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
Related
this sounds pretty basic but every time I try to make a histogram, my code is saying x needs to be numeric. I've been looking everywhere but can't find one relating to my problem. I have data with 240 obs with 5 variables.
Nipper length
Number of Whiskers
Crab Carapace
Sex
Estuary location
There is 3 locations and i'm trying to make a histogram with nipper length
I've tried making new factors and levels, with the 80 obs in each location but its not working
Crabs.data <-read.table(pipe("pbpaste"),header = FALSE)##Mac
names(Crabs.data)<-c("Crab Identification","Estuary Location","Sex","Crab Carapace","Length of Nipper","Number of Whiskers")
Crabs.data<-Crabs.data[,-1]
attach(Crabs.data)
hist(`Length of Nipper`~`Estuary Location`)
Error in hist.default(Length of Nipper ~ Estuary Location) :
'x' must be numeric
Instead of correct result
hist() doesn't seem to like taking more than one variable.
I think you'd have the best luck subsetting the data, that is, making a vector of nipper lengths for all crabs in a given estuary.
crabs.data<-read.table("whatever you're calling it")
names<-(as you have it)
Estuary1<-as.vector(unlist(subset(crabs.data, `Estuary Loc`=="Location", select = `Length of Nipper`)))
hist(Estuary1)
Repeat the last two lines for your other two estuaries. You may not need the unlist() command, depending on your table. I've tended to need it for Excel files, but I don't know what format your table is in (that would've been helpful).
Background:
On this combinatorics question, the issue is how to determine the sample space: the ways 8 different soccer teams can be paired up for the next round of competition. Two different answers have been advanced for that part of the problem: 28 (see comments OP) and 105 (see edit within OP and answer).
I'd like to do this manually to try to hone down on the mistake in whichever answer is incorrect.
What I have tried:
teams = 1:8
names(teams) = c("RM", "BCN", "SEV", "JUV", "ROM", "MC", "LIV", "BYN")
split(sample(teams), rep(1:(length(teams)/2), each=2))
Unfortunately, the output is a list, and I wanted a vector to be able to run something like:
unique(...,MARGIN=2)
Is there a way of doing this in an elegant manner?
After a now erased answer (thank you), I would go with
a <- replicate(1e5, unlist(split(sample(teams), rep(1:(length(teams)/2), each=2))))
to simulate 100,000 random samples, and later run
unique(a, MARGIN = 2).
But how can I account for the fact that the order of the 4 pairings of opponents doesn't matter, and that LIV-BYN and BYN-LIV, for example, is the same pairing (field advantage notwithstanding)?
> u = ncol(unique(replicate(1e6, unlist(split(sample(teams), rep(1:(length(teams)/2), each=2)))), MARGIN = 2))
> u / (factorial(4) * 2^4)
[1] 105
The idea of unlist is from #Song Zhengyi, and if his answer is un-deleted, I'll accept it. The complete answer is in the lines above.
u needs to be divided by 4! because
BCN-RM, BYN-SEV, JUV-ROM, LIV-MC
is exactly the same as
LIV-MC, BCN-RM, BYN-SEV, JUV-ROM
or
BCN-RM, LIV-MC, BYN-SEV, JUV-ROM
etc.
The term 2^4 is to avoid over-counting since for every possible unique draw, each one of the pairings can be flipped without loss (discarding field advantage): BCN-RM is the same as RM-BCN, and there are 4 pairs in each draw.
If field advantage is a consideration (real life)...
> u/factorial(4)
[1] 1680
we end up with 1,680 possible draws.
I'm not sure if my title is properly expressing what I'm asking. Once I'm done writing, it'll make sense. Firstly, I just started learning R, so I am a newbie. I've been reading through tutorial series and PDF's I've found online.
I'm working on a data set and I created a data frame of just the year 2001 and the DAM value Bon. Here's a picture.
What I want to do now is create a matrix with 3 columns: Coho Adults, Coho Jacks and the third column the ratio of Coho Jacks to Adults. This is what I'm having trouble with. The ratio between Coho Jacks to Adults.
If I do a line of code like this I get a normal output.
(cohoPassage <- matrix(fishPassage1995BON[c(5,6, 7)], ncol = 3))
The values are 259756, 6780 114934.
I'm figuring in order to get the ratio, I should divide column 5 and column 6's values. So basically 259756/6780 = 38.31
I've tried many things like:
(cohoPassage <- matrix(fishPassage1995BON[c(5,6, 5/6)], ncol = 3))
This just outputs the value of the fifth column instead of dividing for some reason
I've tried this:
matrix(fishPassage1995BON[c(5,6)],fishPassage1995BON[,5]/fishPassage1995BON[,6], ncol = 3)
Which gives me an incorrect output
I decided to break down the problem and divide the fifth and sixth columns separately and it gave the correct ratio.
If I create a matrix like this
matrix(fishPassage1995BON[,5]/fishPassage1995BON[,6])
It outputs the correct ratio of 38.31209. But when I try to combine everything, I just keep getting errors.
What can I do? Any help would be appreciated. Thank you.
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)]
Based off of a previous question I asked, which #Andrie answered, I have a question about the usage of the cut function and labels.
I'd like get summary statistics based on the range of number of times a user logs in.
Here is my data:
# Get random numbers
NumLogin <- round(runif(100,1,50))
# Set the login range
LoginRange <- cut(NumLogin,
c(0,1,3,5,10,15,20,Inf),
labels=c('1','2','3-5','6-10','11-15','16-20','20+')
)
Now I have my LoginRange, but I'm unsure how the cut function actually works. I want to find users who have logged in 1 time, 2 times, 3-5 times, etc, while only including the user if they are in that range. Is the cut function including 3 twice (In the 2 bucket and the 3-5 bucket)? If I look in my example, I can see a user who logged in 3 times, but they are cut as '2'. I've looked at the documentation and every R book I own, but no luck. What am I doing wrong?
Also - As a usage question - should I attach the LoginRange to my data frame? If so, what's the best way to do so?
DF <- data.frame(NumLogin, LoginRange)
?
Thanks
The intervals defined by the cut() function are (by default) closed on the right. To see what that means, try this:
cut(1:2, breaks=c(0,1,2))
# [1] (0,1] (1,2]
As you can see, the integer 1 gets included in the range (0,1], not in the range (1,2]. It doesn't get double-counted, and for any input value falling outside of the bins you define, cut() will return a value of NA.
When dealing with integer-valued data, I tend to set break points between the integers, just to avoid tripping myself up. In fact, doing this with your data (as shown below), reveals that the 2nd and 3rd bins were actually incorrectly named, which illustrates the point quite nicely!
LoginRange <- cut(NumLogin,
c(0.5, 1.5, 3.5, 5.5, 10.5, 15.5, 20.5, Inf),
# c(0,1,3,5,10,15,20,Inf) + 0.5,
labels=c('1','2-3','4-5','6-10','11-15','16-20','20+')
)