node_degree_out() (graph function)

2025-04-02

Applies to: ✅ Microsoft Fabric ✅ Azure Data Explorer

The node_degree_out function calculates the out-degree, or number of outgoing edges, from a node in a directed graph.

Note

This function is used with the graph-match and graph-shortest-paths operators.

Syntax

node_degree_out([node])

Learn more about syntax conventions.

Parameters

Name	Type	Required	Description
node	`string`	✔️	The reference to a graph node variable in a graph pattern. No parameters should be passed when used inside all(), any() and map() graph functions, in conjunction with inner_nodes().

Returns

Returns the out-degree of the input node or of all inner nodes, when used inside all(), any() and map() functions in conjunction with inner_nodes().

Examples

The examples in this section show how to use the syntax to help you get started.

Find paths between locations and transportation modes

The following example uses the Locations and Routes data tables to construct a graph that finds paths from a source location to a destination location through a route. It returns the source location name, destination location name, transportation methods along the route, the node_degree_out, which is the number of outgoing edges from the source node (location), and the route_nodes_degree_out, which are the number of outgoing edges from the inner nodes (stopover locations) along the route.

Run the query

// Locations table (nodes)
let Locations = datatable(LocationName: string, LocationType: string) [
    "New York", "City",
    "San Francisco", "City",
    "Chicago", "City",
    "Los Angeles", "City",
    "Seattle", "Warehouse"
];
// Routes table (edges)
let Routes = datatable(OriginLocationID: string, DestinationLocationID: string, TransportMode: string) [
    "New York", "San Francisco", "Truck",
    "New York", "Chicago", "Train",
    "San Francisco", "Los Angeles", "Truck",
    "Chicago", "Seattle", "Train",
    "Los Angeles", "New York", "Truck",
    "Seattle", "San Francisco", "Train"
];
Routes
| make-graph OriginLocationID --> DestinationLocationID with Locations on LocationName
| graph-match (src)-[route*1..2]->(dest)
project src.LocationName, 
        dest.LocationName, 
        node_degree_out(src),
        route_TransportModes = map(route, TransportMode),
        route_nodes_degree_out = map(inner_nodes(route), node_degree_out())

Output

src_LocationName	dest_LocationName	node_degree_out	route_TransportModes	route_nodes_degree_out
Chicago	Seattle	1	["Train"]	[]
New York	Chicago	2	["Train"]	[]
Los Angeles	New York	1	["Truck"]	[]
San Francisco	Los Angeles	1	["Truck"]	[]
Seattle	San Francisco	1	["Train"]	[]
New York	San Francisco	2	["Truck"]	[]
Chicago	San Francisco	1	["Train","Train"]	[1]
New York	Seattle	2	["Train","Train"]	[1]
New York	Los Angeles	2	["Truck","Truck"]	[1]
San Francisco	New York	1	["Truck","Truck"]	[1]
Seattle	Los Angeles	1	["Train","Truck"]	[1]
Los Angeles	San Francisco	1	["Truck","Truck"]	[2]
Los Angeles	Chicago	1	["Truck","Train"]	[2]

Find employee with no managers

The following example creates a graph to represent the hierarchical relationships between employees and their managers. It uses the graph-match operator to find employees who report to a top-level manager who doesn't report to anyone else. It uses the node_degree_out function to identify the managers who don't report to any other manager.

Run the query

let employees = datatable(name:string, age:long)
[
"Alice", 32,
"Bob", 31,
"Eve", 27,
"Joe", 29,
"Chris", 45,
"Alex", 35,
"Ben", 23,
"Richard", 39,
"Jim", 42,
];
let reports = datatable(employee:string, manager:string)
[
"Bob", "Alice",
"Chris", "Alice",
"Eve", "Bob",
"Ben", "Chris",
"Joe", "Alice",
"Richard", "Bob",
"Alice", "Jim"
];
reports
| make-graph employee --> manager with employees on name
| graph-match (manager)<-[reports]-(employee)
where node_degree_out(manager) == 0
project manager.name, employee.name, di_m=node_degree_in(manager), do_m=node_degree_out(manager), di_e=node_degree_in(employee), do_e=node_degree_out(employee)

Output

manager_name	employee_name	degree_in_m	degree_out_m
Jim	Alice	1	0

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