CODEWITHSUNDEEP
rcorrelationpearson
Ahdee
Ahdee

Reputation: 4949

Why does spearman produce different result on zscore?

Hi it seems that spearman correlation should produce the same result regardless if its zscore or raw. Here are two examples.

https://stats.stackexchange.com/questions/77562/why-does-correlation-come-out-the-same-on-raw-data-and-z-scored-standardized-d

https://stats.stackexchange.com/questions/13952/can-spearmans-correlation-be-run-on-z-scores

However for this example here the two correlation are different and I'm wondering what is going on. 

df = read.csv("https://www.dropbox.com/s/jdktw9jugzm97v3/test.csv?dl=1", head=F)

cor(df[, 1], df[,2], method="spearman")
cor(scale(df[, 1]), scale(df[,2]), method="spearman")

# 0.8462699 vs 0.8905341

Interestingly pearson gives the same result. I'm wondering what I'm doing or thinking incorrectly here?

edit: so in addition I thought may be this is due to ties so I also use kendall which should handle ties however it also gives different results. 

cor(as.matrix ( df[, 1] ) , as.matrix ( df[,2] ), method="kendall" )
cor(scale(as.matrix ( df[, 1] )), scale(as.matrix ( df[,2] )),  method="kendall")

thanks.

Upvotes: 1

Views: 100

Answers (1)

Ahdee
Ahdee

Reputation: 4949

Hi as mentioned above in the comments this was due to a rounding error. No one answered but I wanted to add this in case someone else stumble on a similar issue. So when I round to 15-16 digits the results are the same. 

df = read.csv("https://www.dropbox.com/s/jdktw9jugzm97v3/test.csv?dl=1", head=F)

df = round(df, digits = 15)

cor(as.matrix ( df[, 1] ) , as.matrix ( df[,2] ), method="spearman" )
cor(scale(df[, 1] ), scale(df[,2] ),  method="spearman")

thanks everyone for helping with this.

Upvotes: 1

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