Looping through XYZ coordinats

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04-25-2022 01:34 PM
Daniel4
New Contributor II

Hi, I working on a road dataset and found all the roads that intersect, like the to roads below:

dabrso_0-1650917273690.png

I want to investigate if the z-value from the two roads are the same. I used arcpy.FeatureVerticesToPoints_management the get the XYZ values from each road, and load them into two python list, like in the example below:

road1 = [(1,2,3),(3,4,5),(4,6,8)]
road2 = [(3,4,6),(1,2,4),(4,5,3)]

The first think I want to do, is to check whatever the roads share vertices (Same XY value) and if the do, whats the difference in their z-value. 

I tried with this function:

def compareLineVertiex(list_orginid,list_destid):
           sameXY = []
           for (x,y,z) in list_orginid:
                 xy = x,y
                if xy in list_destid:

                sameXY.append(xyz)
               return sameXY

I only managed to compare their XY values. I am missing the z-value (which will be difference from the two list)

 

 

 

 

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Accepted Solutions
EricEagle
Occasional Contributor III

Maybe a little long but how's this

def compare_vertices(linea, lineb):
    matches = []
    for i_idx, a in enumerate(linea):
        for b_idx, b in enumerate(lineb):
            if [a[0], a[1]] == [b[0], b[1]]:
                if a[2] - b[2] > 0:
                    zdiff = a[2] - b[2]
                else:
                    zdiff = b[2] - a[2]
                matches.append({
                    "Point A": {
                        "XYZ": a,
                        "Index": a_idx,
                    },
                    "Point B": {
                        "XYZ": b,
                        "Index": b_idx,
                    },
                    "Z diff": zdiff
                })
    return matches

road1 = [(1, 2, 3), (3, 4, 5), (4, 6, 8)]
road2 = [(3, 4, 6), (1, 2, 4), (4, 5, 3)]

x = compare_vertices(road1, road2)

for i in x:
    print(i)

{'Point A': {'XYZ': (1, 2, 3), 'Index': 0}, 'Point B': {'XYZ': (1, 2, 4), 'Index': 1}, 'Z diff': 1}
{'Point A': {'XYZ': (3, 4, 5), 'Index': 1}, 'Point B': {'XYZ': (3, 4, 6), 'Index': 0}, 'Z diff': 1}

 

I only enumerated to know where the points are that match in the array, and returned as a dict because I'm fussy like that and want more info than less usually

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4 Replies
EricEagle
Occasional Contributor III

Maybe a little long but how's this

def compare_vertices(linea, lineb):
    matches = []
    for i_idx, a in enumerate(linea):
        for b_idx, b in enumerate(lineb):
            if [a[0], a[1]] == [b[0], b[1]]:
                if a[2] - b[2] > 0:
                    zdiff = a[2] - b[2]
                else:
                    zdiff = b[2] - a[2]
                matches.append({
                    "Point A": {
                        "XYZ": a,
                        "Index": a_idx,
                    },
                    "Point B": {
                        "XYZ": b,
                        "Index": b_idx,
                    },
                    "Z diff": zdiff
                })
    return matches

road1 = [(1, 2, 3), (3, 4, 5), (4, 6, 8)]
road2 = [(3, 4, 6), (1, 2, 4), (4, 5, 3)]

x = compare_vertices(road1, road2)

for i in x:
    print(i)

{'Point A': {'XYZ': (1, 2, 3), 'Index': 0}, 'Point B': {'XYZ': (1, 2, 4), 'Index': 1}, 'Z diff': 1}
{'Point A': {'XYZ': (3, 4, 5), 'Index': 1}, 'Point B': {'XYZ': (3, 4, 6), 'Index': 0}, 'Z diff': 1}

 

I only enumerated to know where the points are that match in the array, and returned as a dict because I'm fussy like that and want more info than less usually

Daniel4
New Contributor II

Thanks 🙂 this well get the job done. I did a test and its working on the dataset. 

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DanPatterson
MVP Esteemed Contributor

numpy anyone?

 

 

road1 = np.array([(1, 2, 3), (3, 4, 5), (4, 6, 8)])
road2 = np.array([(3, 4, 6), (1, 2, 4), (4, 5, 3)])

xyz = []
for row in road1:
    x, y, z1 = row
    chk = (road2[:, :2][:, None] == row[:2]).all(-1)
    if chk.any():
        w = np.nonzero(chk)[0]
        z2 = road2[w].squeeze()
        vals = [x, y, z1, z2[-1]]
        xyz.append(vals)
    xyz1z2 = np.array(xyz)
    

xyz1z2
array([[1, 2, 3, 4],
       [3, 4, 5, 6]])

 

To bring it back into Pro

 

dt = [('X', 'f8'), ('Y', 'f8'), ('Z1', 'f8'), ('Z2', 'f8')]

# -- do it the slow way
out = np.zeros(xyz1z2.shape[0], dtype=dt)
out['X'] = xyz1z2[:, 0]
out['Y'] = xyz1z2[:, 1]
out['Z1'] = xyz1z2[:, 2]
out['Z2'] = xyz1z2[:, 3]
out
array([(  1.00,   2.00,   3.00,   4.00), (  3.00,   4.00,   5.00,   6.00)],
      dtype=[('X', '<f8'), ('Y', '<f8'), ('Z1', '<f8'), ('Z2', '<f8')])
# --- then
arcpy.da.NumPyArrayToTable(out, path_and_tbl_name)

 

 


... sort of retired...
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Daniel4
New Contributor II

Thanks 🙂 I will try it out. 

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