Add contents of two columns to new columns with a delimiter between them

I have a matrix of 11693682x3 uint8. I would like to create so called 'keys' for my data in a new column. For example:

1, 2, 3 
2, 5, 7
3, 3, 6,
4, 29, 0
Output:
1, 2, 3, 1|2
2, 5, 7, 2|5
3, 3, 6, 3|3
4, 29, 0, 4|29

Could anyone point me at the best way to solve this?

Answer 1

[dpb]

Well, it's simply impossible to add something like that as another column in an int8 array.

To do something similar you could create a corollary variable

>> key=categorical(cellstr(num2str(rb(:,1:2),'%d|%d')))
key = 
   1|2 
   2|5 
   3|3 
   4|29 
>>

I'd suggest more conventional hashing techniques might be more useful, though. What's the end use here?

[Rogier Busscher]

I have travel time data on people biking from all district centers (origin) to all metro stations (station) and from all metro stations to all jobs (destinations)

So to get from every origin to every destination, I need to be able to link home to station and station to job:

A|B, B|C, A|C

If you get what i mean.

[dpb]

Well, it'll work...

> find(key=='3|3')
ans =
   3

but, you don't need it, either, unless you just want the appearance of the combined locations. The categorical is kinda' good about that.

But, get to same place with

> find(ismember(rb(:,[1:2]),[3 3],'rows'))
ans =
   3
>>

directly from the two values.

[Rogier Busscher]

I do not think i've made myself fully clear. I got a 11318560x3 matrix of all possible combination of Origin, Station and Destination.

Furthermore, i have travel time data on:

Any origin (A) to any station(B).

Any station(B) to any destination(C).

So i want to link the trips to that begin and end at the same station. That will result into a matrix of data where 'people' go from every origin to every destination, by using every station possible (A|B|C)

Then, I must find the lowest value for every origin-destination (A|C), so that i know what the lowest possible travel time is between each origin and destination.

So that is where i want to use the keys for, because it should be possible to find the lowest value in column K for each unique value (the keys) in Column J

[Rogier Busscher]

But if you have a faster/better solution for this problem, i'm open to suggestions.

[dpb]

I thought you were looking for a way to find the given pair; the above does that by locating where that pair is in the vector.

If it's two sets, then just combination of

key1==val1 & key2==val2

is the same result.

Answer 2

[Walter Roberson]

If you only need to know what the distance is for each combination, but you do not need the path for each combination, then you can use https://www.mathworks.com/help/matlab/ref/graph.distances.html

If you need to know the path, then you can loop through all sources and use https://www.mathworks.com/help/matlab/ref/graph.shortestpath.html to all destinations.