I have 49 datasets which include different values like this.
inPUT=read.table(file="TEST.csv", sep=",", header=T, row.names=1)
class(inPUT)
[1] "data.frame"
length(inPUT)
[1] 49
head(inPUT)
GO.1 GO.2 ... GO.49
1 811 811 ... 811
2 813 814 ... 814
3 814 819 ... 817
length(inPUT$GO.1)
[1]191
length(inPUT$GO.49)
[1]170
I'd like to calculate overlap between two different datasets among total 49 datasets (all possible pairwise calculation). Is there any R package to calculate how two sets overlap (I'm still new but..). Any ideas?
Is it so that every dataset only represents one column, as in your example?
One possible option is to use the '%in%' operator function, e.g.
mean(GO.1 %in% GO.2)
will tell you the percentage of observation in GO.1 are also present in GO.2. If you want to calculate the total overlap, you could use it in a function.
You can use outer to compute pairwise combinations, and intersect to find the members that are in both sets. So something like:
outer(inPUT, inPUT, intersect)
may help.
Related
I have a database where some people have multiple diagnoses. I posted a similar question in the past, but now have some more nuances I need to work through:
R- How to test multiple 100s of similar variables against a condition
I have this dataset (which was an import of a SAS file)
ID dx1 dx2 dx3 dx4 dx5 dx6 .... dx200
1 343 432 873 129 12 123 3445
2 34 12 44
3 12
4 34 56
Initially, I wanted to be able to create a new variable if any of the "dxs" equals a certain number without using hundreds of if statements? All the different variables have the same format (dx#). So I used the following code:
Ex:
dataset$highbloodpressure <- rowSums(screen[0:832] == "410") > 0
This worked great. However, there are many different codes for the same diagnosis. For example, a heart attack can be defined as:
410.1,
410.71,
410.62,
410.42,
...this goes on for 20 additional codes. BUT! They all start with 410.
I thought about using stringr (the variable is a string), to identify the common code components (410, for the example above), but am not sure how to use it in the context of rowsums.
If anyone has any suggestions for this, please let me know!
Thanks for all the help!
You can use the grepl() function that returns TRUE if a value is present. In order to check all columns simultaneously, just collapse all of them to one character per row:
df$dx.410 = NA
for(i in 1:dim(df)[1]){
if(grepl('410',paste(df[i,2:200],collapse=' '))){
df$dx.410[i]="Present"
}
}
This will loop through all lines, create one large character containing all diagnoses for this case and write "Present" in column dx.410 if any column contains a 410-diagnosis.
(The solution expects the data structure you have here with the dx-variables in columns 2 to 200. If there are some other columns, just adjust these numbers)
Big picture explanation is I am trying to do a sliding window analysis on environmental data in R. I have PAR (photosynthetically active radiation) data for a select number of sequential dates (pre-determined based off other biological factors) for two years (2014 and 2015) with one value of PAR per day. See below the few first lines of the data frame (data frame name is "rollingpar").
par14 par15
1356.3242 1306.7725
NaN 1232.5637
1349.3519 505.4832
NaN 1350.4282
1344.9306 1344.6508
NaN 1277.9051
989.5620 NaN
I would like to create a loop (or any other way possible) to subset the data frame (both columns!) into two week windows (14 rows) from start to finish sliding from one window to the next by a week (7 rows). So the first window would include rows 1 to 14 and the second window would include rows 8 to 21 and so forth. After subsetting, the data needs to be flipped in structure (currently using the melt function in the reshape2 package) so that the values of the PAR data are in one column and the variable of par14 or par15 is in the other column. Then I need to get rid of the NaN data and finally perform a wilcox rank sum test on each window comparing PAR by the variable year (par14 or par15). Below is the code I wrote to prove the concept of what I wanted and for the first subsetted window it gives me exactly what I want.
library(reshape2)
par.sub=rollingpar[1:14, ]
par.sub=melt(par.sub)
par.sub=na.omit(par.sub)
par.sub$variable=as.factor(par.sub$variable)
wilcox.test(value~variable, par.sub)
#when melt flips a data frame the columns become value and variable...
#for this case value holds the PAR data and variable holds the year
#information
When I tried to write a for loop to iterate the process through the whole data frame (total rows = 139) I got errors every which way I ran it. Additionally, this loop doesn't even take into account the sliding by one week aspect. I figured if I could just figure out how to get windows and run analysis via a loop first then I could try to parse through the sliding part. Basically I realize that what I explained I wanted and what I wrote this for loop to do are slightly different. The code below is sliding row by row or on a one day basis. I would greatly appreciate if the solution encompassed the sliding by a week aspect. I am fairly new to R and do not have extensive experience with for loops so I feel like there is probably an easy fix to make this work.
wilcoxvalues=data.frame(p.values=numeric(0))
Upar=rollingpar$par14
for (i in 1:length(Upar)){
par.sub=rollingpar[[i]:[i]+13, ]
par.sub=melt(par.sub)
par.sub=na.omit(par.sub)
par.sub$variable=as.factor(par.sub$variable)
save.sub=wilcox.test(value~variable, par.sub)
for (j in 1:length(save.sub)){
wilcoxvalues$p.value[j]=save.sub$p.value
}
}
If anyone has a much better way to do this through a different package or function that I am unaware of I would love to be enlightened. I did try roll apply but ran into problems with finding a way to apply it to an entire data frame and not just one column. I have searched for assistance from the many other questions regarding subsetting, for loops, and rolling analysis, but can't quite seem to find exactly what I need. Any help would be appreciated to a frustrated grad student :) and if I did not provide enough information please let me know.
Consider an lapply using a sequence of every 7 values through 365 days of year (last day not included to avoid single day in last grouping), all to return a dataframe list of Wilcox test p-values with Week indicator. Then later row bind each list item into final, single dataframe:
library(reshape2)
slidingWindow <- seq(1,364,by=7)
slidingWindow
# [1] 1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106 113 120 127
# [20] 134 141 148 155 162 169 176 183 190 197 204 211 218 225 232 239 246 253 260
# [39] 267 274 281 288 295 302 309 316 323 330 337 344 351 358
# LIST OF WILCOX P VALUES DFs FOR EACH SLIDING WINDOW (TWO-WEEK PERIODS)
wilcoxvalues <- lapply(slidingWindow, function(i) {
par.sub=rollingpar[i:(i+13), ]
par.sub=melt(par.sub)
par.sub=na.omit(par.sub)
par.sub$variable=as.factor(par.sub$variable)
data.frame(week=paste0("Week: ", i%/%7+1, "-", i%/%7+2),
p.values=wilcox.test(value~variable, par.sub)$p.value)
})
# SINGLE DF OF ALL P-VALUES
wilcoxdf <- do.call(rbind, wilcoxvalues)
I am trying to draw a boxplot in R:
I have a dataset with 70 attributes:
The format is
patient number medical_speciality number_of_procedures
111 Ortho 21
232 Emergency 16
878 Pediatrics 20
981 OBGYN 31
232 Care of Elderly 15
211 Ortho 32
238 Care of Elderly 11
219 Care of Elderly 6
189 Emergency 67
323 Emergency 23
189 Pediatrics 1
289 Ortho 34
I have been trying to get a subset to only include emergency, pediatrics in a boxplot (there are 10000+ datapoints in reality)
I thought that I could just do this:
newdata<-subset(olddata[ms$medical_specialty=='emergency'|olddata$medical_specialty=='pediatrics',])
plot(newdata)
Since if I do a summary of newdata, all it has is the pediatrics and emergency results. But when it comes to plotting it still includes the ortho, OBGYN, care of elderly in the x axis with no boxplot.
I presume that there is a way to do this in ggplot by doing
ggplot(newdata, aes(x=medical_speciality, y=num_of_procedures, fill=cond)) + geom_boxplot()
but this gives me the error:
Don't know how to automatically pick scale for object of type data.frame.
Defaulting to continuous
Error: Aesthetics must either be length one, or the same length as the dataProblems:cond
Can someone help me out?
I believe your problem comes from the fact that the column medical_speciality is a factor.
So, even though you subset your data the right way, you still get all the levels (including "Ortho", "OBGYN", etc...).
You can get rid of them by using the function droplevels:
newdata<-subset(olddata[ms$medical_specialty=='emergency'|olddata$medical_specialty=='pediatrics',])
newdata <- droplevels(newdata) ## THIS IS THE NEW ADDITION
plot(newdata)
Does this help?
I have a data frame (451 obs of 8 variables) that has two columns (6&7) that look like this:
Major Minor
C:726 T:2
A:687 G:41
T:3 C:725
I want to create one column that summarises this. To do this, I don't care about the letters in each cell, but I want the larger number to remain, whatever row it's in. i.e. I want it to look like this:
Summary_column
726
687
725
Not necessary, but for those that wonder what Im doing, this is the output from a programme called VCFtools; it has a count function that counts alleles in a VCF, but sometimes it names the allele as "Minor" when it is clearly more common.
Thanks for your help!
I would do something like this :
extract <- function(v) {
gsub("^.*:", "", v)
}
within(d, Summary_column <- pmax(extract(Major), extract(Minor)))
Which gives :
Major Minor Summary_column
1 C:726 T:2 726
2 A:687 G:41 687
3 T:3 C:725 725
I am thinking of writing a data dictionary function in R which, taking a data frame as an argument, will do the following:
1) Create a text file which:
a. Summarises the data frame by listing the number of variables by class, number of observations, number of complete observations … etc
b. For each variable, summarise the key facts about that variable: mean, min, max, mode, number of missing observations … etc
2) Creates a pdf containing a histogram for each numeric or integer variable and a bar chart for each attribute variable.
The basic idea is to create a data dictionary of a data frame with one function.
My question is: is there a package which already does this? And if not, do people think this would be a useful function?
Thanks
There are a variety of describe functions in various packages. The one I am most familiar with is Hmisc::describe. Here's its description from its help page:
" This function determines whether the variable is character, factor, category, binary, discrete numeric, and continuous numeric, and prints a concise statistical summary according to each. A numeric variable is deemed discrete if it has <= 10 unique values. In this case, quantiles are not printed. A frequency table is printed for any non-binary variable if it has no more than 20 unique values. For any variable with at least 20 unique values, the 5 lowest and highest values are printed."
And an example of the output:
Hmisc::describe(work2[, c("CHOLEST","HDL")])
work2[, c("CHOLEST", "HDL")]
2 Variables 5325006 Observations
----------------------------------------------------------------------------------
CHOLEST
n missing unique Mean .05 .10 .25 .50 .75 .90
4410307 914699 689 199.4 141 152 172 196 223 250
.95
268
lowest : 0 10 19 20 31, highest: 1102 1204 1213 1219 1234
----------------------------------------------------------------------------------
HDL
n missing unique Mean .05 .10 .25 .50 .75 .90
4410298 914708 258 54.2 32 36 43 52 63 75
.95
83
lowest : -11.0 0.0 0.2 1.0 2.0, highest: 241.0 243.0 248.0 272.0 275.0
----------------------------------------------------------------------------------
Furthermore, on your point about getting histograms, the Hmisc::latex method for a describe-object will produce histograms interleaved in the output illustrated above. (You do need to have a function LaTeX installation to take advantage of this.) I'm pretty sure you can find an illustration of the output in either Harrell's website or with the Amazon "Look Inside" presentation of his book "Regression Modeling Strategies". The book has a ton of useful material regarding data analysis.