R: Continuous futures working backward - r

I want to create a continuous futures series, that is to eliminate a gap between two series.
First thing I want is to download all individual contracts from the beginning to the now, the syntax is always the same:
Quandl("CME/INSTRUMENT_MONTHCODE_YEAR")
1.INSTRUMENT is GC (gold) in this case
2.MONTHCODE is G J M Q V Z
3.YEAR is from 1975 to 2017 (the actual contract)
With the data, I start working from the last contract, in this case "CME/GCG1975" and with the next contract "CME/GCJ1975". Then I see the last 6 values (are the more recent because date is descending) of the first contract GCG1975
require(Quandl)
GCG1975 = Quandl("CME/GCG1975",order="asc", type="raw")
tail(GCG1975,6)
order can be asc desc (ascending or descending), type can be : raw (data frame) ts xts zoo
And it outputs:
Image: quandl-1.png = Last values of GCG1975
Then I just want the 6th row starting from the final, and I want to eliminate the columns "Last" "Change" (this could be before starting processing each individual contract):
Image: quandl-2.png = Last 6th value GCG1975
Then I want to find the row with date 1975-02-18 (last 6th value GCG1975) in the next contract (GCJ1975):
Image: quandl-3.png = 1975-02-18 on GCJ1975
Then I compute the difference between the "Settle" of the G contract and the "Settle" of the J contract.
Difference_contract = 183.6 - 185.4
Difference_contract = -1.8
So that means that the next or J contract is 1.5 points up respect the before contract so we have to sum -1.8 to all the following numbers of the J contract (Open, High, Low, Settle), including the row 1975-02-18. This:
Image: quandl-4.png = Differences between contracts
And then we have a continuous series like this:
Image: quandl-5.png = Continuous series
All this differences and sums to make a continuous series is done since the last contract until the actual contract.
I think I can't post this because I don't have 10 points of reputation and I can just post 2 image-links.
Any guidance would help me, any question you have ask me.
Thanks and hope everything is well.
RTA
Edit: I have uploaded the photos and its links on post to my dropbox so you must look into it because Stackoverflow don't allow to post more than 2 links without 10 points of reputation.
Dropbox file

Related

Vectorizing R custom calculation with dynamic day range

I have a big dataset (around 100k rows) with 2 columns referencing a device_id and a date and the rest of the columns being attributes (e.g. device_repaired, device_replaced).
I'm building a ML algorithm to predict when a device will have to be maintained. To do so, I want to calculate certain features (e.g. device_reparations_on_last_3days, device_replacements_on_last_5days).
I have a function that subsets my dataset and returns a calculation:
For the specified device,
That happened before the day in question,
As long as there's enough data (e.g. if I want last 3 days, but only 2 records exist this returns NA).
Here's a sample of the data and the function outlined above:
data = data.frame(device_id=c(rep(1,5),rep(2,10))
,day=c(1:5,1:10)
,device_repaired=sample(0:1,15,replace=TRUE)
,device_replaced=sample(0:1,15,replace=TRUE))
# Exaxmple: How many times the device 1 was repaired over the last 2 days before day 3
# => getCalculation(3,1,data,"device_repaired",2)
getCalculation <- function(fday,fdeviceid,fdata,fattribute,fpreviousdays){
# Subset dataset
df = subset(fdata,day<fday & day>(fday-fpreviousdays-1) & device_id==fdeviceid)
# Make sure there's enough data; if so, make calculation
if(nrow(df)<fpreviousdays){
calculation = NA
} else {
calculation = sum(df[,fattribute])
}
return(calculation)
}
My problem is that the amount of attributes available (e.g. device_repaired) and the features to calculate (e.g. device_reparations_on_last_3days) has grown exponentially and my script takes around 4 hours to execute, since I need to loop over each row and calculate all these features.
I'd like to vectorize this logic using some apply approach which would also allow me to parallelize its execution, but I don't know if/how it's possible to add these arguments to a lapply function.

How to manage factors with mixed data types

I'm afraid this question has two sub parts. My project is to determine which insurance carrier has the lowest cost based on CPT Codes. Since there are so many CPT Codes I wanted to group them using cut like this:
uCPTCode<- unique(data$CPTCode)
uCPTCode <- cut(uCPTCode,
breaks = c(-Inf, "01999", "69979", "79999", "89398", "99091", "99499", Inf),
labels = c("NA","Anesthesia", "Surgery", "Radiology", "Pathology&Laboratory", "Medicine","Evaluation&Management", "Temp"),
right = FALSE)
Not sure unique is required or wise, but seemed to make sense to me. The issue is that some codes have leading zeros and terminating letters like this
2608 Levels: 0014F 0159T 0164T 0191T 0195T 0232T 0319T 0326T 0513F 0517F 0518F
So question 1 is what is the process to convert these ranges into integers corresponding to the labels I have in the cut function so I can graph the grouped results the x axis?
Question 2 is that I expected the ranges to be continuous, but they are not. How to I manage what happens around code 99000 through 99216 where previous groups (Medicine, Anesthesiology and Evaluation and Management) get combined? Here is a link to the CPT grouper file https://www.dropbox.com/s/wm55n17pufoacww/CPTGrouper.xlsx?dl=0
Here is a smattering of results to see where I am going with it
https://www.dropbox.com/s/h6sdnvm9yew6jdg/SampleStudyResults.xlsx?dl=0
Thanks very much for your time and attention

Summarized huge data, How to handle it with R?

I am working on EBS, Forex market Limit Order Book(LOB): here is an example of LOB in a 100 millisecond time slice:
datetime|side(0=Bid,1=Ask)| distance(1:best price, 2: 2nd best, etc.)| price
2008/01/28,09:11:28.000,0,1,1.6066
2008/01/28,09:11:28.000,0,2,1.6065
2008/01/28,09:11:28.000,0,3,1.6064
2008/01/28,09:11:28.000,0,4,1.6063
2008/01/28,09:11:28.000,0,5,1.6062
2008/01/28,09:11:28.000,1,1,1.6067
2008/01/28,09:11:28.000,1,2,1.6068
2008/01/28,09:11:28.000,1,3,1.6069
2008/01/28,09:11:28.000,1,4,1.6070
2008/01/28,09:11:28.000,1,5,1.6071
2008/01/28,09:11:28.500,0,1,1.6065 (I skip the rest)
To summarize the data, They have two rules(I have changed it a bit for simplicity):
If there is no change in LOB in Bid or Ask side, they will not record that side. Look at the last line of the data, millisecond was 000 and now is 500 which means there was no change at LOB in either side for 100, 200, 300 and 400 milliseconds(but those information are important for any calculation).
The last price (only the last) is removed from a given side of the order book. In this case, a single record with nothing in the price field. Again there will be no record for whole LOB at that time.
Example:2008/01/28,09:11:28.800,0,1,
I want to calculate minAsk-maxBid(1.6067-1.6066) or weighted average price (using sizes of all distances as weights, there is size column in my real data). I want to do for my whole data. But as you see the data has been summarized and this is not routine. I have written a code to produce the whole data (not just summary). This is fine for small data set but for a large one I am creating a huge file. I was wondering if you have any tips how to handle the data? How to fill the gaps while it is efficient.
You did not give a great reproducible example so this will be pseudo/untested code. Read the docs carefully and make adjustments as needed.
I'd suggest you first filter and split your data into two data.frames:
best.bid <- subset(data, side == 0 & distance == 1)
best.ask <- subset(data, side == 1 & distance == 1)
Then, for each of these two data.frames, use findInterval to compute the corresponding best ask or best bid:
best.bid$ask <- best.ask$price[findInterval(best.bid$time, best.ask$time)]
best.ask$bid <- best.bid$price[findInterval(best.ask$time, best.bid$time)]
(for this to work you might have to transform date/time into a linear measure, e.g. time in seconds since market opening.)
Then it should be easy:
min.spread <- min(c(best.bid$ask - best.bid$price,
best.ask$bid - best.ask$price))
I'm not sure I understand the end of day particularity but I bet you could just compute the spread at market close and add it to the final min call.
For the weighted average prices, use the same idea but instead of the two best.bid and best.ask data.frames, you should start with two weighted.avg.bid and weighted.avg.ask data.frames.

Calculate percentage over time on very large data frames

I'm new to R and my problem is I know what I need to do, just not how to do it in R. I have an very large data frame from a web services load test, ~20M observations. I has the following variables:
epochtime, uri, cache (hit or miss)
I'm thinking I need to do a coule of things. I need to subset my data frame for the top 50 distinct URIs then for each observation in each subset calculate the % cache hit at that point in time. The end goal is a plot of cache hit/miss % over time by URI
I have read, and am still reading various posts here on this topic but R is pretty new and I have a deadline. I'd appreciate any help I can get
EDIT:
I can't provide exact data but it looks like this, its at least 20M observations I'm retrieving from a Mongo database. Time is epoch and we're recording many thousands per second so time has a lot of dupes, thats expected. There could be more than 50 uri, I only care about the top 50. The end result would be a line plot over time of % TCP_HIT to the total occurrences by URI. Hope thats clearer
time uri action
1355683900 /some/uri TCP_HIT
1355683900 /some/other/uri TCP_HIT
1355683905 /some/other/uri TCP_MISS
1355683906 /some/uri TCP_MISS
You are looking for the aggregate function.
Call your data frame u:
> u
time uri action
1 1355683900 /some/uri TCP_HIT
2 1355683900 /some/other/uri TCP_HIT
3 1355683905 /some/other/uri TCP_MISS
4 1355683906 /some/uri TCP_MISS
Here is the ratio of hits for a subset (using the order of factor levels, TCP_HIT=1, TCP_MISS=2 as alphabetical order is used by default), with ten-second intervals:
ratio <- function(u) aggregate(u$action ~ u$time %/% 10,
FUN=function(x) sum((2-as.numeric(x))/length(x)))
Now use lapply to get the final result:
lapply(seq_along(levels(u$uri)),
function(l) list(uri=levels(u$uri)[l],
hits=ratio(u[as.numeric(u$uri) == l,])))
[[1]]
[[1]]$uri
[1] "/some/other/uri"
[[1]]$hits
u$time%/%10 u$action
1 135568390 0.5
[[2]]
[[2]]$uri
[1] "/some/uri"
[[2]]$hits
u$time%/%10 u$action
1 135568390 0.5
Or otherwise filter the data frame by URI before computing the ratio.
#MatthewLundberg's code is the right idea. Specifically, you want something that utilizes the split-apply-combine strategy.
Given the size of your data, though, I'd take a look at the data.table package.
You can see why visually here--data.table is just faster.
Thought it would be useful to share my solution to the plotting part of them problem.
My R "noobness" my shine here but this is what I came up with. It makes a basic line plot. Its plotting the actual value, I haven't done any conversions.
for ( i in 1:length(h)) {
name <- unlist(h[[i]][1])
dftemp <- as.data.frame(do.call(rbind,h[[i]][2]))
names(dftemp) <- c("time", "cache")
plot(dftemp$time,dftemp$cache, type="o")
title(main=name)
}

R Accumulate equity data - add time and price

I have some data formatted as below. I have done some analysis on this and would like to be able to plot the price development in the same graph as the analyzed data.
This requires me to have the same x-axes for the data.
So I would like to aggregate the "shares" column in say 150 increments, and add the "finalprice" and "time" to this.
The aggregation should include the latest time and price, so if the aggregation needs to occur over two or more rows of data then the last row should provide the price and time data.
My question is how to create a new vector with 150 shares per row.
The length of the vector will equal sum(shares)/150.
Is there an easy way to do this? Thanks in advance.
Edit:
I thought about expanding the observations using rep(finalprice, shares) and then getting each 150th value of the expanded vector.
Data sample:
"date","ord","shares","finalprice","time","stock"
20120702,E,2000,99.35,540.84753333,500
20120702,E,28000,99.35,540.84753333,500
20120702,E,50,99.5,542.03073333,500
20120702,E,13874,99.5,542.29411667,500
20120702,E,292,99.5,542.30191667,500
20120702,E,784,99.5,542.30193333,500
20120702,E,13300,99.35,543.04805,500
20120702,E,16658,99.35,543.04805,500
20120702,E,42,99.5,543.04805,500
20120702,E,400,99.4,546.17173333,500
20120702,E,100,99.4,547.07,500
20120702,E,2219,99.3,549.47988333,500
20120702,E,781,99.3,549.5238,500
20120702,E,50,99.3,553.4052,500
20120702,E,1500,99.35,559.86275,500
20120702,E,103,99.5,567.56726667,500
20120702,E,1105,99.7,573.93326667,500
20120702,E,4100,99.5,582.2657,500
20120702,E,900,99.5,582.2657,500
20120702,E,1024,99.45,582.43891667,500
20120702,E,8214,99.45,582.43891667,500
20120702,E,10762,99.45,582.43895,500
20120702,E,1250,99.6,586.86446667,500
20120702,E,5000,99.45,594.39061667,500
20120702,E,20000,99.45,594.39061667,500
20120702,E,15000,99.45,594.39061667,500
20120702,E,4000,99.45,601.34491667,500
20120702,E,8700,99.45,603.53608333,500
20120702,E,3290,99.6,609.23213333,500
I think I got it solved.
expand <- rep(finalprice, shares)
Increment <- expand[seq(from = 1, to = length(expand), by = 150)]

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