I have a dataset like this:
contingency_table<-tibble::tibble(
x1_not_happy = c(1,4),
x1_happy = c(19,31),
x2_not_happy = c(1,4),
x2_happy= c(19,28),
x3_not_happy=c(14,21),
X3_happy=c(0,9),
x4_not_happy=c(3,13),
X4_happy=c(17,22)
)
in fact, there are many other variables that come from a poll aplied in two different years.
Then, I apply a Fisher test in each 2X2 contingency matrix, using this code:
matrix1_prueba <- contingency_table[1:2,1:2]
matrix2_prueba<- contingency_table[1:2,3:4]
fisher1<-fisher.test(matrix1_prueba,alternative="two.sided",conf.level=0.9)
fisher2<-fisher.test(matrix2_prueba,alternative="two.sided",conf.level=0.9)
I would like to run this task using a short code by mean of a function or a loop. The output must be a vector with the p_values of each questions.
Thanks,
Frederick
So this was a bit of fun to do. The main thing that you need to recognize is that you want combinations of your data. There are a number of functions in R that can do that for you. The main workhorse is combn() Link
So in the language of the problem, we want all combinations of your tibble taken 2 at a time link2
From there, you just need to do some looping structure to get your tests to work, and extract the p-values from the object.
list_tables <- lapply(combn(contingency_table,2,simplify=F), fisher.test)
unlist(lapply(list_tables, `[`, 'p.value'))
This should produce your answer.
EDIT
Given the updated requirements for just adjacement data.frame columns, the following modifications should work.
full_list <- combn(contingency_table,2,simplify=F)
full_list <- full_list[sapply(
full_list, function(x) all(startsWith(names(x), substr(names(x)[1], 1,2))))]
full_list <- lapply(full_list, fisher.test)
unlist(lapply(full_list, `[`, 'p.value'))
This is approximately the same code as before, but now we have to find the subsets of the data that have the same question prefix name. This only works if the prefixes are exactly the same (X3 != x3). I think this is a better solution than trying to work with column indexes, and without the guarantee of always being next to one another. The sapply code does just that. The final output should be what you need for the problem.
Related
I have a dataset "res.sav" that I read in via haven. It contains 20 columns, called "Genes1_Acc4", "Genes2_Acc4" etc. I am trying to find a correlation coefficient between those and another column called "Condition". I want to separately list all coefficients.
I created two functions, cor.condition.cols and cor.func to do that. The first iterates through the filenames and works just fine. The second was supposed to give me my correlations which didn't work at all. I also created a new "cor.condition.Genes" which I would like to fill with the correlations, ideally as a matrix or dataframe.
I have tried to iterate through the columns with two functions. However, when I try to pass it, I get the error: "NAs introduced by conversion". This wouldn't be the end of the world (I tried also suppressWarning()). But the bigger problem I have that it seems like my function does not convert said columns into the numeric type I need for my cor() function. I receive the "y must be numeric" error when trying to run the cor() function. I tried to put several arguments within and without '' or "" without success.
When I ran str(cor.condition.cols) I only receive character strings, which makes me think that my function somehow messes up with the as.numeric function. Any suggestions of how else I could iter through these columns and transfer them?
Thanks guys :)
cor.condition.cols <- lapply(1:20, function(x){paste0("res$Genes", x, "_Acc4")})
#save acc_4 columns as numeric columns and calculate correlations
res <- (as.numeric("cor.condition.cols"))
cor.func <- function(x){
cor(res$Condition, x, use="complete.obs", method="pearson")
}
cor.condition.Genes <- cor.func(cor.condition.cols)
You can do:
cor.condition.cols <- paste0("Genes", 1:20, "_Acc4")
res2 <- as.numeric(as.matrix(res[cor.condition.cols]))
cor.condition.Genes <- cor(res2, res$Condition, use="complete.obs", method="pearson")
eventually the short variant:
cor.condition.cols <- paste0("Genes", 1:20, "_Acc4")
cor.condition.Genes <- cor(res[cor.condition.cols], res$Condition, use="complete.obs")
Here is an example with other data:
cor(iris[-(4:5)], iris[[4]])
Hear me out. Consider an arbitrary case where the new column's elements do not require any information from other columns (which I frustrates base $ and mutate assignment), and not every element in the new column is the same. Here is what I've tried:
df$rand<-rep(sample(1:100,1),nrow(df))
unique(df$rand)
[1] 58
and rest assured, nrow(df)>1. I think the correct solution might have to do with an apply function?
Your code repeats one single random number nrow(df) times. Try instead:
df$rand<-sample(1:100, nrow(df))
This samples without replacement from 1:100 nrow(df) times. Now this would give you an error if nrow(df)>100 because you would run out of numbers from 1:100 to sample. To make sure you don't get this error, you can instead sample with replacement:
df$rand<-sample(1:100, nrow(df), replace = TRUE)
If, however, you don't want any random numbers to repeat but would also like to prevent the error, you can do something like this:
df$rand<-sample(1:nrow(df), nrow(df))
if I understand this correctly ,I think this is pretty easily doable in dplyr or data.table .
for e.g dplyr soln on iris
iris%>%mutate(sample(n()))
I need to do a quality control in a dataset with more than 3000 variables (columns). However, I only want to apply some conditions in a couple of them. A first step would be to replace outliers by NA. I want to replace the observations that are greater or smaller than 3 standard deviations from the mean by NA. I got it, doing column by column:
height = ifelse(abs(height-mean(height,na.rm=TRUE)) <
3*sd(height,na.rm=TRUE),height,NA)
And I also want to create other variables based on different columns. For example:
data$CGmark = ifelse(!is.na(data$mark) & !is.na(data$height) ,
paste(data$age, data$mark,sep=""),NA)
An example of my dataset would be:
name = factor(c("A","B","C","D","E","F","G","H","H"))
height = c(120,NA,150,170,NA,146,132,210,NA)
age = c(10,20,0,30,40,50,60,NA,130)
mark = c(100,0.5,100,50,90,100,NA,50,210)
data = data.frame(name=name,mark=mark,age=age,height=height)
data
I have tried this (for one condition):
d1=names(data)
list = c("age","height","mark")
ntraits=length(list)
nrows=dim(data)[1]
for(i in 1:ntraits){
a=list[i]
b=which(d1==a)
d2=data[,b]
for (j in 1:nrows){
d2[j] = ifelse(abs(d2[j]-mean(d2,na.rm=TRUE)) < 3*sd(d2,na.rm=TRUE),d2[j],NA)
}
}
Someone told me that I am not storing d2. How can I create for loops to apply the conditions I want? I know that there are similar questions but i didnt get it yet. Thanks in advance.
You pretty much wrote the answer in your first line. You're overthinking this one.
First, it's good practice to encapsulate this kind of operation in a function. Yes, function dispatch is a tiny bit slower than otherwise, but the code is often easier to read and debug. Same goes for assigning "helper" variables like mean_x: the cost of assigning the variable is very, very small and absolutely not worth worrying about.
NA_outside_3s <- function(x) {
mean_x <- mean(x)
sd_x <- sd(x,na.rm=TRUE)
x_outside_3s <- abs(x - mean(x)) < 3 * sd_x
x[x_outside_3s] <- NA # no need for ifelse here
x
}
of course, you can choose any function name you want. More descriptive is better.
Then if you want to apply the function to very column, just loop over the columns. That function NA_outside_3s is already vectorized, i.e. it takes a logical vector as an argument and returns a vector of the same length.
cols_to_loop_over <- 1:ncol(my_data) # or, some subset of columns.
for (j in cols_to_loop_over) {
my_data[, j] <- NA_if_3_sd(my_data[, j])
}
I'm not sure why you wrote your code the way you did (and it took me a minute to even understand what you were trying to do), but looping over columns is usually straightforward.
In my comment I said not to worry about efficiency, but once you understand how the loop works, you should rewrite it using lapply:
my_data[cols_to_loop_over] <- lapply(my_data[cols_to_loop_over], NA_outside_3s)
Once you know how the apply family of functions works, they are very easy to read if written properly. And yes, they are somewhat faster than looping, but not as much as they used to be. It's more a matter of style and readability.
Also: do NOT name a variable list! This masks the function list, which is an R built-in function and a fairly important one at that. You also shouldn't generally name variables data because there is also a data function for loading built-in data sets.
I'm trying to conduct certain statistics such as t-tests on a table of data containing hundreds to thousands of columns. The data is formatted in a way that the two groups of values I'm comparing are in the same column.
So, basically my first attempt was to cut and paste like the following;
NN <-read.delim("E:/output.txt")
View(NN)
attach(NN)
#output p-values of 100 t-tests
sink(file="E:/ttest.txt", append=TRUE, split=FALSE)
t.test(Tree1[1:13],Tree1[14:34])$p.value
t.test(Tree2[1:13],Tree2[14:34])$p.value
t.test(Tree3[1:13],Tree3[14:34])$p.value
....
...
..
.
As my data grows, this is becoming more and more impractical. Is there a way to loop these t-tests through each column sequentially and save the ouput to file?
Thanks in advance.
lapply will get you there I think with an anonymous function:
> test <- data.frame(a=1:100,b=101:200)
> lapply(test,function(x) t.test(x[1:50],x[51:100])$p.value)
$a
[1] 2.876776e-31
$b
[1] 2.876776e-31
I should do my part for good practice and also note that running 100 t-tests in a single go is fraught with the potential for type-1 errors and other badness.
Extracting the p-value in isolation is also probably a really bad move.
Not sure if this is a wise approach or if it even works correctly but try mapply with the indexed parts as in:
test <- data.frame(a=1:100,b=101:200)
testa <- test[1:50, ]
testb <- test[51:100, ]
t.test2 <- function(x, y) t.test(x, y)[["p.value"]]
mapply(t.test2, testa, testb)
EDIT: I used thelatemail's data so it's comparable. His warning is right on.
Thanks for all the input. Just a few clarifications; while I AM running hundreds of t-tests at once, they are comparing independent sets of data each time. So for example, the values in column 1 (Tree1), rows 1:50 would only be compared once to rows 51:100 in the same column, and never used again. The same for column 2 (Tree2), and so on. Would type-1 error still be a problem? the way I see it I'm basically doing t-tests on separate data sets one at a time.
That being said, I've come up with a way to do this with a for-loop, and the results correspond to those when t-testing each column individually.
for (i in 1:100)
print (t.test(mydata[1:50, i],mydata[51:100, i])$p.value)
end;
The only problem being that my output always has a [1] in front of it.
I am trying to run some Monte Carlo simulations on animal position data. So far, I have sampled 100 X and Y coordinates, 100 times. This results in a list of 200. I then convert this list into a dataframe that is more condusive to eventual functions I want to run for each sample (kernel.area).
Now I have a data frame with 200 columns, and I would like to perform the kernel.area function using each successive pair of columns.
I can't reproduce my own data here very well, so I've tried to give a basic example just to show the structure of the data frame I'm working with. I've included the for loop I've tried so far, but I am still an R novice and would appreciate any suggestions.
# generate dataframe representing X and Y positions
df <- data.frame(x=seq(1:200),y=seq(1:200))
# 100 replications of sampling 100 "positions"
resamp <- replicate(100,df[sample(nrow(df),100),])
# convert to data frame (kernel.area needs an xy dataframe)
df2 <- do.call("rbind", resamp[1:2,])
# xy positions need to be in columns for kernel.area
df3 <- t(df2)
#edit: kernel.area requires you have an id field, but I am only dealing with one individual, so I'll construct a fake one of the same length as the positions
id=replicate(100,c("id"))
id=data.frame(id)
Here is the structure of the for loop I've tried (edited since first post):
for (j in seq(1,ncol(df3)-1,2)) {
kud <- kernel.area(df3[,j:(j+1)],id=id,kern="bivnorm",unin=c("m"),unout=c("km2"))
print(kud)
}
My end goal is to calculate kernel.area for each resampling event (ie rows 1:100 for every pair of columns up to 200), and be able to combine the results in a dataframe. However, after running the loop, I get this error message:
Error in df[, 1] : incorrect number of dimensions
Edit: I realised my id format was not the same as my data frame, so I change it and now have the error:
Error in kernelUD(xy, id, h, grid, same4all, hlim, kern, extent) :
id should have the same length as xy
First, a disclaimer: I have never worked with the package adehabitat, which has a function kernel.area, which I assume you are using. Perhaps you could confirm which package contains the function in question.
I think there are a couple suggestions I can make that are independent of knowledge of the specific package, though.
The first lies in the creation of df3. This should probably be
df3 <- t(df2), but this is most likely correct in your actual code
and just a typo in your post.
The second suggestion has to do with the way you subset df3 in the
loop. j:j+1 is just a single number, since the : has a higher
precedence than + (see ?Syntax for the order in which
mathematical operations are conducted in R). To get the desired two
columns, use j:(j+1) instead.
EDIT:
When loading adehabitat, I was warned to "Be careful" and use the related new packages, among which is adehabitatHR, which also contains a function kernel.area. This function has slightly different syntax and behavior, but perhaps it would be worthwhile examining. Using adehabitatHR (I had to install from source since the package is not available for R 2.15.0), I was able to do the following.
library(adehabitatHR)
for (j in seq(1,ncol(df3)-1,2)) {
kud <-kernelUD(SpatialPoints(df3[,j:(j+1)]),kern="bivnorm")
kernAr<-kernel.area(kud,unin=c("m"),unout=c("km2"))
print(kernAr)
}
detach(package:adehabitatHR, unload=TRUE)
This prints something, and as is mentioned in a comment below, kernelUD() is called before kernel.area().