Loop over data frame comparing pairs - r

Question

I have created the following dataframe:
set.seed(42)
df1 = data.frame(pair = rep(c(1:26),2), size = rnorm(52,5.4,1.89))
It represents random pairs of individuals of a certain size, as assigned by the 'pair' column.
The random distribution (5.4, 1.89) is based on observed data from the group that I sampled in my study (N=26 pairs).
I now want to ask a very basic question that I am unable to code my way to:
Imagine a horizontal line at the mean (5.4), severing the population in two:
What proportion of individuals are paired with another individual from the same side of the line? i.e. is there a tendency for small to be with small and big to be with big?
I want to compare the proportion I observed with the proportion generated from 'asking' the above question a lot of times (e.g. 1000 repetitions).
In my study 18/26 individuals were together with a similar sized partner, and so I want to ask 'out of a 1000 repetitions, how many times was the proportion of similar individuals equal to or greater than 18/26?' this will be my 'p-value'.
I have no clue how to code this, but in my head it goes like this:
For each value in column 'size': when pair value are equal, do this:
is the larger individual equal to or bigger than 5.4? is the smaller
individual equal to or bigger than 5.4?
if so, return a "yes"
OR
is the larger individual equal to or smaller than 5.4? is the smaller
individual equal to or smaller than 5.4?
if so, return a "yes"
if none of the above are true, return a 0
provide an output of the proportion of yes and no. store this in a data.frame repeat
this process 1000 times, adding all the outputs to the mentioned data
frame:
run1 24/26
run2 4/26
...
run999 13/26
I really hope someone can show me the start to this, or the relevant code/arguments/structure.

Answer 1

Is this what you want
#Create empty output, for 10 iterations
same_group_list = replicate(10,0)
diff_group_list = replicate(10,0)
for (j in 1:10){ #For 10 iterations
df1 = data.frame(pair = rep(c(1:26),2), size = rnorm(52,5.4,1.89))
#Sort by 'pair'
df1 = df1[with(df1, order(pair)), ]
#Assign a group based on if 'size' is > or < than mean(size)
for (i in 1:nrow(df1)){
if (df1$size[i] <= mean(df1$size)){ #Use 5.4 explicitly instead of mean(df1$size) if you want
df1$Group[i] = -1
} else {
df1$Group[i] = 1
}
}
df1$Group = as.numeric(df1$Group) #Convert to numeric
output2 = tapply(df1$Group, df1$pair, mean) #Carry out groupwise mean
diff_group_list[j] = sum(output2 == 0) #A mean of 0 means pair grouped with another group
same_group_list[j] = length(output2) - diff_group_list[j] #Everything else is the same group
}
output = data.frame("Same groupout of 26" = same_group_list, "Different Group out of 26" = diff_group_list)

Answer 2

I created a data frame with pairs side by side and then compared which of them where higher than 5.4. Then compared pairs. The pairs with both sizes higher than 5.4 were summed, and then everything was divided by 26.
The data frame proportions shows the proportion for each run.
proportions <- data.frame(run = (1:1000), prop = rep(NA,1000))
for (i in 1:1000) {
df = data.frame(pair = c(1:26),
size1 = rnorm(26,5.4,1.89),
size2 = rnorm(26,5.4,1.89)
)
greaterPairs <- sum(df[,2] > 5.4 & df[,3]>5.4)
proportions[i,2] = greaterPairs/26
}
head(proportions)
I did not keep the proportion in the string format "18/26" because later, if you want to sum the total of them which follows some condition, you will have to do it visually, one by one. So, for example, if you want to know how many of them are greater than or equal 18/26:
sum(proportions$prop >= (18/26))