新手求助：如何用OSMnx获取最大入度节点及其经纬度？

Hey there! Let's work through your problem step by step to get you the node with the highest in-degree centrality, plus its latitude and longitude.

1. Do you need a directed graph for in-degree centrality? #

Absolutely! In-degree only makes sense for directed graphs (where edges have a direction, like one-way streets). Good news: when you use network_type='drive_service', OSMnx automatically creates a directed DiGraph for you, since motorized roads often have direction restrictions. You can double-check this with:

print(type(G))  # Should output <class 'networkx.classes.digraph.DiGraph'>

If for some reason you end up with an undirected graph, you can convert it to directed with G = G.to_directed(), but this shouldn't be needed for drive networks.

2. Why isn't in_degree_centrality in extended_stats? #

OSMnx's extended_stats function doesn't include in-degree centrality by default—no big deal though! We can just calculate it directly using NetworkX's built-in nx.in_degree_centrality() function, which works perfectly on directed graphs.

3. Fixing your code to get the max in-degree node #

Your current code has a couple of small bugs when calculating in-degree centrality. Let's fix those and add the coordinate retrieval:

First, calculate the in-degree centrality correctly:

# Calculate in-degree centrality on the original directed graph
in_degree_centrality = nx.in_degree_centrality(G)

# Find the node with the highest value
max_node, max_in_deg = max(in_degree_centrality.items(), key=lambda x: x[1])

Next, grab the node's latitude and longitude. Important: use the original unprojected graph G for this—G_projected uses UTM (metric) coordinates, not the geographic lat/lon you need. OSMnx stores unprojected coordinates as x (longitude) and y (latitude) in node attributes:

node_lat = G.nodes[max_node]['y']
node_lon = G.nodes[max_node]['x']

4. Full adjusted code #

Here's the complete, working version of your code with all fixes and the coordinate output:

import osmnx as ox, networkx as nx, matplotlib.cm as cm, pandas as pd, numpy as np

place = 'City of Lethbridge, Alberta, Canada'
gdf = ox.gdf_from_place(place)
area = ox.project_gdf(gdf).unary_union.area

# Create directed drive-service graph
G = ox.graph_from_place(place, network_type='drive_service')

# Calculate and merge basic/extended stats
stats = ox.basic_stats(G, area=area)
extended_stats = ox.extended_stats(G, ecc=True, bc=True, cc=True)
for key, value in extended_stats.items():
    stats[key] = value

# Unpack street node counts/proportions into individual stats
for k, count in stats['streets_per_node_counts'].items():
    stats[f'int_{k}_count'] = count
for k, proportion in stats['streets_per_node_proportion'].items():
    stats[f'int_{k}_prop'] = proportion
del stats['streets_per_node_counts']
del stats['streets_per_node_proportion']

# Convert stats to DataFrame (optional)
pd.DataFrame(pd.Series(stats)).T

# Calculate in-degree centrality and find the top node
in_degree_centrality = nx.in_degree_centrality(G)
max_node, max_in_deg = max(in_degree_centrality.items(), key=lambda x: x[1])

# Get the node's geographic coordinates
node_lat = G.nodes[max_node]['y']
node_lon = G.nodes[max_node]['x']

# Print your results
print(f"Node ID with highest in-degree centrality: {max_node}")
print(f"In-degree centrality value: {max_in_deg:.4f}")
print(f"Latitude: {node_lat:.6f}, Longitude: {node_lon:.6f}")

# Optional: Plot the highlighted node on the map
G_projected = ox.project_graph(G)
node_colors = ['red' if node == max_node else '#336699' for node in G_projected.nodes()]
node_sizes = [50 if node == max_node else 8 for node in G_projected.nodes()]
fig, ax = ox.plot_graph(G_projected, node_size=node_sizes, node_color=node_colors, node_zorder=2)

Quick reminders: #

• Stick to the original unprojected graph G when pulling lat/lon—G_projected uses metric coordinates that aren't useful for geographic location.
• If you ever work with an undirected graph (like a walking network), in-degree centrality will be identical to regular degree centrality, since there's no direction to edges.

内容的提问来源于stack exchange，提问作者sola