CODEWITHSUNDEEP
rcorrelationrank
user299791
user299791

Reputation: 2070

How to rank data frame columns

I have a dataframe which holds two variables for a list of users. Such variables represents the number of posts and the number of threads opened by each user.

I'd like to test for a correlation between the two variables, and since the point is to test whether the more you posts the more you also open threads - plus the variables are not normally distributed, I opted for the Spearman correlation in order to assess the relationship between the two variables.

In order to do this, I need to rank my users according to how many posts and threads they have done, and I am stuck at this point. My dataset is a data frame like:

> data
USER SUM(POSTS) SUM(THREADS)
u0          2            2
u1          4            2
u10        212          25
u100         7           1
u102       226          23
u103         1           1
u104         3           1
u105         7           1
u107       234          28

What I have tried so far is to order and find the mean for repeated values with:

p<-ave(order(data[,2]), data[,2]) 
t<-ave(order(data[,3]), data[,3])

If I got the procedure right, which I may not, I expect threads to be ranked like:

4.5 4.5 2 7.5 3 7.5 7.5 7.5 1

but my code produces this ranking:

5.500000 5.500000 6.000000 4.333333 1.000000 4.333333 5.000000 4.333333 9.000000

Any help more than welcome!

Best, Simone

Upvotes: 2

Views: 2458

Answers (1)

Jack Ryan
Jack Ryan

Reputation: 2144

Per droopy's comments, you can try something like:

data[,-1] <- apply(data[,-1], 2, function (x) {rank(1/rank(x))})
data
# USER SUM.POSTS SUM.THREADS
# 1   u0       8.0         4.5
# 2   u1       6.0         4.5
# 3  u10       3.0         2.0
# 4 u100       4.5         7.5
# 5 u102       2.0         3.0
# 6 u103       9.0         7.5
# 7 u104       7.0         7.5
# 8 u105       4.5         7.5
# 9 u107       1.0         1.0

As you see rank() creates golf-style ranks, where lower ranks higher. I ranked the inverse, which appears to give the result you request. Hope this helps.

Upvotes: 2

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