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There are two Boolean vectors, which contain 0 and 1 only. If I calculate the Pearson or Spearman correlation, are they meaningful or reasonable?

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    $\begingroup$ If both variables are dichotomous, Pearson = Spearman = Kendall's tau. Yes it may have sence. With truly binary (boolean) data it also make sence to compute "Pearson" on data without centering, that would be cosine. $\endgroup$ – ttnphns Jun 18 '14 at 8:14
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    $\begingroup$ ... and = Phi (standardized Chi-square) which brings us from scale to contingency table. $\endgroup$ – ttnphns Jun 18 '14 at 8:20
  • $\begingroup$ Just to clarify on what Nick Cox detailed above: People here haven't noted the Phi Coefficient. The Pearson's correlation for binary variables (which can be returned directly in a programming language like r with cor()) is defined for non-uniform vectors and has a name. It also has equivalent interpretability for the Pearson's correlation: en.wikipedia.org/wiki/Phi_coefficient $\endgroup$ – Carlos R. Mercado Dec 9 '20 at 17:04
  • $\begingroup$ A comment on the comment above. The word "above" is a hangover from Carlos' comment being first posted as an answer. For "above" read "below". $\endgroup$ – Nick Cox Feb 22 at 9:32
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The Pearson and Spearman correlation are defined as long as you have some $0$s and some $1$s for both of two binary variables, say $y$ and $x$. It is easy to get a good qualitative idea of what they mean by thinking of a scatter plot of the two variables. Clearly, there are only four possibilities $(0,0), (0,1), (1, 0), (1,1)$ (so that jittering to shake identical points apart for visualization is a good idea). For example, in any situation where the two vectors are identical, subject to having some 0s and some 1s in each, then by definition $y = x$ and the correlation is necessarily $1$. Similarly, it is possible that $y = 1 -x$ and then the correlation is $-1$.

For this set-up, there is no scope for monotonic relations that are not linear. When taking ranks of $0$s and $1$s under the usual midrank convention the ranks are just a linear transformation of the original $0$s and $1$s and the Spearman correlation is necessarily identical to the Pearson correlation. Hence there is no reason to consider Spearman correlation separately here, or indeed at all.

Correlations arise naturally for some problems involving $0$s and $1$s, e.g. in the study of binary processes in time or space. On the whole, however, there will be better ways of thinking about such data, depending largely on the main motive for such a study. For example, the fact that correlations make much sense does not mean that linear regression is a good way to model a binary response. If one of the binary variables is a response, then most statistical people would start by considering a logit model.

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    $\begingroup$ Does that mean in this situation, Pearson or Spearman correlation coefficient is not a good similarity metric for this two binary vectors? $\endgroup$ – Zhilong Jia Jun 23 '14 at 11:33
  • $\begingroup$ Yes in the sense that it doesn't measure similarity and is undefined for all 0s or all 1s for either vector. $\endgroup$ – Nick Cox Jun 23 '14 at 12:26
  • $\begingroup$ The 2 identical or 'opposite' vectors case is not clear to me. If x=c(1,1,1,1,1) and y=(0,0,0,0,0) then y=1-x and it sounds like you're saying this must be the case by definition, implying correlation of -1. Equally y=x-1 implying correlation of +1. There is only 1 point (5 replicates) on a scatterplot so any straight line could be drawn through it. It feels like the correlation is undefined in this instance. Sorry if I misunderstood what you meant. @NickCox $\endgroup$ – PM. Oct 1 '19 at 16:51
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    $\begingroup$ No; I am not saying that, as I do point out in my first sentence that you must have a mix of 0s and 1s for the correlation to be defined. Otherwise if the SD of either variable is 0 then the correlation is undefined. But I have edited my answer to mention that twice. $\endgroup$ – Nick Cox Oct 2 '19 at 7:40
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There are specialised similarity metrics for binary vectors, such as:

  • Jaccard-Needham
  • Dice
  • Yule
  • Russell-Rao
  • Sokal-Michener
  • Rogers-Tanimoto
  • Kulzinsky

etc.

For details, see here.

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    $\begingroup$ Surely there are many more reliable and comprehensive references. Even on the level of getting authors' names right, note Kulczyński and Tanimoto. See e.g. Hubálek, Z. 1982. Coefficients of association and similarity, based on binary (presence-absence) data: An evaluation. Biological Reviews 57: 669–689. $\endgroup$ – Nick Cox Jul 27 '15 at 15:43
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    $\begingroup$ They have obviously misspelled 'Tanimoto' but 'Kulzinsky' has been purposely simplified. Your reference is more credible without a doubt but it's not accessible to everybody. $\endgroup$ – Digio Jul 28 '15 at 8:38
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    $\begingroup$ scipy contains many of these, see this page, under "Distance functions between two boolean vectors (representing sets) u and v." $\endgroup$ – Alex Moore-Niemi Dec 23 '19 at 2:31
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I would not advise to use Pearson's correlation coefficient for binary data, see the following counter-example:

set.seed(10) 
a = rbinom(n=100, size=1, prob=0.9) 
b = rbinom(n=100, size=1, prob=0.9)

in most cases both give a 1

table(a,b)

> table(a,b)
   b
a    0  1
  0  0  3
  1  9 88

but the correlation does not show this

cor(a, b, method="pearson")

> cor(a, b, method="pearson")
[1] -0.05530639

A binary similarity measure such as Jaccard index shows however a much higher association:

install.packages("clusteval")
library('clusteval')
cluster_similarity(a,b, similarity="jaccard", method="independence")

> cluster_similarity(a,b, similarity="jaccard", method="independence")
[1] 0.7854966

Why is this? See here the simple bivariate regression

plot(jitter(a, factor = .25), jitter(b, factor = .25), xlab="a", ylab="b", pch=15, col="blue", ylim=c(-0.05,1.05), xlim=c(-0.05,1.05))
abline(lm(a~b), lwd=2, col="blue")
text(.5,.9,expression(paste(rho, " = -0.055")))

plot below (small noise added to make the number of points clearer) Bivariate regression line

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    $\begingroup$ This should be upvoted more. Pearson for binary data can be misleading. $\endgroup$ – abudis Apr 9 '20 at 6:26
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    $\begingroup$ The Pearson correlation is doing its job here and the scatter plot makes plain why the correlation is weak. Whether other measures or analyses make more sense for other goals are important but different questions. . $\endgroup$ – Nick Cox Jun 16 '20 at 12:10
  • $\begingroup$ @NickCox, I disagree. The question is whether it is meaningful or reasonable to use the Pearson's correlation coefficient (not wether it can be applied on this data on general). I would advise against it because there are more reasonable measures for this kind of data. You can also use OLS for binary data (and often it makes sense) but it is not always reasonable since you can get fitted values outside of [0, 1] among others. $\endgroup$ – Arne Jonas Warnke Feb 20 at 9:24
  • $\begingroup$ Once again: Correlation here has a meaning so long as both variables are genuinely variable; the same reservation applies in general, as correlations are not defined if even one variable is constant. Despite the root of reason what is said to be reasonable or not is usually a matter of rhetoric. not logic. Many economists and some others have an odd habit of using OLS (an estimation method) as a short-hand for the simplest kind of linear regression. The objection that fits aren't confined to $[0, 1]$ is one I have made many times myself in discussion, and there are other objections too. $\endgroup$ – Nick Cox Feb 20 at 9:51
  • $\begingroup$ So we disagree here on the meaning of reasonable, that is fine I guess. And btw I agree that there are other objections too which I tried to imply by adding "among others" $\endgroup$ – Arne Jonas Warnke Feb 20 at 9:57
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Arne's response above isn't quite right. Correlation is a measure of dependence between variables. The samples A and B are both independent draws, although they are from the same distribution, so we should expect ~0 correlation.

Running a similar simulation and creating a new variable c that is dependent on the value of a:

from scipy import stats
a = stats.bernoulli(p=.9).rvs(10000)
b = stats.bernoulli(p=.9).rvs(10000)

dep = .9
c = []
for i in a:
    if i ==0:
        # note this would be quicker with an np.random.choice()
        c.append(stats.bernoulli(p=1-dep).rvs(1)[0])
    else:
        c.append(stats.bernoulli(p=dep).rvs(1)[0])

We can see that the

stas.pearsonr(a,b) ~= 0
stas.pearsonr(a,c) ~= 0.6
stats.spearmanr(a,c) ~=0.6
stats.kendalltau(a,c) ~=0.6
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A possible issue with using the Pearson correlation for two dichotomous variables is that the correlation may be sensitive to the "levels" of the variables, i.e. the rates at which the variables are 1. Specifically, suppose that you think the two dichotomous variables (X,Y) are generated by underlying latent continuous variables (X*,Y*). Then it is possible to construct a sequence of examples where the underlying variables (X*,Y*) have the same Pearson correlation in each case, but the Pearson correlation between (X,Y) changes. The example below in R shows such a sequence. The example shifts the continuous latent (X*,Y*) distribution to the right along the x-axis (not changing the shape of the latent distribution at all), and finds that the Pearson correlation between (X,Y) decreases as we do so.

For this reason, you might consider using the tetrachoric correlation for dichotomous data, if it is feasible to estimate. This question has more details on the polychoric correlation, which is a generalization of the tetrachoric.


# consider two dichotomous variables x and y that are each generated by an 
# underlying common standard normal normal factor and a unique standard normal 
# normal factor, plus x has a shift u that makes it more common than 50:50
set.seed(12345)
library(polycor)
N <- 10000
U <- seq(0,1.2,0.1)
dout <- list()
for(u in U) {
  print(u) # u is the shift
  common <- rnorm(N) # common factor
  xunderlying <- common*0.7 + rnorm(N)*0.3 + u
  yunderlying <- common*0.7 + rnorm(N)*0.3 
  plot(xunderlying,yunderlying)
  abline(v = mean(xunderlying),col='red')
  abline(h = mean(yunderlying),col='red')
  x <- xunderlying > 0 # would be 50:50 chance if u = 0
  y <- yunderlying > 0
  print(table(x,y))

  # obtain tetrachoric correlation using polycor package
  p <- polycor::polychor(x,y,ML=TRUE,std.err = TRUE)

  dout <- rbind(dout,
                data.frame(U=u,
                           pctx = mean(x), # percent of x that is TRUE, used below
                           pcty=mean(y),
                           cor=cor(x,y), # pearson correlation, used below
                           polychor_rho=p$rho, # tetrachoric correlation, used below
                       underlying_cor = cor(xunderlying,yunderlying), # underlying correlation, used below
                       polychor_xthresh = p$row.cuts,
                           polychor_ythresh = p$col.cuts))

}

# plot underlying cor as a function of pctx. 
# does not depend on pctx
plot(dout$pctx,dout$underlying_cor,ylim = c(0,1))

# plot pearson correlation as a function of pctx (which is determined by u). 
# decreasing in pctx!
plot(dout$pctx,dout$cor,ylim = c(0,1))

# plot estimated tetrachoric correlation as a function of pctx. 
# does not depend on pctx
plot(dout$pctx,dout$polychor_rho,ylim = c(0,1))

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    $\begingroup$ This is a red herring. Sure, if the (0, 1) values are based on degrading other scales then the Pearson correlation between indicator variables isn't capturing information it was never shown. That is not a fault of the method, The implication is to use other methods. $\endgroup$ – Nick Cox Jun 16 '20 at 12:15
  • $\begingroup$ Good point. Economists (at least) often view indicator variables as degraded versions of continuous variables, but there may be cases where viewing them in another way motivates pearson rather than tetrachoric. $\endgroup$ – Richard DiSalvo Jun 17 '20 at 15:00

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