minDiff is an R-package that can be used to assign elements to a specified number of groups and minimize differences between created groups.
Note that this package is no longer maintained, but there is a successor package that is under active developement: anticlust. The anticlust package builds on the theory of cluster analysis to create sets of elements that are as similar as possible. Because the minDiff package still seems to attract some users, I inserted R code below to reproduce the README examples using the anticlust package; the anticlust code runs faster and the results are superior.
Install the devtools package first via install.packages("devtools").
To install the minDiff package, type in your R console
devtools::install_github("m-Py/minDiff")
Load the packages minDiff and anticlust via
library(anticlust) # this is better and you should use it, see below library(minDiff)
To reproduce the following example, install the "MASS" package that contains the data set that is used.
library("MASS")
data(survey) # load data set
head(survey, n=10) # look at the data
The survey data set contains some demographic information on a student sample. Imagine you wish to assign students to three different dormitories and want to create a similar groups of students in each house. As a first step, we might want to match average student age.
To do so, we pass our data set to the function create_groups and specify which variable should be made equal in how many sets.
equal <- create_groups(survey, criteria_scale = "Age",
sets_n = 3, repetitions = 100)
By passing the column Age to the argument criteria_scale we inform create_groups that age is a continuous variable, for which we want to minimize the differences between groups. create_groups returns a data.frame that is saved into the object equal. equal is actually the same as the input data survey, but it has one additional column: newSet - this is the group assignment variable that was created.
Let's have a look at this:
table(equal$newSet) # 1 2 3 # 79 79 79
The survey data set has 237 entries, which can be assigned to three groups of equal size. If the data set had not been a multiplier of the group number, create_groups would have created groups that are of similar size.
Let's see how successful we were in creating groups of age:
tapply(equal$Age, equal$newSet, mean) # 1 2 3 # 20.35449 20.57806 20.19099
Not so bad! But how did it work? In the function call above, we specified another parameter, repetitions=100 (which is also the default value if we do not specify a value for repetition). This means that the function randomly assigned all cases (i.e. students) to three groups 100 times, and returned the most equal group assignment. In the default case, what is considered most equal is the assignment that has the minimum difference in group means; but we can specify different criteria if we want to (see below). By varying the parameter repetitions we can increase our chances of creating equal groups. If we conduct 10,000 repetitions (which is still very fast if we only consider one variable), the groups will be very similar with regards to age. Note that random repetitions performs quite okay, but more sophisticated such as used in anticlust improve the results (see below).
Note that it is possible to pass a data set that has been optimized previously; in this case, the program does not start all over, but only tries find more similar groups than the previous best assignment:
equal <- create_groups(equal, criteria_scale = "Age",
sets_n = 3, repetitions = 10000)
> tapply(equal$Age, equal$newSet, mean)
1 2 3
20.37028 20.38194 20.37133
Using the anticlust package, we can obtain a better partitioning using the anticlustering() function, and the code runs much faster:
library(anticlust) survey$anticlustering_groups <- anticlustering( survey$Age, K = 3, # 3 groups as above objective = "variance" # equalizes mean values ) # The output is a grouping vector, not data frame tapply(survey$Age, survey$anticlustering_groups, mean) # 1 2 3 # 20.37453 20.37446 20.37456Considering more than one criterion
We can pass more than one criterion to create_groups. Let's imagine we also want students to be of equal heights in all dormitories:
equal <- create_groups(survey, criteria_scale = c("Age", "Height"),
sets_n = 3, repetitions = 10000)
tapply(equal$Age, equal$newSet, mean)
# 1 2 3
# 20.37877 20.35233 20.39244
tapply(equal$Height, equal$newSet, mean, na.rm=TRUE)
# 1 2 3
# 172.5658 172.3694 172.2292
Note that there were missing values in the variable survey$Height. This is given out as a warning by create_groups, but it will still return a result (and simply disregards the missing value in the variable height).
Using the anticlust package, we can do better, as follows:
# anticlust needs some prior NA handling, simply impute mean for age
survey$imputed_height <- survey$Height
survey$imputed_height[is.na(survey$imputed_height)] <- mean(survey$Height, na.rm = TRUE)
survey$anticlustering_groups <- anticlustering(
survey[, c("Age", "imputed_height")],
K = 3, # 3 groups as above
objective = "variance", # equalizes mean values
standardize = TRUE # because age and height are different in their value range
)
tapply(survey$Age, survey$anticlustering_groups, mean)
# 1 2 3
# 20.37449 20.37451 20.37454
tapply(survey$Height, survey$anticlustering_groups, mean, na.rm=TRUE)
# 1 2 3
# 172.3782 172.3824 172.3818
Generally, I would not recommend only equalizing the means between groups, but also the spread (i.e., the standard deviations), an example of how to do this in minDiff and anticlust is given below.
We may not only wish to minimize differences with regards to age or height, but we might want to create equal gender ratios in all dormitories. Let's check in what ratios our previous group assignment resulted:
table(equal$newSet, equal$Sex) # Female Male # 1 36 43 # 2 38 41 # 3 44 34
We can see that gender ratios are very different between dormitories. create_groups offers the possibility to consider categorical variables when creating groups. These are passed via the criteria_nominal parameter. Let's try this out:
equal <- create_groups(survey, criteria_scale = "Age",
criteria_nominal = "Sex", tolerance_nominal = 2,
sets_n = 3, repetitions = 100)
By specifying the parameter tolerance_nominal = 2, we tell create_groups that we tolerate deviations between dormitories in the frequency of female and male students of no more than 2. Did that work?
table(equal$newSet, equal$Sex) # Female Male # 1 40 39 # 2 38 40 # 3 40 39
Using the anticlust package, we can use the categories argument to consider nominal variables:
survey$anticlustering_groups <- anticlustering( survey$Age, K = 3, objective = "variance", categories = survey$Sex ) table(survey$anticlustering_groups, survey$Sex) # Female Male # 1 40 39 # 2 39 40 # 3 39 39
Note that using minDiff a tolerance_nominal of 1 will most likely not return a feasible solution; anticlust by default finds the best possible balance (so there is no need to specify a tolerance level).
It is possible to pass two categorical criteria create_groups(). There is no limit for scale criteria, but only two categorical variables can be passed. Note that anticlust can incoorporate an arbitrary number of categorical variables.
Here is an example where we want to create dormitories that are similar with regard to smoker status and gender ratio:
equal <- create_groups(survey, criteria_scale = c("Age", "Height"),
criteria_nominal = c("Sex", "Smoke"),
tolerance_nominal = c(2, 3, Inf), sets_n = 3,
repetitions=100)
table(equal$newSet, equal$Sex)
# Female Male
# 1 39 40
# 2 40 38
# 3 39 40
table(equal$newSet, equal$Smoke)
# Heavy Never Occas Regul
# 1 4 63 6 6
# 2 4 62 6 6
# 3 3 64 7 5
Note that the parameter tolerance_nominal expects a vector of length 3 if two categorical variables are passed to criteria_nominal. These values indicate tolerances for deviations in the first, the second and the combined categorical levels. In doubt, use large tolerance values when starting to optimize your groups and see how well the function is running. In the upper case I was not interested to assign an equal number of smoking females and males to each dormitories and considered both categorical variables independently from each other. So, I set the tolerance for deviations in the combined groups to infinity.
Using the anticlust package, I do not need to worry about tolerance levels when using more than one categorical criterion (and the results, again, are better and the code runs much faster):
survey$anticlustering_groups <- anticlustering(
survey$Age,
K = 3,
objective = "variance",
categories = survey[, c("Sex", "Smoke")]
)
table(survey$anticlustering_groups, survey$Sex)
# Female Male
# 1 39 39
# 2 40 39
# 3 39 40
table(survey$anticlustering_groups, survey$Smoke)
# Heavy Never Occas Regul
# 1 4 63 7 5
# 2 4 63 6 6
# 3 3 63 6 6
Use other equalizing functions
I could be interested not only in equalizing mean age between groups, but also the standard deviation of ages to achieve similar distributions of age between dormitories. This is possible by passing another function to the equalize parameter.
equal <- create_groups(survey, criteria_scale = "Age",
criteria_nominal = "Smoke",
tolerance_nominal = 2,
sets_n = 3, repetitions = 500,
equalize = list(mean, sd))
tapply(equal$Age, equal$newSet, mean)
# 1 2 3
# 20.21943 20.41986 20.48425
tapply(equal$Age, equal$newSet, sd)
# 1 2 3
# 6.599697 6.995727 5.855852
In anticlust, the "k-plus" objective can be used to equalize means as well as standard deviations between groups. Again, the results are strongly improved as compared to minDiff.
survey$anticlustering_groups <- anticlustering( survey$Age, K = 3, objective = "kplus", categories = survey$Smoke, method = "local-maximum" # better algorithm to get similar groups ) tapply(survey$Age, survey$anticlustering_groups, mean) # 1 2 3 # 20.42829 20.34710 20.34815 tapply(survey$Age, survey$anticlustering_groups, sd) # 1 2 3 # 6.502561 6.501436 6.501497
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