Hi all, I am working on a project that requires to find the co-ordinates of street intersections. Are there any ways I can convert the data I downloaded from OpenstreetMap to a linear network which the nodes are street intersections? Does any one can give me suggestions about tools that I can use or ways that I can pull this off directly from the website? Thanks so much! asked 08 Nov '13, 00:28 ShengLin lynn edited 13 Nov '13, 17:09 aseerel4c26 ♦ |
One Answer:
See these existing answers: answered 08 Nov '13, 13:03 Frederik Ramm ♦ showing 5 of 7 show 2 more comments |
Thank you!
I just saw two programs called "Osmfilter" and "Osmcorvert." Is that possible for me to get the linear network dataset in a csv file by using these two programs?
I am using R (a software for statistics analysis) for my project, and there are not that many things I can do with an osm file. Could you give some more suggestions if it's possible?
Thanks so much!
As far as I know, osmfilter and osmconvert can not extract intersections easily.
Hello Shenglin, please tell us in what format you have your linear network data set now. Is it already in a kind of database? What about loading it in a database and then form an output you can process with R ?
Right now, I only have the dataset which I downloaded directly from OpenStreetMap, and it's a osm file. I can use R to extract the latitudes and longitudes of all nodes, but I am unable to find a way to filter out the street intersections points.
So I am wandering if there are any other tools can help me to get the co-ordinates of street intersections from the dataset I downloaded from OpenStreetMap Website.
I am not that good at programming, so it's kind of hard for me to make the algorithm to work.
Thanks!
then I really can only recommend that you get more familiar with Routing on OSM based data in general.
See https://wiki.openstreetmap.org/wiki/Routing ... there are links to some opensource projects dealing with that topic. Have a look how they do their basic data preparation.
Or ask at http://gis.stackexchange.com additionaly.
Thank you! I will do.