1
$\begingroup$

I have a signed bipartite graph in which the nodes are (1) students and (2) topics. An edge is drawn between a student and topic node if the student mentions their opinion about the topic in a short answer (i.e., some students have an opinion about one topic but not another). The valence of the edge indicates whether the opinion is positive or negative.

My question is: how do I find out how many other students agree with a particular student? In terms of not just which topics they have an opinion on, but also what that opinion is (positive/negative).

EDIT: Based on a comment below

1) What exactly do you mean by agree? Should all existing opinions align or only those on a specific topic? What if one student has additional opinions about a topic?

All existing opinions (including topics and the valence) should align. If a student gives an same opinion about the same topics as another, but happens to talk about another topic as well, that would not count as full agreement. Perhaps there is a way to calculate partial agreement as well?

2) What exactly is your problem? Defining the characteristics is straightforward: Just count. Do you perhaps struggle with an algorithmic implementation?

Algorithmic implementation is perhaps what I am going for. Since I have 100 students, it would be difficult to hand-count the number of peers that agree with them. So if there is a way to calculate a value for each student, that would be helpful.

$\endgroup$
4
  • $\begingroup$ Thank you for your edit. With respect to the remaining question, all I could answer right now is: Write a computer program. $\endgroup$
    – Wrzlprmft
    Commented Sep 14, 2017 at 8:38
  • $\begingroup$ @Wrzlprmft Thanks for the very helpful feedback. What kind of computer program would that be? $\endgroup$
    – iamnarra
    Commented Sep 14, 2017 at 8:41
  • $\begingroup$ Well, what programming experience do you have? $\endgroup$
    – Wrzlprmft
    Commented Sep 14, 2017 at 8:51
  • $\begingroup$ @Wrzlprmft All I am asking for here is if there an algorithm or formula that could calculate what I am looking for. Obviously I know that a script has to be written but I haven't the slightest clue what that script is. $\endgroup$
    – iamnarra
    Commented Sep 14, 2017 at 8:53

1 Answer 1

2
$\begingroup$

As a fast answer, you can represent each student as a vector with $K$ elements (where $K$ is the number of topics) and values $\{+1, 0, -1\}$, denoting positive/non-existent/negative opinion about this topic.

Then, a simple measure of agreement between two students is the element-wise product between two student-vectors. That is the product will be: $similarity = \sum_{i=1}^{K}st_1[i]*st_2[i]$, where $st_1,st_2$ are the student-vectors. Obviously, only the topics where both students have aligned opinions will boost the total [e.g. $1*1=1$ and $(-1)*(-1)=1]$, while misaligned opinions will decrease the sum. If any of the two students haven't expressed an opinion about a topic, then this topic won't matter in the sum.

In that sense, you can find the most like-minded students to a specific student, as the ones with the highest $similarity$. If what you really need is a number of agreeing students for each unique student, then a threshold on the $similarity$ score can be set. The value of the threshold can be decided empirically from your data.

This is easily implemented and if you are comfortable with coding, I could post a sample script in python. One thing to consider though, is in what format is the bipartite graph (a .csv, a graph file of some kind etc.).

EDIT: MINOR EXAMPLE. Fetch example .csv file used from here.

import pandas as pd
import numpy as np

# Change location of file according to your needs
with open('students_example.csv', 'r') as f:
    df = pd.read_csv(f)
# Print for visualization
print(df.head())
print("~"*25)

# Delete column containing the student_id
del df['Student_ID']
# Parse the pandas DataFrame as matrix
student_vectors = df.as_matrix()
# The number of students at hand, let it be N.
N_students = student_vectors.shape[0]
# Initialize empty matrix of similarity between students
# Its size will be NxN (each student with each other)
similarity_scores = np.zeros((N_students, N_students))
# Iterate over each student vector and calculate the
# similarity with all students
for i, student in enumerate(student_vectors):
    # Reshaping and transposing to get the dot product between each student
    # And all the student vectors
    similarity_scores[i,:] = np.dot(student.reshape(1,-1), student_vectors.T)
# Fill the diagonal (that is the similarity of each student with him/herself)
# with low similarity scores so as not to confuse them with other possibly
# agreeing students
np.fill_diagonal(similarity_scores, -1000)

# Random wanted student for example purposes
wanted_id = 3
# Print Students Opinion
print("Wanted Students Opinion:")
print(df.loc[wanted_id].to_string())
print("~"*25)
print("Most similar:(Student ID = %d)"% np.argsort(similarity_scores[wanted_id,:])[::-1][0])
print df.loc[np.argsort(similarity_scores[wanted_id,:])[::-1][0]].to_string()
print("~"*25)
print("Second most similar:(Student ID = %d)"% np.argsort(similarity_scores[wanted_id,:])[::-1][1])
print df.loc[np.argsort(similarity_scores[wanted_id,:])[::-1][1]].to_string()
print("~"*25)

If you follow the example, the output for the wanted student (with $student_{ID}=3$) with opinions: {Trump -1, Net Neutrality -1,Vaccination 1, Obamacare -1}

will give you two other students with the same opinions and their ids.

You can modify the script to fit your needs accordingly.

P.S.: Sorry for the messy code, it was written rather hastily. Also, i tried it with Python 2.7.

$\endgroup$
2
  • $\begingroup$ Thanks a lot for this. I am still learning Python so it would be great if you could post a samples script. The data is in .csv. $\endgroup$
    – iamnarra
    Commented Sep 14, 2017 at 9:42
  • 1
    $\begingroup$ @iamnarra Ok. Edited the answer and added an example script, alongside a toy dataset in .csv. Hope this is a good starting point and helps you achieve what you want. $\endgroup$
    – Bogas
    Commented Sep 14, 2017 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.