## Calculating the sum of connections for all possible pairings in a network

find the sum of all possible pairs in an array of n elements

find a pair of elements from an array whose sum equals a given number

find all pairs in an array

find all possible pairs in array

find all pairs of given sum

product of difference of all pairs

print all possible pairs in array

Based on a Data Frame listing connections between a source and a destination

import pandas as pd df = pd.DataFrame({'source':['A','B','B'],'destination':['B','C','C']}) print(df) source destination 0 A B 1 B C 2 B C

I want to calculate a square matrix containing the number of connections for all pairings, i.e. the resulting DataFrame should be

A B C A 0 1 0 B 0 0 2 C 0 0 0

where the indices represent the sources and the column labels the destinations.

How can I get there?

Use `crosstab`

with `DataFrame.reindex`

:

v = np.unique(df.values) df1 = pd.crosstab(df.source, df.destination).reindex(index=v, columns=v, fill_value=0) print (df1) destination A B C source A 0 1 0 B 0 0 2 C 0 0 0

**How to prove max number of connection between n nodes is n*(n-1 ,** Every single node (of which there are n ) has to be connected to every one to determine the maximum number of links possible from N network nodes is The total number of possible combinations/connections where a link� because connection (x,y) and (y,x) is the same (for all connections), we end up with n*(n-1)/2. Thank you, the explanation of that intuition was like a breath of fresh air – Monarch Wadia Nov 19 '17 at 0:16

Use `pivot_table`

. `locations`

collects all unique entries to expand the final index and columns to include the zero rows and columns.

import numpy as np locations = np.unique(df.values) df.pivot_table(index='source', columns='destination', aggfunc=len, dropna=False ).loc[locations, locations].fillna(0) destination A B C source A 0.0 1.0 0.0 B 0.0 0.0 2.0 C 0.0 0.0 0.0

**Sum of f(a[i], a[j]) over all pairs in an array of n integers ,** A naive approach is to iterate through all pairs and calculate f(a[i], a[j]) and to get the sum of difference of all pairs possible with that number. In network science, the configuration model is a method for generating random networks from given degree sequence. It is widely used as a reference model for real-life social networks, because it allows the user to incorporate arbitrary degree distributions.

Here is my solution in which I count connections after converting the letters to integers (indices):

import pandas as pd import numpy as np df = pd.DataFrame({'source':['A','B','b'],'destination':['B','C','C']}) print(df) nodes = np.unique(df) n_nodes = len(nodes) # assuming you have no letters missing adj = np.zeros((n_nodes, n_nodes)) lett2num = lambda letter : ord(letter.lower()) - 96 # convert letter to number, case insensitive for index, row in df.iterrows(): i = lett2num(row.source) - 1 j = lett2num(row.destination) - 1 adj[i,j] += 1

It outputs for `adj`

:

array([[0., 1., 0.], [0., 0., 2.], [0., 0., 0.]])

**Introduction to Social Network Methods: Chapter 7: Basic Properties ,** But, clearly not every possible connection is present, and there are "structural holes" (or at To get more precise, and to use computers to apply algorithms to calculate In any network there are (k * k-1) unique ordered pairs of actors (that is AB is The sum of the connections from the actor to others (e.g. actor #1 sends� Will allow if there is an a and b, or a and c, or b and c, or all three a,b and c. In other words, it insists there be at least 2 of a or b or c in the result. So {a,b,f} is accepted, but {a,e,f} is rejected.

**Sum of all paths between all pairs of nodes in a tree,** So you could consider these the best and worst possible test cases. Similarly, finding and removing edges can be done in amortized O(n) time if you handle� Count of pairs is 3. Time Complexity : O(n 2) Auxiliary Space : O(1) A better solution is possible in O(n) time.. Below is the Algorithm. Create a map to store frequency of each number in the array.

**List of measures - Brain Connectivity Toolbox,** Measures for binary and undirected networks will often be faster to compute than Strength: Node strength is the sum of weights of links connected to the node. Density: Density is the fraction of present connections to possible connections. components: Connected components are subnetworks in which all pairs of� Sum-of-Pairs scoring Limits: ~ k = 8 - 10 sequences ~ n = 500 residues Observations 1. A multiple alignment induces pairwise alignments 2. A column in the induced pairwise alignment may contain all gaps, even though no column in the MSA contains all gaps. 3. The pairwise alignments induced by the optimal multiple

**All help headers - Brain Connectivity Toolbox,** Also, there is no good way to determine optimal annealing parameters in NOTE: CIJclus will have a total average degree exactly equal to (or very close to) The distance matrix contains lengths of shortest paths between all pairs of nodes. of the shortest possible path connecting every pair of nodes in the network. The number of connections in a full mesh = n(n - 1) / 2. Ike any network Ethernet must have an algorithm for determining when each network node is allowed to communicate?

##### Comments

- the dataframe you print looks nothing like the one you enter the code for?
- Thanks for noting - please see the corrected version.
- You can do it by maintaining a dictionary with source as the key and its value would be a dictionary of dest as key and no of occurance as its value. Ex. {A:{B:1},B:{C:2}} and finally make a matrix out of this relevant info.