and fix them manually, you could add a function in your script to repair them before the conversion (this might be a good idea if there is a common issue between all of those values, or if you expect the same issue to happen again as new data comes in, if indeed you need to allow for that).
The other option is simply to exclude those rows from your analysis. If you go this route, make sure it's appropriate to the analysis you're running. If it is appropriate in your case, you can add a step to clean up the dataframe before running the steps in your question:
dataframe <- subset( dataframe, !is.na( chron::times( variable ) ) )
NOTE: there's a good chance this will come up with a warning. If you run the same line twice, and the warning goes away the second time (after the offending rows have been removed), you may need to look further into it.
That should drop the offending values, leaving only values that are properly converting to the times format, which should help with the steps you're trying to run. Check how your dataframe dimensions change before and after that step; that'll tell you how many rows you're dropping.
You could do the same thing with POSIXct if that's what you're comfortable with, I'm just personally more comfortable with times for what you're doing.