# Performant cartesian product (CROSS JOIN) of two pandas DataFrames

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The contents of this post were originally meant to be a part of Pandas Merging 101, but due to the nature and size of the content required to fully do justice to this topic, it has been moved to its own QnA.

Given two simple DataFrames;

``left = pd.DataFrame({'col1' : ['A', 'B', 'C'], 'col2' : [1, 2, 3]}) right = pd.DataFrame({'col1' : ['X', 'Y', 'Z'], 'col2' : [20, 30, 50]})  left    col1  col2 0    A     1 1    B     2 2    C     3  right    col1  col2 0    X    20 1    Y    30 2    Z    50 ``

The cross product of these frames can be computed, and will look something like:

``A       1      X      20 A       1      Y      30 A       1      Z      50 B       2      X      20 B       2      Y      30 B       2      Z      50 C       3      X      20 C       3      Y      30 C       3      Z      50 ``

What is the most performant method of computing this result?

Let's start by establishing a benchmark. The easiest method for solving this is using a temporary "key" column:

``def cartesian_product_basic(left, right):     return (        left.assign(key=1).merge(right.assign(key=1), on='key').drop('key', 1))  cartesian_product_basic(left, right)    col1_x  col2_x col1_y  col2_y 0      A       1      X      20 1      A       1      Y      30 2      A       1      Z      50 3      B       2      X      20 4      B       2      Y      30 5      B       2      Z      50 6      C       3      X      20 7      C       3      Y      30 8      C       3      Z      50 ``

How this works is that both DataFrames are assigned a temporary "key" column with the same value (say, 1). `merge` then performs a many-to-many JOIN on "key".

While the many-to-many JOIN trick works for reasonably sized DataFrames, you will see relatively lower performance on larger data.

A faster implementation will require NumPy. Here are some famous NumPy implementations of 1D cartesian product. We can build on some of these performant solutions to get our desired output. My favourite, however, is @senderle's first implementation.

``def cartesian_product(*arrays):     la = len(arrays)     dtype = np.result_type(*arrays)     arr = np.empty([len(a) for a in arrays] + [la], dtype=dtype)     for i, a in enumerate(np.ix_(*arrays)):         arr[...,i] = a     return arr.reshape(-1, la)   ``

### Generalizing: CROSS JOIN on Unique or Non-Unique Indexed DataFrames

This trick will work on any kind of DataFrame. We compute the cartesian product of the DataFrames' numeric indices using the aforementioned `cartesian_product`, use this to reindex the DataFrames, and

``def cartesian_product_generalized(left, right):     la, lb = len(left), len(right)     idx = cartesian_product(np.ogrid[:la], np.ogrid[:lb])     return pd.DataFrame(         np.column_stack([left.values[idx[:,0]], right.values[idx[:,1]]]))  cartesian_product_generalized(left, right)     0  1  2   3 0  A  1  X  20 1  A  1  Y  30 2  A  1  Z  50 3  B  2  X  20 4  B  2  Y  30 5  B  2  Z  50 6  C  3  X  20 7  C  3  Y  30 8  C  3  Z  50  np.array_equal(cartesian_product_generalized(left, right),                cartesian_product_basic(left, right)) True ``

And, along similar lines,

``left2 = left.copy() left2.index = ['s1', 's2', 's1']  right2 = right.copy() right2.index = ['x', 'y', 'y']   left2    col1  col2 s1    A     1 s2    B     2 s1    C     3  right2   col1  col2 x    X    20 y    Y    30 y    Z    50  np.array_equal(cartesian_product_generalized(left, right),                cartesian_product_basic(left2, right2)) True ``

This solution can generalise to multiple DataFrames. For example,

``def cartesian_product_multi(*dfs):     idx = cartesian_product(*[np.ogrid[:len(df)] for df in dfs])     return pd.DataFrame(         np.column_stack([df.values[idx[:,i]] for i,df in enumerate(dfs)]))  cartesian_product_multi(*[left, right, left]).head()     0  1  2   3  4  5 0  A  1  X  20  A  1 1  A  1  X  20  B  2 2  A  1  X  20  C  3 3  A  1  X  20  D  4 4  A  1  Y  30  A  1 ``

### Further Simplification

A simpler solution not involving @senderle's `cartesian_product` is possible when dealing with just two DataFrames. Using `np.broadcast_arrays`, we can achieve almost the same level of performance.

``def cartesian_product_simplified(left, right):     la, lb = len(left), len(right)     ia2, ib2 = np.broadcast_arrays(*np.ogrid[:la,:lb])      return pd.DataFrame(         np.column_stack([left.values[ia2.ravel()], right.values[ib2.ravel()]]))  np.array_equal(cartesian_product_simplified(left, right),                cartesian_product_basic(left2, right2)) True ``

### Performance Comparison

Benchmarking these solutions on some contrived DataFrames with unique indices, we have Do note that timings may vary based on your setup, data, and choice of `cartesian_product` helper function as applicable.

``# Wen's answer: https://stackoverflow.com/a/53699198/4909087 # I've put my own spin on this to make it as fast as possible. def cartesian_product_itertools(left, right):     return pd.DataFrame([            [*x, *y] for x, y in itertools.product(                left.values.tolist(), right.values.tolist())]) ``
``from timeit import timeit import pandas as pd import matplotlib.pyplot as plt  res = pd.DataFrame(        index=['cartesian_product_basic', 'cartesian_product_generalized',                'cartesian_product_multi', 'cartesian_product_simplified',               'cartesian_product_itertools'],        columns=[1, 10, 50, 100, 200, 300, 400, 500, 600, 800, 1000, 2000],        dtype=float )  for f in res.index:      for c in res.columns:         # print(f,c)         if f in {'cartesian_product_itertools'} and c > 600:             continue         left2 = pd.concat([left] * c, ignore_index=True)         right2 = pd.concat([right] * c, ignore_index=True)         stmt = '{}(left2, right2)'.format(f)         setp = 'from __main__ import left2, right2, {}'.format(f)         res.at[f, c] = timeit(stmt, setp, number=5)  ax = res.div(res.min()).T.plot(loglog=True)  ax.set_xlabel("N");  ax.set_ylabel("time (relative)");  plt.show() ``