I have one dataset which contain many categorical variables. The target variable is also categorical. I tried to find correlation between each categorical variable with target variable, using chi squared test and also got Cramer's V value. Here is a part of output-

column  Cramer's V      p value
Col1     0.065430   0.000000e+00
Col2     0.084450   0.000000e+00
Col3     0.059535   0.000000e+00
Col4     0.119343   0.000000e+00
Col5     0.108018   0.000000e+00
Col6     0.040086  1.842584e-218
Col7     0.021307   7.523901e-61
Col8     0.012404   3.865421e-20
Col9     0.009400   2.183289e-11
Col10    0.010728   6.082550e-15

I found p value and chi-sq statistic using python's function scipy.stats.chi2_contingency and then found Cramer's V by taking the square root of the chi-squared statistic divided by the sample size and the minimum dimension minus 1. (according to wikipedia page)

What I got is that all p values belonging to chi squared tests are nearly 0, which means there is very high correlation. But the Cramer's V value is also very less (<<1), which (again from wikipedia page) suggests that there is no strong association. So how to interpret these conflicting results? Is there a strong correlation or there is no correlation? Or, if my approach is wrong, please suggest the correct way.


Following is the contingency table for "col1" column and target column.

TARGET_COl               Grade 1  Grade 2  Grade 3  Grade 4  Grade 5
0                           290      392      932     1812     2854
1                           522      421      574      917     1247
2                         56789    81296   117971   147811   204480
3                          3719     2975     2811     1704     2244

And this is for "col2" column and target column.

TARGET_COl         Grade 1  Grade 2  Grade 3  Grade 4  Grade 5
0                    50867    73899   107101   135400   193526
1                    10453    11185    15187    16844    17299
  • $\begingroup$ Maybe share the contingency table for one of the tests so we can be sure you didn't make some kind of error? $\endgroup$ – Sal Mangiafico Jun 22 '18 at 20:08
  • $\begingroup$ @SalMangiafico I have added contingency tables for 2 columns. $\endgroup$ – Ankit Seth Jun 23 '18 at 5:39

Because your sample size is large, the Chi-square test is likely to return a low p-value even for a table with small differences from the expected proportions.

To get a sense of the effect size being reported by Cramer's v, it is helpful to look at the proportions in the table. For example, for column 1, you can see that there is not much difference in the proportions for each grade within the rows. Row 0 is about one-half to 1 percent of the observations in each grade. Row 2 is, say, 93 to 97 percent of observations in each grade. And so on.

Whether these kind of differences in proportions are meaningful in your context is up to you. The p-value and Cramer's v give you certain information. The practical importance of your results is something you will have decide.

The following is code for R.

I am getting a slightly different Cramer's v than you, so that's something that you might want to look into.

Input =("
Col1               Grade1  Grade2   Grade3   Grade4   Grade5
0                   290      392      932     1812     2854
1                   522      421      574      917     1247
2                 56789    81296   117971   147811   204480
3                  3719     2975     2811     1704     2244

Matrix = as.matrix(read.table(textConnection(Input),



   ### Pearson's Chi-squared test
   ### X-squared =  8113.9, df = 12, p-value < 2.2e-16



   ### Cramer's V        : 0.065 

prop.table(Matrix, margin=2)

   ###        Grade1      Grade2      Grade3      Grade4      Grade5
   ### 0 0.004729289 0.004607212 0.007621353 0.011901947 0.013537294
   ### 1 0.008512720 0.004948051 0.004693837 0.006023226 0.005914858
   ### 2 0.926108937 0.955479291 0.964698090 0.970882268 0.969903949
   ### 3 0.060649054 0.034965446 0.022986720 0.011192559 0.010643899
  • $\begingroup$ Yes your Cramer's V is right, I forgot to take square root at last step. I will update the values. And as you said that p value will be low for this data, so should I go with Cramer's V? I am creating a classification model and need to get those variables which are more correlated to output. $\endgroup$ – Ankit Seth Jun 23 '18 at 17:59
  • $\begingroup$ If you are looking for which variables are more strongly associated, then using Cramer's V makes sense. This approach may not be the best way to choose which variables to include your model, though. $\endgroup$ – Sal Mangiafico Jun 23 '18 at 20:27

As Sal said, seeing your data would be useful but generally speaking a low p-value says that there is an association and a small Cramer's V says that even though there is association the strength of association is not very high.

An example in R, as sample size goes up, p-value gets smaller but Cramer's V stays the same:


df <- as.matrix(data.frame(x = c(10,20,30,40), y = c(30,60,70,90), z = c(40,20,30,100)))

Output: X-squared = 34.301, df = 6, p-value = 0.000005884

Output: 0.1782144

Output: X-squared = 343.01, df = 6, p-value < 0.00000000000000022

Output:  0.1782144
  • $\begingroup$ I have added contingency tables for 2 columns. $\endgroup$ – Ankit Seth Jun 23 '18 at 5:39

Using association-metrics python package to calculate Cramér's coefficient matrix from a pandas.DataFrame object it's quite simple, let me show you:

first, install association_metrics using:

pip install association-metrics

Then, you can use the following pseudocode

# Import association_metrics  
import association_metrics as am
# Convert you str columns to Category columns
df = df.apply(
        lambda x: x.astype("category") if x.dtype == "O" else x)

# Initialize a CamresV object using you pandas.DataFrame
cramersv = am.CramersV(df) 
# will return a pairwise matrix filled with Cramer's V, where columns and index are 
# the categorical variables of the passed pandas.DataFrame

Package info


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