I would like to know how to use Euclidean distance to find similarity between two multivariate time series.

Suppose, I have two $N$-variate time series $u$ and $v$, with $u_i(t)$ denoting the $i$-th component of time series $u$ at time $t$.

Would it make sense to calculate distance between values of same parameters and then sum that distances for all parameters to get the final distance?

$$ d_i = \sqrt{(u_i(0)-v_i(0))^2 + (u_i(1)-v_i(1))^2 + … + (u_i(T)-v_i(T))^2}$$ $$ d = d_0 + d_1 + … + d_N$$

Or would it make sense to treat all the values equally, no matter which parameter is considered?

$$\check{d} = \sqrt{ \hphantom{+\,}(u_1(0)-v_1(0))^2 + (u_1(1)-v_1(1))^2 + … + (u_1(T)-v_1(T))^2 \\ + (u_2(0)-v_2(0))^2 + (u_2(1)-v_2(1))^2 + … + (u_2(T)-v_2(T))^2 \\ + … \\ + (u_N(0)-v_N(0))^2 + (u_N(1)-v_N(1))^2 + … + (u_N(T)-v_N(T))^2 }$$

Or to consider distance between each record (values of each parameter at certain point in time):

$$ \hat{d}_0 = \sqrt{(u_1(0)-v_1(0))^2 + (u_2(0)-v_2(0))^2 + … + (u_m(0)-v_m(0))^2}$$

$$ \hat{d}_1 = \sqrt{(u_1(1)-v_1(1))^2 + (u_2(1)-v_2(1))^2 + … + (u_m(1)-v_m(1))^2}$$

$$ \hat{d}_T = \sqrt{(u_1(T)-v_1(T))^2 + (u_2(T)-v_2(T))^2 + … + (u_m(T)-v_m(T))^2}$$

$$ \hat{d} = \hat{d}_0 + \hat{d}_1 + … + \hat{d}_T$$

Or something else?


Multivariate series 1:

power, current, voltage

100, 10, 10

400, 20, 20

900, 30, 30

Multivariate series 2

power, current, voltage

600, 20, 30

1000, 50, 20

450, 15, 30

What would the Euclidean distance look like?


2 Answers 2


When focussing on it as a comparitve measure, the difference between the three ways boils down to whether you sum the squares of the component-wise or record-wise distances (second method) or not (first and third method):

$$ \sqrt{d_0^2 + d_1^2 + … + d_N^2}\\ = \sqrt{\hat{d}_0^2 + \hat{d}_1^2 + … + \hat{d}_T^2}\\ = \check{d} \\ = \sqrt{ \hphantom{+\,}(u_1(0)-v_1(0))^2 + (u_1(1)-v_1(1))^2 + … + (u_1(T)-v_1(T))^2 \\ + (u_2(0)-v_2(0))^2 + (u_2(1)-v_2(1))^2 + … + (u_2(T)-v_2(T))^2 \\ + … \\ + (u_N(0)-v_N(0))^2 + (u_N(1)-v_N(1))^2 + … + (u_N(T)-v_N(T))^2 }$$

Thus, the main effect of the first approach is that the effect of large component-wise distances is attenuated, while with the third approach, the effect of large record-wise distances is attenuated (both in comparison to the second approach). The reason for this is that squaring before adding increases the relative impact of large summands. Whether this is desired or unwanted, depends on your application.

Either way, what you almost certainly want to do, is to normalise your component time series, as otherwise time series with comparably high values will dominate your measure. For instance, in your example, differences in power would dominate differences in the other quantities. As an illustration why this is bad: Your results would depend on your choice of units.

If the measure you use for your components are linearly scaling with what you are actually interested in (e.g., there is no general offset to one component), a reasonable normalisation would be to make all your component time series to have the same mean. Instead of using $u_i$, you use $\tilde{u}_i$ where

$$\tilde{u}_i(t) := \frac{u_i(t)}{μ_i}$$

and $μ_i$ is the mean over $u_i$ and $v_i$ (and possibly other comparable time series at your disposal).

Note that the standard procedure of subtracting the mean or dividing by the standard deviation may have undesired effects. For example, if a component is constant save for measurement errors, this procedure would blow up the error and let the differences of the measurement error play a considerable part in your analysis.

At the end of the day, what is a reasonable normalisation depends on your application and what you want to extract with your analysis.

Note that there are other ways to determine the similarity of time series that may be better suited to your application. For example, the cross-correlation would be a reasonable approach if you are not interested in differences arising due to linear transformations of an entire time series, i.e., you are only interested in a similar (in the geometric sense) temporal evolution. Again, what is best for you depends on your application.

And finally note that it does not make much sense to use a component that is derived from other components (I mention this because in your example, power always is the product of current and voltage).

  • $\begingroup$ Thank you very much for answer and suggestions. "Thus, the main effect of your second approach is that larger component-wise distances play stronger into the final result." - you mean by the first approach? Do you mean to find average of all values for a parameter (for all time series)? Why divide by mean, as I'm aware of it is practice to divide by max value, or to subtract mean and divide by deviation? Do you have a better similarity measure in mind? $\endgroup$ Dec 5, 2015 at 15:48
  • $\begingroup$ Please check my edited answer, I have added another possibility. I would like not only to check if values of two time series are different, but also to capture correlation of parameters in best way. To see not only if the second series has different values, but also if parameters are correlated. The third way might be the best one because it observes all parameters at the same time. What do you think? $\endgroup$ Dec 5, 2015 at 15:51
  • $\begingroup$ @Marko: I addressed most of your comments in an edit to my answer. — To see not only if the second series has different values, but also if parameters are correlated. – You should do a separate analysis for this, first. — The third way might be the best one because it observes all parameters at the same time. – I fail to see how. Also, you did not say, how you want correlations to influence your measure. $\endgroup$
    – Wrzlprmft
    Dec 5, 2015 at 21:03
  • $\begingroup$ @Marko: If this answer solves your problem, you probably also want to upvote it. If you do not want to, it would be nice, if you could elaborate why. $\endgroup$
    – Wrzlprmft
    Dec 8, 2015 at 10:47

You should treat values of same parameters as vectors. E. g. $values_1 = (3, 4, 5)$ and $values_2 = (8, 9, 2)$. The distance between dots that are represented by that vectors is the following: $$distance(values_1, values_2) = \sqrt{(3-8)^2 + (4-9)^2 + (5-2)^2} $$ And general formula: $$distance(u, v) = \sqrt{\sum_{i=0}^{dimensionsCount} (u_i - v_i)^2}$$ That works for any number of dimensions.

As for timeseries, you could measure average distance between corresponding points, squaring is not required as long as you do not intend to penalize one big difference more than several smaller ones.

  • $\begingroup$ Thank you for your answer, but this is only for one point with three parameters, right? But what about series of those? Please check my edited answer. $\endgroup$ Dec 4, 2015 at 10:25
  • $\begingroup$ @Marko, I updated my answer $\endgroup$
    – Yurii
    Dec 4, 2015 at 11:07
  • $\begingroup$ Could you please show what you mean by that on the example I provided? Is this what you suggested a practice (well defined as Euclidean distance), or is it just a way to handle the concrete problem? $\endgroup$ Dec 4, 2015 at 11:18
  • $\begingroup$ @Marko I am not aware of such practice, so it is just my suggestion for this problem $\endgroup$
    – Yurii
    Dec 4, 2015 at 11:21

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