# Binary distance measure

I want a metric - not an answer - for this.

If I have a main binary sequence - 00100 I want a measure to tell me how far away another binary sequences 1's are relating to the sequence of interest.

So comparing my main sequence with these:

A - 01000

B - 10000

I want to say that B is X% further from the main sequence compared to Y% for sequence A. By further I mean the distance of the positive entries of one sequence to the positive entries of another - their index in the sequence.

Any extant metrics for this type of distance-specific binary sequence analysis.

I dont want a measure to say how similar the sequences are, I want to say how much further one is from another.

I am not doing homeworks or jobs or anything - I am 30 just tinkering with protein sequence analysis and I want a metric that does this, that is all.

U.

• Are you looking for Hamming distance, perhaps? May 8 '18 at 9:52
• @JanKukacka - No, Hamming Distance gives me a integer value indicating the number of differences - so if there is a sequence 10000 and 01000 comparing to 00100, then both of these sequences have the same hamming distance, but the first sequence is "further" from the one of interest. May 8 '18 at 10:37
• @urema I'm saying that if you have a similarity measure, then the negative of it is a distance measure.
– Tim
May 8 '18 at 11:00
• @Tim - So from an integer similarity measure, taking the negative is the difference? You have to explain this my friend. SizeSimilarity measure gives me a value of 2 - similarity is 2.... so how is -2 a difference May 8 '18 at 11:13
• @urema things that are more distant are less similar to each other, so if you are able to measure distance, then similarity is opposite of it. That's all I'm saying.
– Tim
May 8 '18 at 11:20

A suggestion.

Why not say that each of your boolean is positioned onto an "indexline", i.e. $01234$. Seeing your reference ($R$), $A$ and $B$ binary sequences as row vectors, you could encode (->) them as

$R$ : $(0,0,1,0,0)$ -> $(0,0,2,0,0)$

$A$ : $(0,1,0,0,0)$ -> $(0,1,0,0,0)$

$B$ : $(1,0,0,0,0)$ -> $(0,0,0,0,0)$

and then compute

$\sum_i^5 \sum_j^5 |R_i-A^T_j| =13$

$\sum_i^5 \sum_j^5 |R_i-B^T_j| =10$

with the $0\%$ "further-value" being

$\sum_i^5 \sum_j^5 |R_i-R^T_j| =16$

Finally, you could, e.g, say things such as

$R$ is (undefined since) $\left(-1+\frac{16-16}{16-16}\right)$ but can be seen as $0\%$ further than itself.

or

$B$ is $\left(-1+\frac{16-10}{16-13}\right)=100\%$ further from $R$ than $A$ is...

You can also get rid of the absolute-value operator if you want to keep informations related to distance "direction".

• Kanak - everyone says to avoid saying thanks but I am not a robot - thank you, this was a great breakdown and gives me an insight into using encoding for my purposes. May 13 '18 at 10:24