I have a time series which is unevenly sampled and I want to downsample it to 20Hz. I have made a moving average by binning the data points when in 0.05s time windows (20Hz) and applying an arithmetic mean to it. The data frame looks like this:
Time Right Left
1 0.000000000 18.21980 30.98789
2 0.009222031 22.15157 37.18590
3 0.022511959 25.63218 42.49231
4 0.029854059 28.43851 46.57811
5 0.039320946 30.43885 49.29414
6 0.052499056 31.60561 50.67852
7 0.059612036 32.01045 50.92879
8 0.076606989 31.80335 50.34975
9 0.082647085 31.18134 49.29151
10 0.090698957 30.35415 48.09110
And the code I used for the moving average was this:
data$group_num <- floor(data$Time/0.05)
data2<-NULL
data2$Right = aggregate(data$Right,
list(group_num=data$group_num), mean)
data2$Left = aggregate(data$Left,
list(group_num=data$group_num), mean)
data2$Time = aggregate(data$Time,
list(group_num=data$group_num), mean)
However, for optimizing it I want to make rather a Gaussian filter so that the data points in the middle of the bin have more weight. I couldnt find any function that could deal with uneven sampling. Thus, I started writting a script, where I managed to give them weights.
data$weight <- ((data$Time-data$group_num*0.05)*((data$group_num+1)*0.05-data$Time))^5
I have to normalize these weights by the mean of the weights in their own bin (for instance). By trying to normalize these weights to the mean of their own group I ran into problems with too slow functions. Could anybody give me a hand with it??
I finally managed to do what I wanted. The key function that helped me to overcome this one was ave(). Thank god. So this is what I basically did:
data$weight <- (abs(data$Time-(data$group_num*0.05+0.025)))^(-1)
data$Norm<-ave(data$weight,data$group_num,FUN=function(x) x/sum(x))
data$Time2<- data$Time*data$Norm
data$Right2<- data$Right*data$Norm
data$Left2<- data$Left*data$Norm
data2$Time<- tapply(data$Time2, data$group_num, sum)
data2$Right<- tapply(data$Right2, data$group_num, sum)
data2$Left<- tapply(data$Left2, data$group_num, sum)
Thanks Marat Talipov for the help. From what I see in your code this could also work. But because this is just working fine and quick enough I will stay with this.
Related
Very new to R here, also very new to the idea of coding and computer stuff.
Second week of class and I need to find some summary statistics from a set of data my professor provided. I downloaded the chart of data and tried to follow along with his verbal instructions during class, but I am one of the only non-computer science backgrounds in my degree program (I am an RN going for degree in Health Informatics), so he went way too fast for me.
I was hoping for some input on just where to start with his list of tasks for me to complete. I downloaded his data into an excel file, and then uploaded it into R and it is now a matrix. However, everything I try for getting the mean and standard deviation of the columns he wants comes up with an error. I am understanding that I need to convert these column names into some sort of vector, but online every website tells me to do these tasks differently. I don't even know where to start with this assignment.
Any help on how to get myself started would be greatly appreciated. Ive included a screenshot of his instructions and of my matrix. and please, excuse my ignorance/lack of familiarity compared to most of you here... this is my second week into my masters I am hoping I begin to pick this up soon I am just not there yet.
the instructions include:
# * Import the dataset
# * summarize the dataset,Compute the mean and standard deviation for the three variables (columns): age, height, weight
# * Tabulate smokers and age.level data with the variable and its frequency. How many smokers in each age category ?
# * Subset dataset by the mothers that smoke and weigh less than 100kg,how many mothers meet this requirements?
# * Compute the mean and standard deviation for the three variables (columns): age, height, weight
# * Plot a histogram
Stack Overflow is not a place for homeworks, but I feel your pain. Let's get piece by piece.
First let's use a package that helps us do those tasks:
library(data.table) # if not installed, install it with install.packages("data.table")
Then, let's load the data:
library(readxl) #again, install it if not installed
dt = setDT(read_excel("path/to/your/file/here.xlsx"))
Now to the calculations:
1 summarize the dataset. Here you'll see the ranges, means, medians and other interesting data of your table.
summary(dt)
1A mean and standard deviation of age, height and weight (replace age with the column name of height and weight to get those)
dt[, .(meanValue = mean(age, na.rm = TRUE), stdDev = sd(age, na.rm = TRUE))]
2 tabulate smokers and age.level. get the counts for each combination:
dt[, .N, by = .(smoke, age.level)]
3 subset smoker mothers with wt < 100 (I'm asuming non-pregnant mothers have NA in the gestation field. Adjust as necessary):
dt[smoke == 1 & weight < 100 & !is.na(gestation), .N]
4 Is the same as 1A.
5 Plot a histogram (but you don't specify of what variable, so let's say it's age):
hist(dt$age)
Keep on studying R, it's not that difficult. The book recommended in the comments is a very good start.
I am a newbie in R programming and seek help in analyzing the Metabolomics data - 118 metabolites with 4 conditions (3 replicates per condition). I would like to know, for each metabolite, which condition(s) is significantly different from which. Here is part of my data
> head(mydata)
Conditions HMDB03331 HMDB00699 HMDB00606 HMDB00707 HMDB00725 HMDB00017 HMDB01173
1 DMSO_BASAL 0.001289121 0.001578235 0.001612297 0.0007772231 3.475837e-06 0.0001221674 0.02691318
2 DMSO_BASAL 0.001158363 0.001413287 0.001541713 0.0007278363 3.345166e-04 0.0001037669 0.03471329
3 DMSO_BASAL 0.001043537 0.002380287 0.001240891 0.0008595932 4.007387e-04 0.0002033625 0.07426482
4 DMSO_G30 0.001195253 0.002338346 0.002133992 0.0007924157 4.189224e-06 0.0002131131 0.05000778
5 DMSO_G30 0.001511538 0.002264779 0.002535853 0.0011580857 3.639661e-06 0.0001700157 0.02657079
6 DMSO_G30 0.001554804 0.001262859 0.002047611 0.0008419137 6.350990e-04 0.0000851638 0.04752020
This is what I have so far.
I learned the first line from this post
kwtest_pvl = apply(mydata[,-1], 2, function(x) kruskal.test(x,as.factor(mydata$Conditions))$p.value)
and this is where I loop through the metabolite that past KW test
tCol = colnames(mydata[,-1])[kwtest_pvl <= 0.05]
for (k in tCol){
output = posthoc.kruskal.dunn.test(mydata[,k],as.factor(mydata$Conditions),p.adjust.method = "BH")
}
I am not sure how to manage my output such that it is easier to manage for all the metabolites that passed KW test. Perhaps saving the output from each iteration appending to excel? I also tried dunn.test package since it has an option of table or list output. However, it still leaves me at the same point. Kinda stuck here.
Moreover, should I also perform some kind of adjusted p-value, i.e FWER, FDR, BH right after KW test - before performing the posthoc test?
Any suggestion(s) would be greatly appreciated.
I'm quite a newbie in R so I was interested in the optimality of my solution. Even if it works it could be (a bit) long and I wanted your advice to see if the "way I solved it" is "the best" and it could help me to learn new techniques and functions in R.
I have a dataset on students identified by their id and I have the school where they are matched and the score they obtained at a specific test (so for short: 3 variables id,match and score).
I need to construct the following table: for students in between two percentiles of score, I need to calculate the average score (between students) of the average score of the students of the school they are matched to (so for each school I take the average score of the students matched to it and then I calculate the average of this average for percentile classes, yes average of a school could appear twice in this calculation). In English it allows me to answer: "A student belonging to the x-th percentile in terms of score will be in average matched to a school with this average quality".
Here is an example in the picture:
So in that case, if I take the median (15) for the split (rather than percentiles) I would like to obtain:
[0,15] : 9.5
(15,24] : 20.25
So for students having a score between 0 and 15 I take the average of the average score of the school they are matched to (note that b average will appears twice but that's ok).
Here how I did it:
match <- c(a,b,a,b,c)
score <- c(18,4,15,8,24)
scoreQuant <- cut(score,quantile(score,probs=seq(0,1,0.1),na.rm=TRUE))
AvgeSchScore <- tapply(score,match,mean,na.rm=TRUE)
AvgScore <- 0
for(i in 1:length(score)) {
AvgScore[i] <- AvgeSchScore[match[i]]
}
results <- tapply(AvgScore,scoreQuant,mean,na.rm = TRUE)
If you have a more direct way of doing it.. Or I think the bad point is 3) using a loop, maybe apply() is better ? But I'm not sure how to use it here (I tried to code my own function but it crashed so I "bruted force it").
Thanks :)
The main fix is to eliminate the for loop with:
AvgScore <- AvgeSchScore[match]
R allows you to subset in ways that you cannot in other languages. The tapply function outputs the names of the factor that you grouped by. We are using those names for match to subset AvgeScore.
data.table
If you would like to try data.table you may see speed improvements.
library(data.table)
match <- c("a","b","a","b","c")
score <- c(18,4,15,8,24)
dt <- data.table(id=1:5, match, score)
scoreQuant <- cut(dt$score,quantile(dt$score,probs=seq(0,1,0.1),na.rm=TRUE))
dt[, AvgeScore := mean(score), match][, mean(AvgeScore), scoreQuant]
# scoreQuant V1
#1: (17.4,19.2] 16.5
#2: NA 6.0
#3: (12.2,15] 16.5
#4: (7.2,9.4] 6.0
#5: (21.6,24] 24.0
It may be faster than base R. If the value in the NA row bothers you, you can delete it after.
is it possible to run an ANOVA in r with only means, standard deviation and n-value? Here is my data frame:
q2data.mean <- c(90,85,92,100,102,106)
q2data.sd <- c(9.035613,11.479667,9.760268,7.662572,9.830258,9.111457)
q2data.n <- c(9,9,9,9,9,9)
q2data.frame <- data.frame(q2data.mean,q2data.sq,q2data.n)
I am trying to find the means square residual, so I want to take a look at the ANOVA table.
Any help would be really appreciated! :)
Here you go, using ind.oneway.second from the rspychi package:
library(rpsychi)
with(q2data.frame, ind.oneway.second(q2data.mean,q2data.sd,q2data.n) )
#$anova.table
# SS df MS F
#Between (A) 2923.5 5 584.70 6.413
#Within 4376.4 48 91.18
#Total 7299.9 53
# etc etc
Update: the rpsychi package was archived in March 2022 but the function is still available here: http://github.com/cran/rpsychi/blob/master/R/ind.oneway.second.R (hat-tip to #jrcalabrese in the comments)
As an unrelated side note, your data could do with some renaming. q2data.frame is a data.frame, no need to put it in the title. Also, no need to specify q2data.mean inside q2data.frame - surely mean would suffice. It just means you end up with complex code like:
q2data.frame$q2data.mean
when:
q2$mean
would give you all the info you need.
I am completely new to R. I tried reading the reference and a couple of good introductions, but I am still quite confused.
I am hoping to do the following:
I have produced a .txt file that looks like the following:
area,energy
1.41155882174e-05,1.0914586287e-11
1.46893363946e-05,5.25011714434e-11
1.39244046855e-05,1.57904991488e-10
1.64155121046e-05,9.0815757601e-12
1.85202830392e-05,8.3207522281e-11
1.5256036289e-05,4.24756620609e-10
1.82107587343e-05,0.0
I have the following command to read the file in R:
tbl <- read.csv("foo.txt",header=TRUE).
producing:
> tbl
area energy
1 1.411559e-05 1.091459e-11
2 1.468934e-05 5.250117e-11
3 1.392440e-05 1.579050e-10
4 1.641551e-05 9.081576e-12
5 1.852028e-05 8.320752e-11
6 1.525604e-05 4.247566e-10
7 1.821076e-05 0.000000e+00
Now I want to store each column in two different vectors, respectively area and energy.
I tried:
area <- c(tbl$first)
energy <- c(tbl$second)
but it does not seem to work.
I need to different vectors (which must include only the numerical data of each column) in order to do so:
> prob(energy, given = area), i.e. the conditional probability P(energy|area).
And then plot it. Can you help me please?
As #Ananda Mahto alluded to, the problem is in the way you are referring to columns.
To 'get' a column of a data frame in R, you have several options:
DataFrameName$ColumnName
DataFrameName[,ColumnNumber]
DataFrameName[["ColumnName"]]
So to get area, you would do:
tbl$area #or
tbl[,1] #or
tbl[["area"]]
With the first option generally being preferred (from what I've seen).
Incidentally, for your 'end goal', you don't need to do any of this:
with(tbl, prob(energy, given = area))
does the trick.