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
    Commented Jun 18, 2014 at 8:14
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    $\begingroup$ ... and = Phi (standardized Chi-square) which brings us from scale to contingency table. $\endgroup$
    – ttnphns
    Commented Jun 18, 2014 at 8:20
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    $\begingroup$ 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 M.
    Commented Dec 9, 2020 at 17:04
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    $\begingroup$ 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
    Commented Feb 22, 2021 at 9:32

7 Answers 7


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? $\endgroup$ Commented Jun 23, 2014 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
    Commented Jun 23, 2014 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.
    Commented Oct 1, 2019 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
    Commented Oct 2, 2019 at 7:40

There are specialised similarity metrics for binary vectors, such as:

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


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
    Commented Jul 27, 2015 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
    Commented Jul 28, 2015 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." $\endgroup$ Commented Dec 23, 2019 at 2:31
  • $\begingroup$ These are NOT correlation metrics! Indeed notice the example below: independent draws leading to high scores (while pearson correctly shows 0 correlation!). Almost everyone on this thread is confusing marginal distributions with correlations. $\endgroup$
    – semola
    Commented Mar 9, 2023 at 15:32
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    $\begingroup$ Also see en.wikipedia.org/wiki/Phi_coefficient $\endgroup$
    – semola
    Commented Mar 9, 2023 at 15:35

I would not advise to use Pearson's correlation coefficient for binary data, see the following counter-example:

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)
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:

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
    Commented Apr 9, 2020 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
    Commented Jun 16, 2020 at 12:10
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    $\begingroup$ @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. $\endgroup$ Commented Feb 20, 2021 at 9:24
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    $\begingroup$ 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
    Commented Feb 20, 2021 at 9:51
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    $\begingroup$ I consider this answer simply wrong. The two variables were generated through completely independent mechanisms, so it's very intuitive that the correlation is ~ 0. Analogously, if one used a similar approach to generate two independent Gaussian variables, and set the mean of both to be 40 with sd of 1, just because the values are 'similar' doesn't mean you would expect a correlation. $\endgroup$
    – Charlie
    Commented Apr 12, 2021 at 19:32

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()

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

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
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 
  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

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

  dout <- rbind(dout,
                           pctx = mean(x), # percent of x that is TRUE, used below
                           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
    Commented Jun 16, 2020 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. $\endgroup$ Commented Jun 17, 2020 at 15:00

Simple answer (building on @Carlos M's comment): Yes, in particular, computing Pearson's correlation is meaningful, in fact when you run Pearson's on binary data it gives you equivalent of the Phi coefficient ( https://en.wikipedia.org/wiki/Phi_coefficient). The awesome thing here is that R, Pandas, and SQL engines already have this as their default correlation. You can either split out the binary fields separately or just pile them in and compute it all at once :)

import pandas as pd

# Sample DataFrame with continuous and binary data
df = pd.DataFrame({
        'cont1': [1.2, 3.4, 2.2, 4.5, 5.2],
        'cont2': [2.4, 3.1, 4.5, 6.1, 5.3],
        'binary1': [0, 1, 0, 1, 1],
        'binary2': [1, 0, 1, 0, 0],
        'binary3': [0, 1, 0, 0, 1]

# Compute correlation for all columns

            cont1     cont2   binary1   binary2   binary3
cont1    1.000000  0.772457  0.893869 -0.893869  0.558668
cont2    0.772457  1.000000  0.496162 -0.496162 -0.047823
binary1  0.893869  0.496162  1.000000 -1.000000  0.666667
binary2 -0.893869 -0.496162 -1.000000  1.000000 -0.666667
binary3  0.558668 -0.047823  0.666667 -0.666667  1.000000

I'd like to add that if the goal of this correlation analysis is feature selection in a machine learning context, I'd suggest looking at mutual information.

From scikit-learn's mutual_info_classif link:

Mutual information (MI) between two random variables is a non-negative value, which measures the dependency between the variables. It is equal to zero if and only if two random variables are independent, and higher values mean higher dependency.

For two random variables, you should use mutual_info_score link


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