I have a huge matrix of this form, with 1000000 rows and 10000 columns. This is a toy example:
A B C Mean
1 3 4 2.66
2 4 3 3
1 3 4 2.66
9 9 9 9
1 3 2 2
2 4 5 3
1 2 6 3
2 3 5 3.33
The rows in column "Mean" represent the mean of A, B and C for each row. On the other hand, the global mean of column "Mean" is 3.58. I would like to know, using a t-test and R, whether the mean in each row is significantly higher from the global mean. How can I get the p-values for comparison?. Comparing means between 2 groups is very simple using t.test(), but I am not able to find how to compare a single value with the mean of a group that includes that value.
I strongly agree with Roman that you should go back to CV, since this seems liable to giving you a number of false positives.
But in terms of your R question, you could try a one-sample t-test here:
global.mean <- 3.58
val.matrix <- matrix(c(...),...)
pvals <- apply(val.matrix,1,function(r) t.test(r,mu=global.mean)$p.value)
### should do a multiple comparison correction here, e.g., pvals*nrow(val.matrix)
This will give you a vector of size nrow(val.matrix) with each element being the p-value from the two-sided t-test testing whether the values of a row are
significantly different from 3.58. I'm not advocating for this statistical approach, but this is how you could implement it.
Related
I have found a lot of information on how to analyze a 2*2 (AB/BA) crossover trial; however, there are fewer materials on how to disentangle the carryover effect when the study is designed in three periods and two sequences (ABB/BAA). It is worth mentioning that A and B are the treatments and there have been wash-out phases between the three periods.
As sample data, I would like to use the bioequivalence data from "daewr" library.
library("daewr")
data(bioequiv)
head(bioequiv)
Group Subject Period Treat Carry y
1 1 2 1 A none 112.25
2 1 2 2 B A 106.36
3 1 2 3 B B 88.59
4 1 3 1 A none 153.71
5 1 3 2 B A 150.13
6 1 3 3 B B 151.31
The variable Carry contains lagged information from the previous period's Treatment.
The model below should be able to disentangle the effects, but I don't know how to replace the none in the Carry column. I am not sure how to specify this, or how to check if the carryover effect is negligible.
If we don't replace the none in the Carry column, the undermentioned model faces a problem of multicollinearity.
fit <- lmer(y ~ Period+Treat+Carry+(1|Subject), bioequiv)
anova(fit)
summary(fit)
I'm trying to compute a grand mean in R.
lets say I had some data like this
mean1 mean2 fire1 fire2
1 1 2 3
2 2 3 4
3 3 4 5
If I wanted to find the grand mean of that dataset is there a function that might handle it or do I need to do it the old fashion way?
mean(c(mean.default(dataset[[1]]), mean.default(dataset[[2]])))
where in c() you have one mean.default(dataset[[n]]) for each n in the range n=1 to n = [number of columns to be used in calculation]
I'm starting with data of scores at the "group-person" level as follows:
group_id person_id score
1 1 3
1 2 1
1 3 5
2 1 3
2 2 3
2 3 6
The goal is to generate data on person-person pairs that looks like the following:
person_id1 person_id2 sumsquarederror
1 2 4
1 3 13
2 3 25
where the "sumsquarederror" variable is defined as the sum across all groups of the squared differences in score values for each possible pair of persons. In mathspeak, this variable would be defined like: for persons i=1 and i=2 and groups j=(1,...,J)
sumsquarederror(i=1,i=2) = sum_j (( score(i=1) - score(i=2) )^2)
Building this data is trivial with small numbers of groups and persons, but I have roughly 1,000 groups and 150,000 persons, so creating matrices/dataframes for all combinations possible quickly becomes computationally burdensome (=150K by 150K by 1K, before collapsing to the sumsquarederror variable)
I'm guessing there might be some linear algebra approaches or regression-type ideas, but am stumped. Any tips or tricks or useful packages would be greatly appreciated!
I will present my question in two ways. First, requesting a solution for a task; and second, as a description of my overall objective (in case I am overthinking this and there is an easier solution).
1) Task Solution
Data context: each row contains four price variables (columns) representing (a) the price at which the respondent feels the product is too cheap; (b) the price that is perceived as a bargain; (c) the price that is perceived as expensive; (d) the price that is too expensive to purchase.
## mock data set
a<-c(1,5,3,4,5)
b<-c(6,6,5,6,8)
c<-c(7,8,8,10,9)
d<-c(8,10,9,11,12)
df<-as.data.frame(cbind(a,b,c,d))
## result
# a b c d
#1 1 6 7 8
#2 5 6 8 10
#3 3 5 8 9
#4 4 6 10 11
#5 5 8 9 12
Task Objective: The goal is to create a single column in a new data frame that lists all of the unique values contained in a, b, c, and d.
price
#1 1
#2 3
#3 4
#4 5
#5 6
...
#12 12
My initial thought was to use rbind() and unique()...
price<-rbind(df$a,df$b,df$c,df$d)
price<-unique(price)
...expecting that a, b, c and d would stack vertically.
[Pseudo illustration]
a[1]
a[2]
a[...]
a[n]
b[1]
b[2]
b[...]
b[n]
etc.
Instead, the "columns" are treated as rows and stacked horizontally.
V1 V2 V3 V4 V5
1 1 5 3 4 5
2 6 6 5 6 8
3 7 8 8 10 9
4 8 10 9 11 12
How may I stack a, b, c and d such that price consists of only one column ("V1") that contains all twenty responses? (The unique part I can handle separately afterwards).
2) Overall Objective: The Bigger Picture
Ultimately, I want to create a cumulative share of population for each price (too cheap, bargain, expensive, too expensive) at each price point (defined by the unique values described above). For example, what percentage of respondents felt $1 was too cheap, what percentage felt $3 or less was too cheap, etc.
The cumulative shares for bargain and expensive are later inverted to become not.bargain and not.expensive and the four vectors reside in a data frame like this:
buckets too.cheap not.bargain not.expensive too.expensive
1 0.01 to 0.50 0.000000000 1 1 0
2 0.51 to 1.00 0.000000000 1 1 0
3 1.01 to 1.50 0.000000000 1 1 0
4 1.51 to 2.00 0.000000000 1 1 0
5 2.01 to 2.50 0.001041667 1 1 0
6 2.51 to 3.00 0.001041667 1 1 0
...
from which I may plot something that looks like this:
Above, I accomplished my plotting objective using defined price buckets ($0.50 ranges) and the hist() function.
However, the intersections of these lines have meanings and I want to calculate the exact price at which any of the lines cross. This is difficult when the x-axis is defined by price range buckets instead of a specific value; hence the desire to switch to exact values and the need to generate the unique price variable.
[Postscript: This analysis is based on Peter Van Westendorp's Price Sensitivity Meter (https://en.wikipedia.org/wiki/Van_Westendorp%27s_Price_Sensitivity_Meter) which has known practical limitations but is relevant in the context of my research which will explore consumer perceptions of value under different treatments rather than defining an actual real-world price. I mention this for two reasons 1) to provide greater insight into my objective in case another approach comes to mind, and 2) to keep the thread focused on the mechanics rather than whether or not the Price Sensitivity Meter should be used.]
We can unlist the data.frame to a vector and get the sorted unique elements
sort(unique(unlist(df)))
When we do an rbind, it creates a matrix and unique of matrix calls the unique.matrix
methods('unique')
#[1] unique.array unique.bibentry* unique.data.frame unique.data.table* unique.default unique.IDate* unique.ITime*
#[8] unique.matrix unique.numeric_version unique.POSIXlt unique.warnings
which loops through the rows as the default MARGIN is 1 and then looks for unique elements. Instead, if we use the 'price', either as.vector or c(price) converts into vector
sort(unique(c(price)))
#[1] 1 3 4 5 6 7 8 9 10 11 12
If we use unique.default
sort(unique.default(price))
#[1] 1 3 4 5 6 7 8 9 10 11 12
I have a data set consisting of 2000 individuals. For each individual, i:2000 , the data set contains n repeated situations. Letting d denote this data set, each row of dis indexed by i and n. Among other variables, d has a variable pid which takes on identical value for an individual across different (situations) rows.
Taking into consideration the panel nature of the data, I want to re-sample d (as in bootstrap):
with replacement,
store each re-sample data as a data frame
I considered using the sample function but could not make it work. I am a new user of r and have no programming skills.
The data set consists of many variables, but all the variables have numeric values. The data set is as follows.
pid x y z
1 10 2 -5
1 12 3 -4.5
1 14 4 -4
1 16 5 -3.5
1 18 6 -3
1 20 7 -2.5
2 22 8 -2
2 24 9 -1.5
2 26 10 -1
2 28 11 -0.5
2 30 12 0
2 32 13 0.5
The first six rows are for the first person, for which pid=1, and the next sex rows, pid=2 are different observations for the second person.
This should work for you:
z <- replicate(100,
d[d$pid %in% sample(unique(d$pid), 2000, replace=TRUE),],
simplify = FALSE)
The result z will be a list of dataframes you can do whatever with.
EDIT: this is a little wordy, but will deal with duplicated rows. replicate has its obvious use of performing a set operation a given number of times (in the example below, 4). I then sample the unique values of pid (in this case 3 of those values, with replacement) and extract the rows of d corresponding to each sampled value. The combination of a do.call to rbind and lapply deal with the duplicates that are not handled well by the above code. Thus, instead of generating dataframes with potentially different lengths, this code generates a dataframe for each sampled pid and then uses do.call("rbind",...) to stick them back together within each iteration of replicate.
z <- replicate(4, do.call("rbind", lapply(sample(unique(d$pid),3,replace=TRUE),
function(x) d[d$pid==x,])),
simplify=FALSE)