Ok straight to the question. I have a database with lots and lots of categorical variable.
Sample database with a few variables as below
gender <- as.factor(sample( letters[6:7], 100, replace=TRUE, prob=c(0.2, 0.8) ))
smoking <- as.factor(sample(c(0,1),size=100,replace=T,prob=c(0.6,0.4)))
alcohol <- as.factor(sample(c(0,1),size=100,replace=T,prob=c(0.3,0.7)))
htn <- as.factor(sample(c(0,1),size=100,replace=T,prob=c(0.2,0.8)))
tertile <- as.factor(sample(c(1,2,3),size=100,replace=T,prob=c(0.3,0.3,0.4)))
df <- as.data.frame(cbind(gender,smoking,alcohol,htn,tertile))
I want to test the hypothesis, using a chi square test, that there is a difference in the portion of smokers, alcohol use, hypertension (htn) etc by tertile (3 factors). I then want to extract the p values for each variable.
Now i know i can test each individual variable using a 2 by 3 cross tabulation but is there a more efficient code to derive the test statistic and p-value across all variables in one go and extract the p value across each variable
Thanks in advance
Anoop
If you want to do all the comparisons in one statement, you can do
mapply(function(x, y) chisq.test(x, y)$p.value, df[, -5], MoreArgs=list(df[,5]))
# gender smoking alcohol htn
# 0.4967724 0.8251178 0.5008898 0.3775083
Of course doing tests this way is somewhat statistically inefficient since you are doing multiple tests here so some correction is required to maintain an appropriate type 1 error rate.
You can run the following code chunk if you want to get the test result in details:
lapply(df[,-5], function(x) chisq.test(table(x,df$tertile), simulate.p.value = TRUE))
You can get just p-values:
lapply(df[,-5], function(x) chisq.test(table(x,df$tertile), simulate.p.value = TRUE)$p.value)
This is to get the p-values in the data frame:
data.frame(lapply(df[,-5], function(x) chisq.test(table(x,df$tertile), simulate.p.value = TRUE)$p.value))
Thanks to RPub for inspiring.
http://www.rpubs.com/kaz_yos/1204
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'm new to R and am trying to run a t-test for two means. I keep getting the error is.atomic is not TRUE. I know I need to make my data atomic, but I haven't found a way online.
I've ran code to check that the data is recursive and did a as.data.frame(mydata).
titanic_summary <- data.frame(Outcome = c("Survived", "Died"),
Mean_Age = c(28.34369, 30.62618),
N = c(342, 549),
Total_Missing = c(52, 125))
titanic_summary
Run a stats test (two sample T-test)
str(titanic_summary)
as.data.frame(titanic_summary)
is.atomic(titanic_summary)
is.recursive(titanic_summary)
titanic_test <- titanic_summary %>%
t.test(Outcome~Mean_Age)
Error in var(x) : is.atomic(x) is not TRUE
t.test does not work the way you seem to think. To avoid that particular error, you could instead use something like titanic_test <- t.test(Mean_Age ~ Outcome, data = titanic_summary) but that would just give you different errors, which comes down to the real question:
You presumably want to see whether there may be a relationship between age and survival, i.e. whether the difference in average ages of 2.28249 is significant but you will need the individual ages or some other additional information about dispersion for this
If you do use the detailed dataset then I suspect that what you really want is something like this:
library(titanic)
titanic_test <- t.test(Age ~ Survived, data = titanic_train)
which would give (for the Kaggle selected training set used in the titanic package)
> titanic_test
Welch Two Sample t-test
data: Age by Survived
t = 2.046, df = 598.84, p-value = 0.04119
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.09158472 4.47339446
sample estimates:
mean in group 0 mean in group 1
30.62618 28.34369
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)$alt)], mean)))
economic
Accessible 5.13
Debate 4.97
Information 5.08
Officials 4.92
Responsive 5.09
Social 4.91
Trade.Offs 4.91
I've got a data frame with 36 variables and 74 observations. I'd like to make a two sample paired ttest of 35 variables by 1 grouping variable (with two levels).
For example: the data frame contains "age" "body weight" and "group" variables.
Now I suppose I can do the ttest for each variable with this code:
t.test(age~group)
But, is there a way to test all the 35 variables with one code, and not one by one?
Sven has provided you with a great way of implementing what you wanted to have implemented. I, however, want to warn you about the statistical aspect of what you are doing.
Recall that if you are using the standard confidence level of 0.05, this means that for each t-test performed, you have a 5% chance of committing Type 1 error (incorrectly rejecting the null hypothesis.) By the laws of probability, running 35 individual t-tests compounds your probability of committing type 1 error by a factor of 35, or more exactly:
Pr(Type 1 Error) = 1 - (0.95)^35 = 0.834
Meaning that you have about an 83.4% chance of falsely rejecting a null hypothesis. Basically what this means is that, by running so many T-tests, there is a very high probability that at least one of your T-tests is going to provide an incorrect result.
Just FYI.
An example data frame:
dat <- data.frame(age = rnorm(10, 30), body = rnorm(10, 30),
weight = rnorm(10, 30), group = gl(2,5))
You can use lapply:
lapply(dat[1:3], function(x)
t.test(x ~ dat$group, paired = TRUE, na.action = na.pass))
In the command above, 1:3 represents the numbers of the columns including the variables. The argument paired = TRUE is necessary to perform a paired t-test.
In the following code I use bootstrapping to calculate the C.I. and the p-value under the null hypothesis that two different fertilizers applied to tomato plants have no effect in plants yields (and the alternative being that the "improved" fertilizer is better). The first random sample (x) comes from plants where a standard fertilizer has been used, while an "improved" one has been used in the plants where the second sample (y) comes from.
x <- c(11.4,25.3,29.9,16.5,21.1)
y <- c(23.7,26.6,28.5,14.2,17.9,24.3)
total <- c(x,y)
library(boot)
diff <- function(x,i) mean(x[i[6:11]]) - mean(x[i[1:5]])
b <- boot(total, diff, R = 10000)
ci <- boot.ci(b)
p.value <- sum(b$t>=b$t0)/b$R
What I don't like about the code above is that resampling is done as if there was only one sample of 11 values (separating the first 5 as belonging to sample x leaving the rest to sample y).
Could you show me how this code should be modified in order to draw resamples of size 5 with replacement from the first sample and separate resamples of size 6 from the second sample, so that bootstrap resampling would mimic the “separate samples” design that produced the original data?
EDIT2 :
Hack deleted as it was a wrong solution. Instead one has to use the argument strata of the boot function :
total <- c(x,y)
id <- as.factor(c(rep("x",length(x)),rep("y",length(y))))
b <- boot(total, diff, strata=id, R = 10000)
...
Be aware you're not going to get even close to a correct estimate of your p.value :
x <- c(1.4,2.3,2.9,1.5,1.1)
y <- c(23.7,26.6,28.5,14.2,17.9,24.3)
total <- c(x,y)
b <- boot(total, diff, strata=id, R = 10000)
ci <- boot.ci(b)
p.value <- sum(b$t>=b$t0)/b$R
> p.value
[1] 0.5162
How would you explain a p-value of 0.51 for two samples where all values of the second are higher than the highest value of the first?
The above code is fine to get a -biased- estimate of the confidence interval, but the significance testing about the difference should be done by permutation over the complete dataset.
Following John, I think the appropriate way to use bootstrap to test if the sums of these two different populations are significantly different is as follows:
x <- c(1.4,2.3,2.9,1.5,1.1)
y <- c(23.7,26.6,28.5,14.2,17.9,24.3)
b_x <- boot(x, sum, R = 10000)
b_y <- boot(y, sum, R = 10000)
z<-(b_x$t0-b_y$t0)/sqrt(var(b_x$t[,1])+var(b_y$t[,1]))
pnorm(z)
So we can clearly reject the null that they are the same population. I may have missed a degree of freedom adjustment, I am not sure how bootstrapping works in that regard, but such an adjustment will not change your results drastically.
While the actual soil beds could be considered a stratified variable in some instances this is not one of them. You only have the one manipulation, between the groups of plants. Therefore, your null hypothesis is that they really do come from the exact same population. Treating the items as if they're from a single set of 11 samples is the correct way to bootstrap in this case.
If you have two plots, and in each plot tried the different fertilizers over different seasons in a counterbalanced fashion then the plots would be statified samples and you'd want to treat them as such. But that isn't the case here.