Symmetrical matrix

[leejayd]

Hi,

I have a vector representing distances between cities.

I have an nxn matrix looking like :

                Manchester  Leeds London
Manchester
Leeds
London

I'm trying to get the distances loaded to the matrix but I only have distances in one direction so I need to transpose them to make a fully symmetrical matrix i.e

                Manchester  Leeds London
Manchester           0       20     200
Leeds               20        0     230
London              200      230    0

What's the easiest way to do this?

 

[Dason]

If you show us how you actually have your data loaded that would help

 

[leejayd]

In an int vector

cities<-c(0,20,200,20,0,230,200,230,0)

I can manipulate this as it's actually coming to me in json format (with lots of other gunk) with all city distances - I have to ignore half of them as the matrix must be symmetrical.

 

[Dason]

Is that what you actually have? I was under the impression you only had the distances one time. If that's really what you have then matrix(cities, 3, 3) should do the trick. But I'm guessing you misunderstood what I was asking for. In what form is the data coming to you in R? Can you give a sample of that?

 

[leejayd]

I realise I can use

matrix(cities, nrow=3, ncol=3)

but I can't assume the reverse distances will be same (data from google gives different distances depending on direction!)

 

[leejayd]

The data is coming to me as a JSON file but I don't expect you to help me with this. I was hoping there was a transpose type function that will overwrite the other half of matrix i.e. Manchester - Leeds = Leeds -Manchester. Apologies, when I said I don't have it I meant I can't use it.

You can see the sample JSON here :

http://maps.googleapis.com/maps/api...destinations=Manchester,UK|Leeds,UK|London,UK

 

[Dason]

If you can create half the matrix and the other half is 0s then just add the matrix to its transpose.

> x <- matrix(c(0, 1, 2, 0,0,3,0,0,0),3,3)
> x
     [,1] [,2] [,3]
[1,]    0    0    0
[2,]    1    0    0
[3,]    2    3    0
> x+t(x)
     [,1] [,2] [,3]
[1,]    0    1    2
[2,]    1    0    3
[3,]    2    3    0

 

[leejayd]

Perfect. Thanks so much. I had an idea there would be a function like this.