I'm trying to figure out how mean average rank (MAR) in graph modelling is calculated. Could someone look through my example and tell me if I'm correct?
Say we have two graphs that look like this:
1-2, 3-4
1, 2, 3-4
In graph 1, and there are two edges (between nodes 1 and 2 and between nodes 3 and 4).
In graph 2, there is only one edge (between nodes 3 and 4).
For graph 1:
To compute the average rank for graph 1 (for edge 1-2):
1) We compute the probability that node 1 would connect with nodes 2, 3 and 4, according to our model. Suppose this gives us [0.09, 0.90, 0.01].
2) The rank of node 2 (the ground truth connection) would be 2 here, because it is the connection with the second highest probability.
Now for edge 3-4 in graph 1:
1) We compute the probability that node 3 would connect with nodes 1, 2 and 4, according to our model. Suppose this gives us [0.21, 0.04, 0.75].
3) The rank of node 4 (the ground truth) is 1.
So the average rank for the first graph is (2+1)/2 = 1.5
For graph 2:
1) We replace node 4 with nodes 1 and 2.
2) We compute the probability that node 3 connects with nodes 1, 2, or 4. Say that gives us [0.05, 0.80, 0.15].
3) The ground truth is node 4, which had probability 0.15, which has rank 2 (is the second highest probability).
So the average rank for the second graph is 2/1 = 2.
The mean average rank (MAR) would be: (1.5 + 2)/2 = 1.75.
Is this correct?
Mean Average Rank is Mean of the Average ranks calculated. See Point 3. Results in https://academic.oup.com/bioinformatics/article/33/7/1031/2571354
IMHO your understanding is correct. You can also go through https://arxiv.org/pdf/1811.04441.pdf for Mean reciprocal rank (MRR).
To be sure you can mail the first author on the email id he has provided.
Related
I need to make quality score for cell in google which automatically give the results.
There are 3 columns
1. high quality
2. medium quality
3. low quality
High quality has 3 points and medium quality has 2 points and low quality has 1 point.
So if high quality column has 2 then it will get 3*2 = 6 and medium quality column has 3 then it will get 3*2 = 6. And low quality column has 2 then it will get 2*1 =2
So total quality score will be 6+6+2 = 14
So quality score column will be 14.
ie quality score = [column e*3 + column f*2 + column g*1]
You have already figure it. Simply use below formula to H2 cell.
=E2*3+F2*2+G2*1
You have many options:
you can use SUM:
H2: =sum(E2*3,F2*2,G2*1)
you can use SUMPRODUCT:
H2: =SUMPRODUCT({3,2,1},E2:G2)
you can simply sum them up:
H2: =E2*3+F2*2+G2*1
Reading your screenshot, it looks to me you probably have more rows further down.
Instead of writing a formula for every single row you can use just one
=ARRAYFORMULA(IF(LEN(D2:D),(E2:E3+F2:F2+G2:G),""))
Functions used:
ArrayFormula
IF
LEN
I have a spreadsheet that works properly in Excel. However, when I import it to Google Sheets it gives me the #DIV/)! error. I am at a loss for how to fix this.
I am trying to rank the items based on the number in column P. I would like for the highest number in column P to be ranked 1, then 2, 3, etc. If two numbers in column P are the same I would like for them to both be ranked the same. However, I don't want the formula to then skip the next number in the ranking order. Also, I am not sure if it matters, but column P displays a number but is technically filled with a formula to obtain that number. Example:
Points column is populated using the following formula:
=SUM(H2,J2,L2,N2,O2)
Points Rank
5 3
3 4
8 1
3 4
6 2
2 5
=SUMPRODUCT((P2 < P$2:P$36)/COUNTIF(P$2:P$36,P$2:P$36))+1
Any ideas?
Add the opposite of the numerator to the denominator to ensure you never receive #DIV/0!.
=SUMPRODUCT((P2 < P$2:P$36)/(COUNTIF(P$2:P$36,P$2:P$36)+(P2 >= P$2:P$36)))+1
When (P2 < P$2:P$36) is false, the numerator will be zero so it doesn't matter what the denominator is as long as it isn't zero.
Can you help me visualise an undirected graph?
I have about 500 strings that look like this:
;javascript;java;tapestry;d;design;jquery;css;html;com;air;testing;events;crm;soa;documentation;.a;email;iso;dynamic;mobile;this;project;resolution;s;automation;web;like;e-commerce;profile;commerce;out;jobs;inventory;operators;environment;system;include;integration;relationship;field;implementation;key;.profile;planning;knockout.js;sun;packaging;collaboration;report;public;virtual;communication;send;state;member;execution;solution;provider;members;continuous;writing;e;cuba;required;transactional;subject;manual;capacity;portfolio;.so;leader;take
;c;python;java;.a;basic;equivalent;cad;requirements;catia;.x;nx;self;communication;selected;base;summary
;javascript;c;python;java;rest;android;security;linux;sql;git;design;perl;css;html;svn;yaml;architecture;ios;json;api;ubuntu;pyramid;deployment;bash;documentation;configuration;frameworks;module;object;.a;multitasking;centos;hosting;project;fluent;administrator;monitoring;control;specifications;web;version;platform;admin;components;out;minimum;environment;system;include;using;key;falcon;communication;migrate;deadlines;ansible;back;cycle;production;red;analysis;administration;graphic;maintenance;autonomy;french;required;environments;hat;lead;arch;take
and what I would like to do with them is calculate and visualise the edges between the shared elements of the strings. Like if in the first two strings we find javascript and python, then the edge would be thicker between them for every match occurence in all strings in the final graph.
What I've done so far is to parse out the strings and separate each one in a 1/0 matrix, with the string names as column names (in a csv file) but that did not seem to work because I don't know if the labels could be seen in Gelphi as column names.
javascript java tapestry
---------------------------------------
Row 1 1 0 1
Row 2 0 1 0
Row 3 1 1 1
So I transposed the matrix to get the strings all in a column, but those columns enumerated by number don't mean much to me.
name Col1 Col2 Col3 Col4
------- ------------------------
javascript 1 0 0 0 1
java 0 1 1 0 0
tapestry 1 0 1 0 1
I am thinking a matrix multiplied by its transverse might help, although Im not sure how the math works for the result interpretation.
what I would like to do with them is calculate and visualise the edges between the shared elements of the strings.
Gephi just visualizes what is created manually, or selected for import.
Can you help me visualise an undirected graph?
Graph as .csv
Turning associations into a static graph (as Gephi-compatible .csv file):
Nodes
List unique names, save to .csv like:
id,label
0,"node #1"
1,"node #2"
2,"node #3"
Optionally, add additional columns as required.
Edges
Enumerate associations, add to weight for each occurrence, save to .csv, like:
id,source,target,label,weight
0,0,1,"edge #1",1
1,0,2,"edge #2",3
Alternatively, create new edge per association-occurrence and merge them after import.
Do not create edges for 0 -weight values (no connection).
source and target -column reference node id.
label -column is optional.
May contain type -column (allows for Directed or Undirected -value).
Optionally, convert to weight (0.0 - 1.0 as opposed to amount) by recalculating weight as:
weight = weight / highest_weight
I need some help defining a custom similarity measure.
I have a dataset whose elements are defined by 4 attributes.
As an example, consider the following two items:
Element 1:
A1: "R1", "R3", "R4", "R7"
A2: "H1"
A3 "F1", "F2"
A4 "aaa" "bbb"
Element 2:
A1: "R1", "R2"
A2: "H1"
A3 "F1", "F2"
A4 "aaa" "bbb" "ccc" "ddd" "eee" "fff"
I have to implement a similarity measure which should satisfies the following conditions:
1 - If A2 value is the same, the two elements must belong to the same cluster
2 - If two elements have at least one common value on A4, the who elements must belong to the same cluster.
I need to use a sort of weighted Jaccard measure. Is it mathematically correct to define a similarity measure that sums the jaccard distance of each attribute and then to add a sort of high weigth if condition 1 and 2 are satisfied for A2 and A4?
If so, how can I transform the similarity matrix into a distance matrix?
(1) Distance = 1 - similarity. This is a common characteristic.
(2) Summing the distances of the attributes is valid, although you may wish to scale it back to the [0, 1] range.
(3) Putting a high weight is not correct for what you've described. If the A2 or A4 values show a match, simply set the distance to 0. The clustering is a requirement, not merely strong advice. Is there some other semantic to your distance function, that you didn't want to take this route?
FYI, the basics for being a topological metric's distance function, D are:
D(a, a) = 0
D(a,b) = D(b,a)
D(a,b) + D(b,c) >= D(a,c)
Here is my word vector :
google
test
stackoverflow
yahoo
I have assigned a value for these words as follows :
google : 1
test : 2
stackoverflow : 3
yahoo : 4
Here are some sample users and their words :
user1 google, test , stackoverflow
user2 test , google
user3 test , yahoo
user4 stackoverflow , yahoo
user5 stackoverflow , google
user6
To cater for users which do not have value contained in the
word vector I assign '0'
Based on this, this corresponds to :
user1 1, 2 , 3
user2 2 , 1 , 0
user3 2 , 4 , 0
user4 3 , 4 , 0
user5 3 , 1, 0
user6 0 , 0 , 0
I am unsure if these are the correct values or even is correct approach for applying values to each word vector value so can apply 'Eucludeian distance' and 'correlation'. I'm basing this on snippet from book 'Programming Collective Intelligence' :
"Collecting Preferences The first thing you need is a way to represent
different people and their preferences. If you were building a
shopping site, you might use a value of 1 to indicate that someone had
bought an item in the past and a value of 0 to indicate that they had
not. "
For my dataset I do not have preference values so I am just using a unique numerical value to represent if a user contains a word in word vector or not.
Are these the correct values to set for my word vector ? How should I determine what these values should be ?
To make distance and similarity metrics work out, you need one column per word in the vocabulary, then fill those columns with booleans zero and one as the corresponding words occur in samples. E.g.
G T SO Y!
google, test, stackoverflow => 1, 1, 1, 0
test, google => 1, 1, 0, 0
stackoverflow, yahoo => 0, 0, 1, 1
etc.
The squared Euclidean distance between the first two vectors is now
(1 - 1)² + (1 - 1)² + (1 - 0)² + (0 - 0)² = 1
which makes intuitive sense as the vectors differ in exactly one position. Similarly, the squared distance between the final two vectors is four, which is the maximal squared distance in this space.
This encoding is an extension of the "one-hot" or "one-of-K" coding, and it's a staple of machine learning on text (although few textbooks care to spell it out).