# Contingency tables¶

statsmodels supports a variety of approaches for analyzing contingency tables, including methods for assessing independence, symmetry, homogeneity, and methods for working with collections of tables from a stratified population.

The methods described here are mainly for two-way tables. Multi-way tables can be analyzed using log-linear models. statsmodels does not currently have a dedicated API for loglinear modeling, but Poisson regression in statsmodels.genmod.GLM can be used for this purpose.

A contingency table is a multi-way table that describes a data set in which each observation belongs to one category for each of several variables. For example, if there are two variables, one with $$r$$ levels and one with $$c$$ levels, then we have a $$r \times c$$ contingency table. The table can be described in terms of the number of observations that fall into a given cell of the table, e.g. $$T_{ij}$$ is the number of observations that have level $$i$$ for the first variable and level $$j$$ for the second variable. Note that each variable must have a finite number of levels (or categories), which can be either ordered or unordered. In different contexts, the variables defining the axes of a contingency table may be called categorical variables or factor variables. They may be either nominal (if their levels are unordered) or ordinal (if their levels are ordered).

The underlying population for a contingency table is described by a distribution table $$P_{i, j}$$. The elements of $$P$$ are probabilities, and the sum of all elements in $$P$$ is 1. Methods for analyzing contingency tables use the data in $$T$$ to learn about properties of $$P$$.

The statsmodels.stats.Table is the most basic class for working with contingency tables. We can create a Table object directly from any rectangular array-like object containing the contingency table cell counts:

In [1]: import numpy as np

In [2]: import pandas as pd

In [3]: import statsmodels.api as sm

In [4]: df = sm.datasets.get_rdataset("Arthritis", "vcd").data

In [5]: tab = pd.crosstab(df['Treatment'], df['Improved'])

In [6]: tab = tab.loc[:, ["None", "Some", "Marked"]]

In [7]: table = sm.stats.Table(tab)


Alternatively, we can pass the raw data and let the Table class construct the array of cell counts for us:

In [8]: data = df[["Treatment", "Improved"]]

In [9]: table = sm.stats.Table.from_data(data)


## Independence¶

Independence is the property that the row and column factors occur independently. Association is the lack of independence. If the joint distribution is independent, it can be written as the outer product of the row and column marginal distributions:

$P_{ij} = \sum_k P_{ij} \cdot \sum_k P_{kj} \quad \text{for all} \quad i, j$

We can obtain the best-fitting independent distribution for our observed data, and then view residuals which identify particular cells that most strongly violate independence:

In [10]: print(table.table_orig)
Improved   Marked  None  Some
Treatment
Placebo         7    29     7
Treated        21    13     7

In [11]: print(table.fittedvalues)
Improved      Marked  None      Some
Treatment
Placebo    14.333333  21.5  7.166667
Treated    13.666667  20.5  6.833333

In [12]: print(table.resid_pearson)
Improved     Marked      None      Some
Treatment
Placebo   -1.936992  1.617492 -0.062257
Treated    1.983673 -1.656473  0.063758


In this example, compared to a sample from a population in which the rows and columns are independent, we have too many observations in the placebo/no improvement and treatment/marked improvement cells, and too few observations in the placebo/marked improvement and treated/no improvement cells. This reflects the apparent benefits of the treatment.

If the rows and columns of a table are unordered (i.e. are nominal factors), then the most common approach for formally assessing independence is using Pearson’s $$\chi^2$$ statistic. It’s often useful to look at the cell-wise contributions to the $$\chi^2$$ statistic to see where the evidence for dependence is coming from.

In [13]: rslt = table.test_nominal_association()

In [14]: print(rslt.pvalue)
0.0014626434089526352

In [15]: print(table.chi2_contribs)
Improved     Marked      None      Some
Treatment
Placebo    3.751938  2.616279  0.003876
Treated    3.934959  2.743902  0.004065


For tables with ordered row and column factors, we can us the linear by linear association test to obtain more power against alternative hypotheses that respect the ordering. The test statistic for the linear by linear association test is

$\sum_k r_i c_j T_{ij}$

where $$r_i$$ and $$c_j$$ are row and column scores. Often these scores are set to the sequences 0, 1, …. This gives the ‘Cochran-Armitage trend test’.

In [16]: rslt = table.test_ordinal_association()

In [17]: print(rslt.pvalue)
0.023644578093923983


We can assess the association in a $$r\times x$$ table by constructing a series of $$2\times 2$$ tables and calculating their odds ratios. There are two ways to do this. The local odds ratios construct $$2\times 2$$ tables from adjacent row and column categories.

In [18]: print(table.local_oddsratios)
Improved     Marked      None  Some
Treatment
Placebo    0.149425  2.230769   NaN
Treated         NaN       NaN   NaN

In [19]: taloc = sm.stats.Table2x2(np.asarray([[7, 29], [21, 13]]))

In [20]: print(taloc.oddsratio)
0.14942528735632185

In [21]: taloc = sm.stats.Table2x2(np.asarray([[29, 7], [13, 7]]))

In [22]: print(taloc.oddsratio)
2.230769230769231


The cumulative odds ratios construct $$2\times 2$$ tables by dichotomizing the row and column factors at each possible point.

In [23]: print(table.cumulative_oddsratios)
Improved     Marked      None  Some
Treatment
Placebo    0.185185  1.058824   NaN
Treated         NaN       NaN   NaN

In [24]: tab1 = np.asarray([[7, 29 + 7], [21, 13 + 7]])

In [25]: tacum = sm.stats.Table2x2(tab1)

In [26]: print(tacum.oddsratio)
0.18518518518518517

In [27]: tab1 = np.asarray([[7 + 29, 7], [21 + 13, 7]])

In [28]: tacum = sm.stats.Table2x2(tab1)

In [29]: print(tacum.oddsratio)
1.0588235294117647


A mosaic plot is a graphical approach to informally assessing dependence in two-way tables.

In [30]: from statsmodels.graphics.mosaicplot import mosaic

In [31]: fig, _ = mosaic(data, index=["Treatment", "Improved"])


## Symmetry and homogeneity¶

Symmetry is the property that $$P_{i, j} = P_{j, i}$$ for every $$i$$ and $$j$$. Homogeneity is the property that the marginal distribution of the row factor and the column factor are identical, meaning that

$\sum_j P_{ij} = \sum_j P_{ji} \forall i$

Note that for these properties to be applicable the table $$P$$ (and $$T$$) must be square, and the row and column categories must be identical and must occur in the same order.

To illustrate, we load a data set, create a contingency table, and calculate the row and column margins. The Table class contains methods for analyzing $$r \times c$$ contingency tables. The data set loaded below contains assessments of visual acuity in people’s left and right eyes. We first load the data and create a contingency table.

In [32]: df = sm.datasets.get_rdataset("VisualAcuity", "vcd").data

In [33]: df = df.loc[df.gender == "female", :]

In [34]: tab = df.set_index(['left', 'right'])

In [35]: del tab["gender"]

In [36]: tab = tab.unstack()

In [37]: tab.columns = tab.columns.get_level_values(1)

In [38]: print(tab)
right     1     2     3    4
left
1      1520   234   117   36
2       266  1512   362   82
3       124   432  1772  179
4        66    78   205  492


Next we create a SquareTable object from the contingency table.

In [39]: sqtab = sm.stats.SquareTable(tab)

In [40]: row, col = sqtab.marginal_probabilities

In [41]: print(row)
right
1    0.255049
2    0.297178
3    0.335295
4    0.112478
dtype: float64

In [42]: print(col)
right
1    0.264277
2    0.301725
3    0.328474
4    0.105524
dtype: float64


The summary method prints results for the symmetry and homogeneity testing procedures.

In [43]: print(sqtab.summary())
Statistic P-value DF
--------------------------------
Symmetry       19.107   0.004  6
Homogeneity    11.957   0.008  3
--------------------------------


If we had the individual case records in a dataframe called data, we could also perform the same analysis by passing the raw data using the SquareTable.from_data class method.

sqtab = sm.stats.SquareTable.from_data(data[['left', 'right']])
print(sqtab.summary())


## A single 2x2 table¶

Several methods for working with individual 2x2 tables are provided in the sm.stats.Table2x2 class. The summary method displays several measures of association between the rows and columns of the table.

In [44]: table = np.asarray([[35, 21], [25, 58]])

In [45]: t22 = sm.stats.Table2x2(table)

In [46]: print(t22.summary())
Estimate   SE   LCB   UCB  p-value
-------------------------------------------------
Odds ratio        3.867       1.890 7.912   0.000
Log odds ratio    1.352 0.365 0.636 2.068   0.000
Risk ratio        2.075       1.411 3.051   0.000
Log risk ratio    0.730 0.197 0.345 1.115   0.000
-------------------------------------------------


Note that the risk ratio is not symmetric so different results will be obtained if the transposed table is analyzed.

In [47]: table = np.asarray([[35, 21], [25, 58]])

In [48]: t22 = sm.stats.Table2x2(table.T)

In [49]: print(t22.summary())
Estimate   SE   LCB   UCB  p-value
-------------------------------------------------
Odds ratio        3.867       1.890 7.912   0.000
Log odds ratio    1.352 0.365 0.636 2.068   0.000
Risk ratio        2.194       1.436 3.354   0.000
Log risk ratio    0.786 0.216 0.362 1.210   0.000
-------------------------------------------------


## Stratified 2x2 tables¶

Stratification occurs when we have a collection of contingency tables defined by the same row and column factors. In the example below, we have a collection of 2x2 tables reflecting the joint distribution of smoking and lung cancer in each of several regions of China. It is possible that the tables all have a common odds ratio, even while the marginal probabilities vary among the strata. The ‘Breslow-Day’ procedure tests whether the data are consistent with a common odds ratio. It appears below as the Test of constant OR. The Mantel-Haenszel procedure tests whether this common odds ratio is equal to one. It appears below as the Test of OR=1. It is also possible to estimate the common odds and risk ratios and obtain confidence intervals for them. The summary method displays all of these results. Individual results can be obtained from the class methods and attributes.

In [50]: data = sm.datasets.china_smoking.load_pandas()

In [51]: mat = np.asarray(data.data)

In [52]: tables = [np.reshape(x.tolist(), (2, 2)) for x in mat]

In [53]: st = sm.stats.StratifiedTable(tables)

In [54]: print(st.summary())
Estimate   LCB    UCB
-----------------------------------------
Pooled odds           2.174   1.984 2.383
Pooled log odds       0.777   0.685 0.868
Pooled risk ratio     1.519

Statistic P-value
-----------------------------------
Test of OR=1       280.138   0.000
Test constant OR     5.200   0.636

-----------------------
Number of tables    8
Min n             213
Max n            2900
Avg n            1052
Total n          8419
-----------------------


## Module Reference¶

 Table(table[, shift_zeros]) A two-way contingency table. Table2x2(table[, shift_zeros]) Analyses that can be performed on a 2x2 contingency table. SquareTable(table[, shift_zeros]) Methods for analyzing a square contingency table. StratifiedTable(tables[, shift_zeros]) Analyses for a collection of 2x2 contingency tables. mcnemar(table[, exact, correction]) McNemar test of homogeneity. cochrans_q(x[, return_object]) Cochran's Q test for identical binomial proportions.