Cartesian Product Data Frame

cartesian product in pandas

In recent versions of Pandas (>= 1.2) this is built into merge so you can do:

from pandas import DataFrame
df1 = DataFrame({'col1':[1,2],'col2':[3,4]})
df2 = DataFrame({'col3':[5,6]})    

df1.merge(df2, how='cross')

This is equivalent to the previous pandas < 1.2 answer but is easier to read.

For pandas < 1.2:

If you have a key that is repeated for each row, then you can produce a cartesian product using merge (like you would in SQL).

from pandas import DataFrame, merge
df1 = DataFrame({'key':[1,1], 'col1':[1,2],'col2':[3,4]})
df2 = DataFrame({'key':[1,1], 'col3':[5,6]})

merge(df1, df2,on='key')[['col1', 'col2', 'col3']]

Output:

   col1  col2  col3
0     1     3     5
1     1     3     6
2     2     4     5
3     2     4     6

See here for the documentation: http://pandas.pydata.org/pandas-docs/stable/merging.html

Cartesian product data frame

You can use expand.grid(A, B, C)

EDIT: an alternative to using do.call to achieve the second part, is the function mdply from the package plyr:

library(plyr)

d = expand.grid(x = A, y = B, z = C)
d = mdply(d, f)

To illustrate its usage using a trivial function 'paste', you can try

d = mdply(d, 'paste', sep = '+');

Create a pandas DataFrame from the cartesian product of lists

itertools.product is the most pythonic way to do this. However, if you want to use the code you already have, you're almost right there

#create a list to append your values into
data=[]

categories = ['small', 'medium', 'big']
parameters = ['p1_5_p2_4_p3_2', 'p1_3_p2_8_p3_3', 'p1_4_p2_3_p3_6']
Blue = [5, 4, 3]

for parameter in parameters:
    for category in categories:
        for x in Blue:
            y = x + 1
            z = x + 2

            #append instead of printing
            row=[category,parameter,y,z]
            data.append(row)

#create your dataframe
my_df=pd.DataFrame(columns=['Category','Parameters','Value_1','Value_2'], data=data)

  Category  Parameters      Value_1 Value_2
0   small   p1_5_p2_4_p3_2  6       7
1   small   p1_5_p2_4_p3_2  5       6
2   small   p1_5_p2_4_p3_2  4       5
3   medium  p1_5_p2_4_p3_2  6       7
4   medium  p1_5_p2_4_p3_2  5       6

Performant cartesian product (CROSS JOIN) with pandas

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

pandas <= 1.1.X

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)

pandas >= 1.2

left.merge(right, how="cross") # implements the technique above

  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

Disclaimer

These solutions are optimised for DataFrames with non-mixed scalar dtypes. If dealing with mixed dtypes, use at your
own risk!

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

Sample Image

Do note that timings may vary based on your setup, data, and choice of cartesian_product helper function as applicable.

Performance Benchmarking Code

This is the timing script. All functions called here are defined above.

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'],
       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)
        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()

Continue Reading

Jump to other topics in Pandas Merging 101 to continue learning:

Merging basics - basic types of joins
Index-based joins
Generalizing to multiple DataFrames
Cross join ^*

_{* you are here}

Join two dataframes to get cartesian product

You can use a dummy column to merge on:

df1.assign(dummy=1).merge(df2.assign(dummy=1), on='dummy', how='outer').drop('dummy', axis=1)

Output:

   values_x  values_y
0         4         7
1         4         8
2         4         9
3         5         7
4         5         8
5         5         9
6         6         7
7         6         8
8         6         9

R: any function for Cartesian Product of two data frames?

merge(A, B), provided there are no columns linking the two, should do this by default, no?

From ?merge (emphasis mine):

If by or both by.x and by.y are of length 0 (a length zero vector or NULL), the result, r, is the Cartesian product of x and y, i.e., dim(r) = c(nrow(x)*nrow(y), ncol(x) + ncol(y)).

Admittedly, this does require one to know to look in ?merge. Context-based searching in R is severely lacking; even rseek doesn't immediately provide this.

Cartesian product of a pandas dataframe with itself

We want to get the indices for the upper and lower triangles of a square matrix. Or in other words, where the identity matrix is zero

np.eye(len(df))

array([[ 1.,  0.,  0.],
       [ 0.,  1.,  0.],
       [ 0.,  0.,  1.]])

So I subtract it from 1 and

array([[ 0.,  1.,  1.],
       [ 1.,  0.,  1.],
       [ 1.,  1.,  0.]])

In a boolean context and passed to np.where I get exactly the upper and lower triangle indices.

i, j = np.where(1 - np.eye(len(df)))
df.iloc[i].reset_index(drop=True).join(
    df.iloc[j].reset_index(drop=True), rsuffix='_2')

   id value  id_2 value_2
0   1     a     2       b
1   1     a     3       c
2   2     b     1       a
3   2     b     3       c
4   3     c     1       a
5   3     c     2       b

Create a pandas DataFrame from a Cartesian product of two large lists

list1 = ["A", "B", "C", "D", "E"]
list2 = ["x1", "x2", "x3", "x4", "x5", "x6", "x7", "x8"]

df1 = pd.DataFrame(list1)
df2 = pd.DataFrame(list2)

df1['key'] = 0
df2['key'] = 0
print( df1.merge(df2, on='key', how='outer').drop(columns='key') )

Prints:

   0_x 0_y
0    A  x1
1    A  x2
2    A  x3
3    A  x4
4    A  x5
5    A  x6
6    A  x7
7    A  x8
8    B  x1
9    B  x2

...

cartesian product with dplyr R

Use crossing from the tidyr package:

x <- data.frame(x=c("a","b","c"))
y <- data.frame(y=c(1,2,3))

crossing(x, y)

Result: