# [R] misclassification matrix

Tom Fletcher tom.fletcher.mp7e at statefarm.com
Wed Oct 6 19:46:59 CEST 2010

```I think what you are looking for is

?table
and/or
?prop.table

So, let's say you have two matrices: ACTUAL and CLASS, you can ...

table(ACTUAL, CLASS)

Or, diag(1-prop.table(table(ACTUAL, CLASS), 1)) to get row percentages
and take the diagonal.

So, using your example:

# table() as above would render the following matrix
mat <- matrix(c(100, 25, 10, 250, 100, 5, 50, 25, 40), 3)

and prop.table() renders:

> prop.table(mat, 1)
[,1]      [,2]      [,3]
[1,] 0.2500000 0.6250000 0.1250000
[2,] 0.1666667 0.6666667 0.1666667
[3,] 0.1818182 0.0909091 0.7272727

In the above, the diagonal is the % correctly classified.

> diag(1-prop.table(mat, 1))
[1] 0.7500000 0.3333333 0.2727273

TF

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On Behalf Of Gregory Ryslik
Sent: Wednesday, October 06, 2010 10:19 AM
To: R Help
Subject: [R] misclassification matrix

HI Everyone,

I am working with the following situation. I have n observations and j
possible outcomes and each one of the n observations is assigned a class
from 1 to j.  Furthermore, this process is done m times (for some large
m > 1000).

What I want to do is create a misclassification matrix which tells me
for each one of the possible classes, how many observations were
classified correctly, and then how many were incorrectly at each level.
Here is an example of what I mean

0 	1	 2	Misclassification
0     100 250  50   .75
1	25	100	25	.333333
2	10	5	40   .2727273

For each one of the 1 to j elements, I can use a nested for loop to
count how many were classified as 0, are in 1, etc and then construct
such matrix. Thus for each element I have j comparisons and then I have
j total rows leading to an O(j^2) running time.

Any way I can avoid such a for loop and perhaps make it run a bit
quicker? If not, any ideas then at least how to avoid the double for
loop and make the code more aesthetically pleasing?

As always, thank you for your help!

Kind regards,
Greg
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