How to group intersecting Shapely geometric objects in a list of tuples

Question

I have one list of data as follows:

from shapely.geometry import box

data = [box(1,2,3,4), box(5,6,7,8), box(1,2,3,4)]
codes = ['A','B','C']

The list 'data' has following elements:

A = box(1,2,3,4)
B = box(5,6,7,8)
C = box(1,2,3,4)

I have to check if an element intersect with any other elements. If intersects, they should put in one tuple; and if not intersect they should put in different tuple. The expected result is:

result = [(A,C), (B)]

How to do it?

I tried it as:

results = []
for p,c in zip(data,codes):
    for x in data:
        if p.intersects(x): ##.intersects return true if they overlap else false
            results.append(c)
print results

@PadraicCunningham they are just to ease out the results to put with names. — Roman, Jun 14 '15 at 23:01
@PadraicCunningham You can also avoid the codes and put the results with elements.. — Roman, Jun 14 '15 at 23:07

Padraic Cunningham · Accepted Answer · 2015-06-14T23:55:53.830

Keep a dict of objects mapped to A,B and C, a set of matched objects and only add the single elements that have no matches after we get to a new letter if they are not in our matched set as all possible combinations will have been tested:

from shapely.geometry import box
from itertools import combinations

codes = ["A", "B", "C"]
d = dict(zip(codes, data))
prev = codes[0]
matched, out = set(), []
for p1, p2 in combinations(codes, 2):
    if d[p1].intersects(d[p2]):
        out.append((p1, p2))
        matched.update([p1, p2])
    # when p1 is a new letter, we have tried all combs for that prev
    # if prev is not in matched it did not intersect any other so
   # add it as a single tuple and add to matched to avoid dupes
    elif p1 != prev and prev not in matched:
        out.append(tuple(prev,))
        matched.add(prev)
    prev = p1
# catch the last letter
if p2 not in matched:
    out.append(tuple(p2,))
print(out)
[('A', 'C'), ('B',)]

@jose, if you want the correct result and in the correct order it does both — Padraic Cunningham, Jun 14 '15 at 23:44
@jose, you can make it less complex but also less efficient , you have to check all the objects against each other no matter what you do so using itertools is efficient. using sets gives you 0(1) lookups, you can simplify it but lose efficiency, if you data is small we could do it in a coupke of lines — Padraic Cunningham, Jun 14 '15 at 23:51

score 1 · Answer 2 · answered Jun 14 '15 at 23:31

from shapely.geometry import box

data = [box(1,2,3,4), box(5,6,7,8), box(1,2,3,4)]
codes = ['A','B','C']

Create a dictionary to map your code to your boxes:

d = dict(zip(codes, data))

Check all combinations:

intersecting = set()

for i, a in enumerate(codes, 1):
    for b in codes[i:]:
        if d[a].intersection(d[b]):
            intersecting |= {a, b}

print(tuple(intersecting), tuple(set(codes)^intersecting))
# ('C', 'A') ('B',)

Tuples will be unordered because sets were used.

hilcharge · Answer 3 · 2015-06-14T23:57:42.650

-1

For each value in data, make a tuple of all elements having intersecting value. Add it to the results list if the list is not in the list already.

results=[]
for b in data
    same_b = tuple([d for d in data if d.intersects(b)])
    if not same_b in results:
        results.append(same_b)

The result is a list of tuples, each of which has all elements with the same value, ie intersecting elements.

You can make this more efficient by not making a tuple if that element has already been extracted into results.

Note that for the given data set, equality with == would work instead of intersects.

If you want codes instead of data, use a dictionary not variable names. codes={'A':box(...),..}

edited Jun 14 '15 at 23:57

answered Jun 14 '15 at 23:19

hilcharge

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I really have to test.intersects in my real problem. Please show me using .intersects. – Roman Jun 14 '15 at 23:31
Do you mean you must use the intersects method? In the question, the data is exactly the same so any equality would work. – hilcharge Jun 14 '15 at 23:35

How to group intersecting Shapely geometric objects in a list of tuples

3 Answers3