Algorithm for Connected Components of Graph

I am looking for the most efficient algorithm in order to find both the number of connected components in a network, and the number of nodes for each connected component.

Example:

Given the following inputs:

no_of_nodes = 8

graph_to = [1,1,3,5,6]

graph_from = [2,6,4,7,3]

I would receive the following output:

[[1, 2, 3, 4, 6], [5, 7], [8]]

This is what I have so far:

def connections(no_of_nodes, graph_from, graph_to):

    nodes = list(range(1, no_of_nodes+1))
    
    singles = []
    
    # removes all unconnected nodes

    for node in nodes:
        if node not in graph_from + graph_to:
            singles.append([node])
    
    conns = [[graph_from[0],graph_to[0]]]
    
    graph_from.pop(0)
    graph_to.pop(0)
    
    n=0
    k = 0
    
    while n < len(graph_from):
        
        x = graph_from[n]
        y = graph_to[n]

        if x in conns[k]:
            conns[k].append(y)
            graph_from.pop(n)
            graph_to.pop(n)
        else:
            conns.append([x,y])
            k += 1
            print(conns)
            n+=1
    
    return conns + singles

I have found a way to iterate through the nodes assuming that all connections are adjacently placed in the graph_from list, but this, of course, will not hold up for all cases.

EDIT: I am looking for a way to do this without importing a module

Answer

Using the summarized process answered above, I was able to formulate the following which after testing looks to check out.

def connected(graph_nodes, graph_from, graph_to):
 
    
    unseen = [i for i in range(1,graph_nodes + 1)]
    
    edges = [[graph_from[i],graph_to[i]] for i in range(len(graph_from))]
    
    
    connects = []
    
    while unseen != []:
        start = unseen[0]
        connection = []

        queue = [start]
        unseen.remove(start)

        while queue != []:

            end = queue[-1]

            if end not in connection:
                connection.append(end)

            queue = queue[:-1]
            
            for k in edges:

                if k[0] == end:
       
                    queue.append(k[1])
                    unseen.remove(k[1])
            
        connects.append(connection)

        
   
    return connects