City of Surrey data projectCan a computer find Surrey’s communities on its own, from playground locations alone?
The City of Surrey maps every park playground it runs and labels each with a community such as Newton or Cloverdale. Our students hide those labels, let the k-means algorithm group the playgrounds by location, then compare its groups with the City’s. The clusters and counts here are our output; the community labels are the City’s.
Our k-means results on City of Surrey Park Playgrounds (167 playgrounds fetched 22 October 2026; k = 6 clusters; 20 random starts of 50 rounds each)| Run | Total squared distance to cluster centres | Playgrounds in a cluster led by their own community | Communities leading the six clusters |
|---|
| Tightest of the 20 runs | 891 million square metres | 137 of 167 | Cloverdale, Guildford, Newton, South Surrey (twice), Whalley |
| Tightest run with six different leaders | 944 million square metres | 144 of 167 | Cloverdale, Fleetwood, Guildford, Newton, South Surrey, Whalley |
This is the full program in standard-library Python. It downloads the playground outlines in the City’s own map grid, where positions are in metres, reduces each outline to a rough centre by averaging its corner points, and runs k-means twenty times from different random starting centres. When we ran it on 22 October 2026 it printed exactly the two rows in the table:import json, random, urllib.request
from collections import Counter
URL = ("https://services5.arcgis.com/YRpe0VKTJytZSSIB/arcgis/rest/services/"
"Park%20Playgrounds/FeatureServer/0/query?where=1%3D1"
"&outFields=COMMUNITY&f=json")
spots = [] # centre of each playground outline, metres east and north (UTM zone 10N)
for f in json.load(urllib.request.urlopen(URL))["features"]:
ring = f["geometry"]["rings"][0]
spots.append((sum(p[0] for p in ring) / len(ring),
sum(p[1] for p in ring) / len(ring), f["attributes"]["COMMUNITY"]))
def nearest(x, y, centres):
gaps = [(x - cx) ** 2 + (y - cy) ** 2 for cx, cy in centres]
return gaps.index(min(gaps)), min(gaps)
def kmeans(k, seed):
random.seed(seed)
centres = [s[:2] for s in random.sample(spots, k)]
for _ in range(50):
groups = [[] for _ in range(k)]
for s in spots:
groups[nearest(s[0], s[1], centres)[0]].append(s)
centres = [(sum(s[0] for s in g) / len(g), sum(s[1] for s in g) / len(g))
if g else centres[i] for i, g in enumerate(groups)]
spread = sum(nearest(s[0], s[1], centres)[1] for s in spots)
return spread, groups
runs = sorted(kmeans(6, seed) for seed in range(1, 21))
def leaders(groups):
return [Counter(s[2] for s in g).most_common(1)[0] for g in groups]
def report(label, spread, groups):
top = leaders(groups)
print(label, "spread", round(spread / 1e6), "million m2;",
sum(n for _, n in top), "of", len(spots), "match their cluster")
print(" clusters led by", sorted(str(name) for name, _ in top))
report("tightest of 20 runs:", *runs[0])
six = next(r for r in runs if len({name for name, _ in leaders(r[1])}) == 6)
report("tightest run with six different leaders:", *six)
1. No answers given
Unlike most school exercises, k-means is never told the right groups. It places six centres, gives each playground to the nearest one, moves each centre to the middle of its group, and repeats. The community names are used only afterwards, to judge the result.
2. Different starts, different answers
Twenty starts did not all agree. The run with the smallest total distance split South Surrey into two clusters and left Fleetwood without one of its own; a slightly looser run matched all six communities and placed 144 playgrounds with their own label.
3. Which one is better?
That is BC Computer Science 11’s own question, and here it has no single answer. The tightest run is better by the algorithm’s own measure; the six-community run is better at matching how the City describes itself. Students write down which they prefer, and why.
What the labels hide
Of the 167 playgrounds, 13 carry no community label and one is tagged City Centre. Newton has the most labelled playgrounds (38), then Cloverdale (32), Whalley (28), South Surrey (24), Guildford (18) and Fleetwood (13). Because South Surrey covers a large area with fewer playgrounds, a method that only cares about distance is tempted to cut it in two. Clusters reflect the data you give them, not the map in a resident’s head.
A note on the numbers: averaging an outline’s corners gives a rough centre, not a true centroid, which is close enough for clustering a city’s worth of small parks. The playground layer is published by the City of Surrey with an update frequency the City gives as monthly, so a later download may differ slightly. The City’s open data licence grants “a worldwide, royalty-free, perpetual, non-exclusive license to use the Information, including for commercial purposes”, and asks for the statement “Contains information licensed under the Open Government License”, with the City of Surrey named after it. Source: City of Surrey, Park Playgrounds layer. The City has not endorsed this page or our classes, and our clusters are not City boundaries.
