crosstab

Cross-tabulation

Syntax

tbl = crosstab(x1,x2)

tbl = crosstab(x1,...,xn)

[tbl,chi2,p] = crosstab(___)

[tbl,chi2,p,labels] = crosstab(___)

Description

tbl = crosstab(x1,x2) returns a cross-tabulation, tbl, of two vectors of the same length, x1 and x2.

tbl = crosstab(x1,...,xn) returns a multi-dimensional cross-tabulation, tbl, of data for multiple input vectors, x1, x2, ..., xn.

[tbl,chi2,p] = crosstab(___) also returns the chi-square statistic, chi2, and its p-value, p, for a test that tbl is independent in each dimension. You can use any of the previous syntaxes.

[tbl,chi2,p,labels] = crosstab(___) also returns a cell array, labels, which contains one column of labels for each input argument, x1 ... xn.

Examples

Cross-Tabulate Two Data Vectors

Create two sample data vectors, containing three and four distinct values, respectively.

x = [1 1 2 3 1];
y = [1 2 5 3 1];

Cross-tabulate x and y.

table = crosstab(x,y)

table = 3×4

     2     1     0     0
     0     0     0     1
     0     0     1     0

The rows in table correspond to the three distinct values in x, and the columns correspond to the four distinct values in y.

Cross-Tabulate Independent Data Vectors

Generate two independent vectors, x1 and x2, each containing 50 discrete uniform random numbers in the range 1:3.

rng default;  % for reproducibility
x1 = unidrnd(3,50,1);
x2 = unidrnd(3,50,1);

Cross-tabulate x1 and x2.

[table,chi2,p] = crosstab(x1,x2)

table = 3×3

     1     6     7
     5     5     2
    11     7     6

chi2 = 7.5449

p = 0.1097

The returned p value of 0.1097 indicates that, at the 5% significance level, crosstab fails to reject the null hypothesis that table is independent in each dimension.

Cross-Tabulate Grouped Data

Load the sample data, which contains measurements of large model cars during the years 1970-1982.

load carbig

Cross-tabulate the data of four-cylinder cars (cyl4) based on model year (when) and country of origin (org).

[table,chi2,p,labels] = crosstab(cyl4,when,org);

Use labels to determine the index location in table for the number of four-cylinder cars made in the USA during the late period of the data.

labels

labels=3×3 cell array
    {'Other'   }    {'Early'}    {'USA'   }
    {'Four'    }    {'Mid'  }    {'Europe'}
    {0x0 double}    {'Late' }    {'Japan' }

The first column of labels corresponds to the data in cyl4, and indicates that row 2 of table contains data on cars with four cylinders. The second column of labels corresponds to the data in when, and indicates that column 3 of table contains data on cars made during the late period. The third column of labels corresponds to the data in org, and indicates that location 1 of the third dimension of table contains data on cars made in the USA.

Therefore, table(2,3,1) contains the number of four-cylinder cars made in the USA during the late period.

table(2,3,1)

ans = 38

The data contains 38 four-cylinder cars made in the USA during the late period.

Generate and Visualize Contingency Table

Create a contingency table from data, and visualize the table in a heatmap chart.

Load the hospital data.

load hospital

The hospital dataset array contains data on 100 hospital patients, including last name, gender, age, weight, smoking status, and systolic and diastolic blood pressure measurements.

Convert the dataset array to a MATLAB® table.

Tbl = dataset2table(hospital);

Determine whether smoking status is independent of gender by creating a 2-by-2 contingency table of smokers and nonsmokers, grouped by gender.

[conttbl,chi2,p,labels] = crosstab(Tbl.Sex,Tbl.Smoker)

conttbl = 2×2

    40    13
    26    21

chi2 = 4.5083

p = 0.0337

labels = 2x2 cell
    {'Female'}    {'0'}
    {'Male'  }    {'1'}

The rows of the resulting contingency table conttbl correspond to patient gender, with row 1 containing data for females and row 2 containing data for males. The columns correspond to patient smoking status, with column 1 containing data for nonsmokers and column 2 containing data for smokers. The returned result chi2 = 4.5083 is the value of the chi-squared test statistic for a Pearson's chi-squared test of independence. The $\mathit{p}$-value for the test p = 0.0337 suggests, at a 5% level of significance, rejection of the null hypothesis that gender and smoking status are independent.

Visualize the contingency table in a heatmap. Plot smoking status on the $\mathit{x}$-axis and gender on the $\mathit{y}$-axis.

heatmap(Tbl,'Smoker','Sex')

Input Arguments

x1 — Input vector

vector of grouping variables

Input vector, specified as a vector of grouping variables. All input vectors, including x1, x2, ..., xn, must be the same length.

Data Types: single | double | char | string | logical | categorical

x2 — Input vector

vector of grouping variables

Input vector, specified as a vector of grouping variables. All input vectors, including x1, x2, ..., xn, must be the same length.

Data Types: single | double | char | string | logical | categorical

x1,...,xn — Input vectors

vectors of grouping variables

Input vectors, specified as vectors of grouping variables. If you use this syntax to specify more than two input vectors, then crosstab generates a multi-dimensional cross-tabulation table. All input vectors, including x1, x2, ..., xn, must be the same length.

Data Types: single | double | char | string | logical | categorical

Output Arguments

tbl — Cross-tabulation table

matrix of integer values

Cross-tabulation table, returned as a matrix of integer values.

If you specify two input vectors, x1 and x2, then tbl is an m-by-n matrix, where m is the number of distinct values in x1 and n is the number of distinct values in x2.

If you specify three or more input vectors, then tbl(i,j,...,n) is a count of indices where grp2idx(x1) is i, grp2idx(x2) is j, grp2idx(x3) is k, and so on.

chi2 — Chi-square statistic

positive scalar value

Chi-square statistic, returned as a positive scalar value. The null hypothesis is that the proportion in any entry of tbl is the product of the proportions in each dimension.

p — p-Value

scalar value in the range [0,1]

p-value for the chi-square test statistic, returned as a scalar value in the range [0,1]. crosstab tests that tbl is independent in each dimension.

labels — Data labels

cell array

Data labels, returned as a cell array. The entries in the first column are labels for the rows of tbl, the entries in the second column are labels for the columns, and so on, for a multi-dimensional tbl.

Algorithms

• crosstab uses grp2idx to assign a positive integer to each distinct value. tbl(i,j) is a count of indices where grp2idx(x1) is i and grp2idx(x2) is j. The numerical order of grp2idx(x1) and grp2idx(x2) order rows and columns of tbl, respectively.

  In this case, the returned value of tbl(i,j,...,n) is a count of indices where grp2idx(x1) is i, grp2idx(x2) is j, grp2idx(x3) is k, and so on.

• crosstab computes the p-value of the chi-square test statistic using a formula that is asymptotically valid for a large sample size. The approximation is less accurate for small samples or samples with uneven marginal distributions. If your sample includes only two variables and each has two levels, you can use fishertest instead. This function performs Fisher’s exact test, which does not depend on large-sample distribution assumptions.

Extended Capabilities

Tall Arrays

Calculate with arrays that have more rows than fit in memory.

This function supports tall arrays for out-of-memory data with the limitation:

The fourth output, labels, is returned as a cell array containing M unevaluated tall cell arrays, where M is the number of input grouping variables. Each unevaluated tall cell array, labels{j}, contains the labels for one grouping variable.

For more information, see Tall Arrays for Out-of-Memory Data.