Firstly I need to explain that I've had MINIMAL training on R and have 0 knowledge of coding languages or programmes like R so please excuse me if I ask silly questions or don't understand something basic.
Also, I have tried to look at past topics/answers on this but I'm having a hard time relating the answers to my data so I apologise if this question has already been answered.
Basically I have a data set and I'm trying to find the mean of two variables (Peak flow before a walk in the cold, and peak flow after a walk in the cold) in this set. This is the entire code I've used so far:
drugs <- read.table(file = "C:\\Users\\Becky\\My Documents\\Asthmadata.txt", header = TRUE)
drugs
str(drugs)
mean.Asthmadata <- tapply (Asthmadata$trial1, list(Asthmadata$PEFR1), mean)
mean.Asthmadata
It works fine until the mean.Asthmadata. The data comes up in R just fine with the other codes but when I get to the mean and do the mean.Asthmadata [...] code, I keep getting the same error: "object 'mean.Asthmadata' not found"
My friend used the same code I did and it worked for him so I'm confused. Am I doing something wrong?
Thanks
EDIT:
#BenBolker
This is my data set
trial1 PEFR1 trial2 PEFR2
Before 310 After 299
Before 242 After 201
Before 340 After 232
Before 388 After 312
Before 294 After 221
Before 251 After 256
Before 391 After 327
Before 401 After 331
Before 287 After 231
And here's all the code I've used:
drugs <- read.table(file = "C:\\Users\\Becky\\My Documents\\Asthmadata.txt", header = TRUE)
drugs
str(drugs)
mean.drugs <- tapply (drugs$trial1, list(drugs$PEFR1), mean)
mean.drugs
The R version I have has two versions: i386 3.1.3, and x64 3.1.3 – I've tried both but neither seem to do what I want. I'm also using Windows 7 Home Premium 64bit. Hope I've included everything you need and I apologise if my formatting is off – I can't quite figure out how to format properly on here yet.
And the error I'm getting NOW is: “Error in split.default(X, group) : first argument must be a vector” when running the code Roland kindly provided. So I'm getting a different error each time I try it – it must be something I'm doing wrong.
Hope I've formatted that all correctly and included everything you need. Thanks :)
drugs <- read.table(header=TRUE,text="
trial1 PEFR1 trial2 PEFR2
Before 310 After 299
Before 242 After 201
Before 340 After 232
Before 388 After 312
Before 294 After 221
Before 251 After 256
Before 391 After 327
Before 401 After 331
Before 287 After 231")
In the current format you can calculate the mean before and after just by doing
mean(drugs$PEFR1)
and
mean(drugs$PEFR2)
What you may have had in mind was this shape:
drugs2 <- with(drugs,
data.frame(trial=c(as.character(trial1),
as.character(trial2)),
PEFR=c(PEFR1,PEFR2)))
I used with() for convenience -- it's a way to temporarily attach a data frame so you can refer directly to the variables therein.)
There's a bit of a pitfall in combining trial1 and trial2, as they get coerced to their numeric codes, which are all 1s in both cases, unless you use as.character() ...
you had the order of the variable to aggregate and the variable to group by backwards (you want to aggregate PEFR by trial, not the other way around)
mean.drugs <- with(drugs2,
tapply (PEFR, list(trial), mean))
## After Before
## 267.7778 322.6667
Related
>nuevos<-(exam[411:510,1])
> [,1]
401 -0.325087210
402 0.576824342
403 0.314110438
404 -0.710141482
405 0.079179458
406 0.876819478
407 -0.563755647
408 -0.024573542
409 0.072860869
410 0.141759722
411 0.645346837
412 -0.178754696
413 -0.745086021
414 0.741761201
415 1.537360962
416 0.478560270
417 -0.721503050
418 -0.136435690
419 -0.264058207
420 1.851815905
421 0.854542022
422 0.055184071
423 0.214454147
424 -0.374941314
425 0.268580192
426 0.458531169
427 0.440158449
428 -1.539627467
429 -0.146698388
430 -0.174844929
This is my data, it's a matrix. The first column is the ID and the second column is the X value. I want to select 10 ID. In the 10 selected, 5 should be from unpair number ID, and the other 5 should be from ood number ID. The 10 ID selection should be in function from the X value (the most negative value is the best). I want to have something like this:
ID X
428 -1.539627467
413 -0.745086021
....
I tried to use sort(data[data%%2==1])[1:5] but I don't understand how can I extract the column ID from the dataset, because this is a result from a linear model, so R give me the positions but I want to work with this positions and the X value. Please, help me!
Thanks.
The numbers in the first "column" are the rownames of the matrix.
Since the objects in your question have differing names, it's not entirely clear to me if the following works like that.
So I would do something like this:
df=data.frame(ID=rownames(exam),X=exam[,1])
Otherwise please post the output of dput(exam) or dput(data)
Based on what I think you want to do, here's a working example, given the following data frame:
# generate random input data
data <- data.frame(ID=1:20, X=rnorm(20))
Tidyverse offers the cleanest solution:
require(tidyverse)
data %>%
arrange(X)
will sort in ascending order according to column x. Check the documentation for arrange for further details; you can do more complex things such as sorting by group, sorting on multiple columns (ie, specify a first column, and break ties based on successively sorted columns, etc). So what I would recommend would be to put your data into a data frame first:
data <- data.frame(ID=rownames(nuevos), X=nuevos[,1])
where you could substitute ID with whatever you want and then do the above. Add a dput of nuevos for more specific feedback. Note there are a million ways under the sun to do this not involving tidyverse (ie, sort as you mentioned, for instance); tidyverse just tends to make for the cleanest, simplest mechanism in my opinion (since it is plug and play with many other useful things, like ggplot, dplyr, etc) and is really a great way of thinking to get accustomed to for working with data frames, such as this.
I am brand new at using R/Rattle and am having difficulty understanding how to interpret the last line of this code output. Here is the function call along with it's output:
> head(weatherRF$model$predicted, 10)
336 342 94 304 227 173 265 44 230 245
No No No No No No No No No No
Levels: No Yes
This code is implementing a weather data set in which we are trying to get predictions for "RainTomorrow". I understand that this function calls for the predictions for the first 10 observations of the data set. What I do NOT understand is what the last line ("Levels: No Yes") means in the output.
It's called a factor variable.
That is the list of permitted values of the factor, here the values No and Yes are permitted.
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 do have next trouble:
I created a decision tree with R based on rpart library, and since I have a broad list of variables, rules are and endeless list.
By using asRules(tree) from rattle library, result is nicer than by just running tree once tree is computed.
The problem is the set of rules is longer than number of lines printeables from console, so I can't copy them by Control + C, and by saving this result into a variable, for instance:
t <- asRules(tree)
I would expect something like
Rule number: 1 [target=0 cover=500 (4%) prob=0.8]
var1 < 10
var2 < 2
var3 >=45
var4 >=5
Eventhough result is
[1] 297 242 295 126 127 124
And obviously this isn't what I am looking for.
So I understand 3 ways of solving:
Increasing limit of printable lines to access from console (I don't know how to do that).
Print in console with a key press to continue, in order to first copy, then paste, and the pressing the button to get next results (I don't know how to do that either).
Being able to save bunch of rules into a txt file or something similar instead of [1] 297 242 295 126 127 124.
Guys, any help is very much appreciated!
Thank you!
For #3 use
sink(file='somefile.txt')
asRules(tree)
sink()
I've been really racking my brain over this problem and with no expedient solution in sight.
I have a dataset over which I am trying to permute one variable (an attribute) within another variable (a location), irrespective of an object (an item).
Here's a snippet of the data:
ID_FIELD SPCD Total
1177 833 428.286591
11383 691 1175.846712
24081 316 137.042979
11383 318 177.335481
1177 71 166.629921
24081 110 1170.012216
1177 12 8.379811
30284 541 585.039300
24081 746 188.808428
24081 531 196.142482
1177 111 47.258113
1177 12 198.443376
11383 827 16.095224
Using ddply() function in the plyr package, with R version 3.2.0, I've submitted the following code:
ddply(data,.(Total,ID_FIELD),sample)
Here, I am trying to permute Total (the attribute) across SPCD (the item) within ID_FIELD (the location), and after running ddply() code twice in sequence, the result is the exact same as before, which is not what I want. I'd like this process randomized at each running of the function (i.e. a new shuffling of Total each submission of ddply()).
Any clues as to how to accomplish this? A speedy process would be appreciated as well, given that the application is with a large dataset. I am at my wit's end.
Many thanks.
Using plyr:
ddply(data, .(ID_FIELD), function(df) df[sample(nrow(df)),])
Using dplyr, which has a sampling function supplied:
library(dplyr)
data %>% group_by(ID_FIELD) %>% sample_frac