I have a dataframe which looks like this:
df <- data.frame(id= rep(seq(1:125),3),
timpoint= c(rep("T1", 125), rep("T2", 125), rep("T3", 125)),
treatment=c(rep("A",25),rep("B",25),rep("C",25),rep("D",25),rep("E",25)))
interaction.col <- paste(df$timpoint, df$treatment, sep = "_")
df <- cbind(df, interaction.col)
I am trying to generate a sum coding scheme for the interaction column which is a combination of the first two columns. According to this paper I should get a matrix of (a−1)×(b−1) columns and n rows(in this case 375)
I have read up on using contrasts:
contrasts(df$interaction.col) <- "contr.sum"
df.c <- contrasts(df$interaction.col)
However, somehow the output is a 15x14 matrix, while it should be a 375 x8.
Also, only the very last row is set to -1, which shouldn't be the case. For all the ID's of the last treatment (E) the interaction column should be set to -1 for the corresponding timepoint.
The last ID in treatment group E should be -1 for all columns. What am i doing wrong here?
Depending on what effects you are interested in and how you will fit the model, you will end up with different number of effects. For example, in the case whereby you fit the main and interaction effects, you should end up with 8 columns for the interactions ie (a-1) x (b-1). In the case you do not fit the main effects you end up with a*b - 1:
Here is how to create your matrix:
With main effects:
model.matrix(~treatment * timpoint, df, list(treatment = contr.sum, timpoint=contr.sum))
In this case, the last 8 columns are the ones you are interested in
Without main effects:
model.matrix(~treatment:timpoint, df, list(treatment = contr.sum, timpoint=contr.sum))
Related
I need to run a 2-sample independent t-test, comparing Column1 to Column2. But Column1 is in DataframeA, and Column2 is in DataframeB. How should I do this?
Just in case relevant (feel free to ignore): I am a true beginner. My experience with R so far has been limited to running 2-sample matched t-tests within the same data frame by doing the following:
t.test(response ~ Column1,
data = (Dataframe1 %>%
gather(key = "Column1", value = "response", "Column1", "Column2")),
paired = TRUE)
TL;DR
t_test_result = t.test(DataframeA$Column1, DataframeB$Column2, paired=TRUE)
Explanation
If the data is paired, I assume that both dataframes will have the same number of observations (same number of rows). You can check this with nrow(DataframeA) == nrow(DataframeB) .
You can think of each column of a dataframe as a vector (an ordered list of values). The way that you have used t.test is by using a formula (y~x), and you were essentially saying: Given the dataframe specified in data, perform a t test to assess the significance in the difference in means of the variable response between the paired groups in Column1.
Another way of thinking about this is by grabbing the data in data and separating it into two vectors: the vector with observations for the first group of Column1, and the one for the second group. Then, for each vector, you compute the mean and stdev and apply the appropriate formula that will give you the t statistic and hence the p value.
Thus, you can just extract those 2 vectors separately and provide them as arguments to the t.test() function. I hope it was beginner-friendly enough ^^ otherwise let me know
EDIT: a few additions
(I was going to reply in the comments but realized I did not have space hehe)
Regarding the what #Ashish did in order to turn it into a Welch's test, I'd say it was to set var.equal = FALSE. The paired parameter controls whether the t-test is run on paired samples or not, and since your data frames have unequal number of rows, I'm suspecting the observations are not matched.
As for the Cohen's d effect size, you can check this stats exchange question, from which I copy the code:
For context, m1 and m2 are the group's means (which you can get with n1 = mean(DataframeA$Column1)), s1 and s2 are the standard deviations (s2 = sd(DataframeB$Column2)) and n1 and n2 the sample sizes (n2 = length(DataframeB$Column2))
lx <- n1- 1 # Number of observations in group 1
ly <- n2- 1 # # Number of observations in group 1
md <- abs(m1-m2) ## mean difference (numerator)
csd <- lx * s1^2 + ly * s2^2
csd <- csd/(lx + ly)
csd <- sqrt(csd) ## common sd computation
cd <- md/csd ## cohen's d
This should work for you
res = t.test(DataFrameA$Column1, DataFrameB$Column2, alternative = "two.sided", var.equal = FALSE)
I have got a text file containing 200 models all compared to eachother and a molecular distance for each 2 models compared. It looks like this:
1 2 1.2323
1 3 6.4862
1 4 4.4789
1 5 3.6476
.
.
All the way down to 200, where the first number is the first model, the second number is the second model, and the third number the corresponding molecular distance when these two models are compared.
I can think of a way to import this into R and create a nice 200x200 matrix to perform some clustering analyses on. I am still new to Stack and R but thanks in advance!
Since you don't have the distance between model1 and itself, you would need to insert that yourself, using the answer from this question:
(you can ignore the wrong numbering of the models compared to your input data, it doesn't serve a purpose, really)
# Create some dummy data that has the same shape as your data:
df <- expand.grid(model1 = 1:120, model2 = 2:120)
df$distance <- runif(n = 119*120, min = 1, max = 10)
head(df)
# model1 model2 distance
# 1 2 7.958746
# 2 2 1.083700
# 3 2 9.211113
# 4 2 5.544380
# 5 2 5.498215
# 6 2 1.520450
inds <- seq(0, 200*119, by = 200)
val <- c(df$distance, rep(0, length(inds)))
inds <- c(seq_along(df$distance), inds + 0.5)
val <- val[order(inds)]
Once that's in place, you can use matrix() with the ncol and nrow to "reshape" your vector of distance in the appropriate way:
matrix(val, ncol = 200, nrow = 200)
Edit:
When your data only contains the distance for one direction, so only between e.g. model1 - model5 and not model5 - model1 , you will have to fill the values in the upper triangular part of a matrix, like they do here. Forget about the data I generated in the first part of this answer. Also, forget about adding the ones to your distance column.
dist_mat <- diag(200)
dist_mat[upper.tri(dist_mat)] <- your_data$distance
To copy the upper-triangular entries to below the diagonal, use:
dist_mat[lower.tri(dist_mat)] <- t(dist_mat)[lower.tri(dist_mat)]
As I do not know from your question what format is your file in, I will assume the most general file format, i.e., CSV.
Then you should look at the reading files, read.csv, or fread.
Example code:
dt <- read.csv(file, sep = "", header = TRUE)
I suggest using data.table package. Then:
setDT(dt)
dt[, id := paste0(as.character(col1), "-", as.character(col2))]
This creates a new variable out of the first and the second model and serves as a unique id.
What I do is then removing this id and scale the numerical input.
After scaling, run clustering algorithms.
Merge the result with the id to analyse your results.
Is that what you are looking for?
I am trying to convert a data frame with categorical variables to a model.matrix but am losing levels of variables.
Here's my code:
df1 <- data.frame(id = 1:200, y =rbinom(200, 1, .5), var1 = factor(rep(c('abc','def','ghi','jkl'),50)))
df1$var2 <- factor(rep(c('ab c','ghi','jkl','def'),50))
df1$var3 <- factor(rep(c('abc','ghi','nop','xyz'),50))
df1$var2 <- as.character(df1$var2)
df1$var2 <- gsub('\\s','',df1$var2)
df1$var2 <- factor(df1$var2)
sapply(df1, levels)
mm1 <- model.matrix(~ 0+.,df1)
head(mm1)
Any suggestions? Is this a matrix non-invertability issue?
The model matrix is perfectly correct. For factors, the model matrix contains one column less than there are factors: this information is already contained in the (Intercept) column. You are missing this column because you have specified +0 in your model term. Try this:
mm2 <- model.matrix(~., df1)
head(mm2)
You will now see the (Intercept) column which encodes "default" information, and now also the first level of var1 is missing in the column names. The (Intercept) represents your observation at the "reference level", which is the combination of first level of each categorical attribute. Any deviation from this reference level is encoded in the var*??? columns, and since your model assumes no interactions between these columns, you get (4 - 1) * 3 var*??? columns plus the (Intercept) column (which is replaced by var1abc in your initial model matrix).
Unfortunately I lack the precise terms to describe this. Anyone help me out?
I received some good help getting my data formatted properly produce a multinomial logistic model with mlogit here (Formatting data for mlogit)
However, I'm trying now to analyze the effects of covariates in my model. I find the help file in mlogit.effects() to be not very informative. One of the problems is that the model appears to produce a lot of rows of NAs (see below, index(mod1) ).
Can anyone clarify why my data is producing those NAs?
Can anyone help me get mlogit.effects to work with the data below?
I would consider shifting the analysis to multinom(). However, I can't figure out how to format the data to fit the formula for use multinom(). My data is a series of rankings of seven different items (Accessible, Information, Trade offs, Debate, Social and Responsive) Would I just model whatever they picked as their first rank and ignore what they chose in other ranks? I can get that information.
Reproducible code is below:
#Loadpackages
library(RCurl)
library(mlogit)
library(tidyr)
library(dplyr)
#URL where data is stored
dat.url <- 'https://raw.githubusercontent.com/sjkiss/Survey/master/mlogit.out.csv'
#Get data
dat <- read.csv(dat.url)
#Complete cases only as it seems mlogit cannot handle missing values or tied data which in this case you might get because of median imputation
dat <- dat[complete.cases(dat),]
#Change the choice index variable (X) to have no interruptions, as a result of removing some incomplete cases
dat$X <- seq(1,nrow(dat),1)
#Tidy data to get it into long format
dat.out <- dat %>%
gather(Open, Rank, -c(1,9:12)) %>%
arrange(X, Open, Rank)
#Create mlogit object
mlogit.out <- mlogit.data(dat.out, shape='long',alt.var='Open',choice='Rank', ranked=TRUE,chid.var='X')
#Fit Model
mod1 <- mlogit(Rank~1|gender+age+economic+Job,data=mlogit.out)
Here is my attempt to set up a data frame similar to the one portrayed in the help file. It doesnt work. I confess although I know the apply family pretty well, tapply is murky to me.
with(mlogit.out, data.frame(economic=tapply(economic, index(mod1)$alt, mean)))
Compare from the help:
data("Fishing", package = "mlogit")
Fish <- mlogit.data(Fishing, varying = c(2:9), shape = "wide", choice = "mode")
m <- mlogit(mode ~ price | income | catch, data = Fish)
# compute a data.frame containing the mean value of the covariates in
# the sample data in the help file for effects
z <- with(Fish, data.frame(price = tapply(price, index(m)$alt, mean),
catch = tapply(catch, index(m)$alt, mean),
income = mean(income)))
# compute the marginal effects (the second one is an elasticity
effects(m, covariate = "income", data = z)
I'll try Option 3 and switch to multinom(). This code will model the log-odds of ranking an item as 1st, compared to a reference item (e.g., "Debate" in the code below). With K = 7 items, if we call the reference item ItemK, then we're modeling
log[ Pr(Itemk is 1st) / Pr(ItemK is 1st) ] = αk + xTβk
for k = 1,...,K-1, where Itemk is one of the other (i.e. non-reference) items. The choice of reference level will affect the coefficients and their interpretation, but it will not affect the predicted probabilities. (Same story for reference levels for the categorical predictor variables.)
I'll also mention that I'm handling missing data a bit differently here than in your original code. Since my model only needs to know which item gets ranked 1st, I only need to throw out records where that info is missing. (E.g., in the original dataset record #43 has "Information" ranked 1st, so we can use this record even though 3 other items are NA.)
# Get data
dat.url <- 'https://raw.githubusercontent.com/sjkiss/Survey/master/mlogit.out.csv'
dat <- read.csv(dat.url)
# dataframe showing which item is ranked #1
ranks <- (dat[,2:8] == 1)
# for each combination of predictor variable values, count
# how many times each item was ranked #1
dat2 <- aggregate(ranks, by=dat[,9:12], sum, na.rm=TRUE)
# remove cases that didn't rank anything as #1 (due to NAs in original data)
dat3 <- dat2[rowSums(dat2[,5:11])>0,]
# (optional) set the reference levels for the categorical predictors
dat3$gender <- relevel(dat3$gender, ref="Female")
dat3$Job <- relevel(dat3$Job, ref="Government backbencher")
# response matrix in format needed for multinom()
response <- as.matrix(dat3[,5:11])
# (optional) set the reference level for the response by changing
# the column order
ref <- "Debate"
ref.index <- match(ref, colnames(response))
response <- response[,c(ref.index,(1:ncol(response))[-ref.index])]
# fit model (note that age & economic are continuous, while gender &
# Job are categorical)
library(nnet)
fit1 <- multinom(response ~ economic + gender + age + Job, data=dat3)
# print some results
summary(fit1)
coef(fit1)
cbind(dat3[,1:4], round(fitted(fit1),3)) # predicted probabilities
I didn't do any diagnostics, so I make no claim that the model used here provides a good fit.
You are working with Ranked Data, not just Multinomial Choice Data. The structure for the Ranked data in mlogit is that first set of records for a person are all options, then the second is all options except the one ranked first, and so on. But the index assumes equal number of options each time. So a bunch of NAs. We just need to get rid of them.
> with(mlogit.out, data.frame(economic=tapply(economic, index(mod1)$alt[complete.cases(index(mod1)$alt)], mean)))
economic
Accessible 5.13
Debate 4.97
Information 5.08
Officials 4.92
Responsive 5.09
Social 4.91
Trade.Offs 4.91
I have decided to learn R and am going through Introduction to Scientific programming in R book (http://www.ms.unimelb.edu.au/spuRs/)
I am currently stuck on chapter 7 question 3 of the book, the question is:
Consider the following very simple genetic model. A population consists of
equal numbers of two sexes: male and female. At each generation men and
women are paired at random, and each pair produces exactly two offspring,
one male and one female. We are interested in the distribution of height
from one generation to the next. Suppose that the height of both children
is just the average of the height of their parents, how will the distribution
of height change across generations?
Represent the heights of the current generation as a dataframe with two
variables, m and f, for the two sexes. The command rnorm(100, 160, 20)
will generate a vector of length 100, according to the normal distribution
with mean 160 and standard deviation 20 (see Section 16.5.1). We use it to
randomly generate the population at generation 1:
pop <- data.frame(m = rnorm(100, 160, 20), f = rnorm(100, 160, 20))
The command sample(x, size = length(x)) will return a random sample
of size size taken from the vector x (without replacement). (It will also
sample with replacement, if the optional argument replace is set to TRUE.)
The following function takes the dataframe pop and randomly permutes the
ordering of the men. Men and women are then paired according to rows,
and heights for the next generation are calculated by taking the mean of
each row. The function returns a dataframe with the same structure, giving
the heights of the next generation.
next.gen <- function(pop) {
pop$m <- sample(pop$m)
pop$m <- apply(pop, 1, mean)
pop$f <- pop$m
return(pop)
}
Use the function next.gen to generate nine generations, then use the lattice
function histogram to plot the distribution of male heights in each
generation, as in Figure 7.7. The phenomenon you see is called regression
to the mean.
Hint: construct a dataframe with variables height and generation, where
each row represents a single man.
I have constructed a blank data frame:
generations <- data.frame(gen="", height="")
For now I am trying to get just the first generation height information into it, so I run:
next.gen(pop)
generations$height <- pop$m
and I get the following error:
Error in `$<-.data.frame`(`*tmp*`, "height", value = c(165.208323681597, :
replacement has 100 rows, data has 1
I understand that I'm trying to squeeze in information from pop$m dataframe into a single row of generations$height and that is causing the problem, I do not know how to fix this? I thought that a blank data frame is flexible enough to add rows as they are being copied from pop data frame?
I tried then to run this code:
generations <- pop$m
And I get 100 values but that just turns my generations dataframe into a vector I think and running
generations
Just lists the values copied in a vector only.
I think I am approaching the first step wrong, is my dataframe definition correct? Why can't I copy row information from 1 data frame into an empty one and just adjust the size of the empty data frame as needed?
Thank you
Unsure the exact output you are looking for. Here is an approach which should be simple enough to follow. ** Note: There are workable approaches aplenty.
pop <- data.frame(m = rnorm(100, 160, 20), f = rnorm(100, 160, 20))
next.gen <- function(pop) {
pop$m <- sample(pop$m)
pop$m <- apply(pop, 1, mean)
pop$f <- pop$m
return(pop)
}
# the code
test <- list()
for (i in 1:9) {
test[[i]] <- next.gen(pop)["m"]
test[[i]]$generation <- paste0("g", i)
}
library(data.table)
test2 <- rbindlist(test)
# result
m generation
1: 174.6558 g1
2: 143.2617 g1
3: 185.2829 g1
4: 168.9719 g1
5: 151.6948 g1
---
896: 159.6091 g9
897: 161.4546 g9
898: 171.8679 g9
899: 138.4982 g9
900: 152.7390 g9
Try:
> generations <- data.frame(gen="", height="", stringsAsFactors=F)
> for(i in 1:length(pop$m)) generations[i,] = c("",pop$m[i])
> generations
gen height
1 136.70042632318
2 153.985392293761
3 122.077485676327
4 166.582538529591
5 170.751368839498
6 190.8894492681
...