# Using NumPy to build an array of all combinations of two arrays

## Question:

I’m trying to run over the parameters space of a six-parameter function to study its numerical behavior before trying to do anything complex with it, so I’m searching for an efficient way to do this.

My function takes float values given in a 6-dim NumPy array as input. What I tried to do initially was this:

First, I created a function that takes two arrays and generate an array with all combinations of values from the two arrays:

```
from numpy import *
def comb(a, b):
c = []
for i in a:
for j in b:
c.append(r_[i,j])
return c
```

Then, I used `reduce()`

to apply that to m copies of the same array:

```
def combs(a, m):
return reduce(comb, [a]*m)
```

Finally, I evaluate my function like this:

```
values = combs(np.arange(0, 1, 0.1), 6)
for val in values:
print F(val)
```

This works, but it’s *way* too slow. I know the space of parameters is huge, but this shouldn’t be so slow. I have only sampled 10^{6} (a million) points in this example and it took more than 15 seconds just to create the array `values`

.

Is there a more efficient way of doing this with NumPy?

I can modify the way the function `F`

takes its arguments if it’s necessary.

## Answers:

itertools.combinations is in general the fastest way to get combinations from a Python container (if you do in fact want combinations, i.e., arrangements *without* repetitions and independent of order; that’s not what your code appears to be doing, but I can’t tell whether that’s because your code is buggy or because you’re using the wrong terminology).

If you want something different than combinations perhaps other iterators in itertools, `product`

or `permutations`

, might serve you better. For example, it looks like your code is roughly the same as:

```
for val in itertools.product(np.arange(0, 1, 0.1), repeat=6):
print F(val)
```

All of these iterators yield tuples, not lists or NumPy arrays, so if your *F* is picky about getting specifically a NumPy array, you’ll have to accept the extra overhead of constructing or clearing and refilling one at each step.

Here’s a pure-NumPy implementation. It’s about 5 times faster than using itertools.

## Python 3:

```
import numpy as np
def cartesian(arrays, out=None):
"""
Generate a Cartesian product of input arrays.
Parameters
----------
arrays : list of array-like
1-D arrays to form the Cartesian product of.
out : ndarray
Array to place the Cartesian product in.
Returns
-------
out : ndarray
2-D array of shape (M, len(arrays)) containing Cartesian products
formed of input arrays.
Examples
--------
>>> cartesian(([1, 2, 3], [4, 5], [6, 7]))
array([[1, 4, 6],
[1, 4, 7],
[1, 5, 6],
[1, 5, 7],
[2, 4, 6],
[2, 4, 7],
[2, 5, 6],
[2, 5, 7],
[3, 4, 6],
[3, 4, 7],
[3, 5, 6],
[3, 5, 7]])
"""
arrays = [np.asarray(x) for x in arrays]
dtype = arrays[0].dtype
n = np.prod([x.size for x in arrays])
if out is None:
out = np.zeros([n, len(arrays)], dtype=dtype)
#m = n / arrays[0].size
m = int(n / arrays[0].size)
out[:,0] = np.repeat(arrays[0], m)
if arrays[1:]:
cartesian(arrays[1:], out=out[0:m, 1:])
for j in range(1, arrays[0].size):
#for j in xrange(1, arrays[0].size):
out[j*m:(j+1)*m, 1:] = out[0:m, 1:]
return out
```

## Python 2:

```
import numpy as np
def cartesian(arrays, out=None):
arrays = [np.asarray(x) for x in arrays]
dtype = arrays[0].dtype
n = np.prod([x.size for x in arrays])
if out is None:
out = np.zeros([n, len(arrays)], dtype=dtype)
m = n / arrays[0].size
out[:,0] = np.repeat(arrays[0], m)
if arrays[1:]:
cartesian(arrays[1:], out=out[0:m, 1:])
for j in xrange(1, arrays[0].size):
out[j*m:(j+1)*m, 1:] = out[0:m, 1:]
return out
```

It looks like you want a grid to evaluate your function, in which case you can use numpy.ogrid (open) or numpy.mgrid (fleshed out):

```
import numpy
my_grid = numpy.mgrid[[slice(0, 1, 0.1)]*6]
```

You can do something like this

```
import numpy as np
def cartesian_coord(*arrays):
grid = np.meshgrid(*arrays)
coord_list = [entry.ravel() for entry in grid]
points = np.vstack(coord_list).T
return points
a = np.arange(4) # Fake data
print(cartesian_coord(*6*[a])
```

which gives

```
array([[0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 2],
...,
[3, 3, 3, 3, 3, 1],
[3, 3, 3, 3, 3, 2],
[3, 3, 3, 3, 3, 3]])
```

The following NumPy implementation should be approximately two times the speed of the given previous answers:

```
def cartesian2(arrays):
arrays = [np.asarray(a) for a in arrays]
shape = (len(x) for x in arrays)
ix = np.indices(shape, dtype=int)
ix = ix.reshape(len(arrays), -1).T
for n, arr in enumerate(arrays):
ix[:, n] = arrays[n][ix[:, n]]
return ix
```

Here’s yet another way, using pure NumPy, no recursion, no list comprehension, and no explicit for loops. It’s about 20% slower than the original answer, and it’s based on np.meshgrid.

```
def cartesian(*arrays):
mesh = np.meshgrid(*arrays) # Standard NumPy meshgrid
dim = len(mesh) # Number of dimensions
elements = mesh[0].size # Number of elements, any index will do
flat = np.concatenate(mesh).ravel() # Flatten the whole meshgrid
reshape = np.reshape(flat, (dim, elements)).T # Reshape and transpose
return reshape
```

For example,

```
x = np.arange(3)
a = cartesian(x, x, x, x, x)
print(a)
```

gives

```
[[0 0 0 0 0]
[0 0 0 0 1]
[0 0 0 0 2]
...,
[2 2 2 2 0]
[2 2 2 2 1]
[2 2 2 2 2]]
```

In newer versions of NumPy (>1.8.x), numpy.meshgrid() provides a much faster implementation:

For pv’s solution:

```
In [113]:
%timeit cartesian(([1, 2, 3], [4, 5], [6, 7]))
10000 loops, best of 3: 135 µs per loop
In [114]:
cartesian(([1, 2, 3], [4, 5], [6, 7]))
Out[114]:
array([[1, 4, 6],
[1, 4, 7],
[1, 5, 6],
[1, 5, 7],
[2, 4, 6],
[2, 4, 7],
[2, 5, 6],
[2, 5, 7],
[3, 4, 6],
[3, 4, 7],
[3, 5, 6],
[3, 5, 7]])
```

numpy.meshgrid() used to be two-dimensional only, but now it is capable of multidimensional. In this case, three-dimensional:

```
In [115]:
%timeit np.array(np.meshgrid([1, 2, 3], [4, 5], [6, 7])).T.reshape(-1,3)
10000 loops, best of 3: 74.1 µs per loop
In [116]:
np.array(np.meshgrid([1, 2, 3], [4, 5], [6, 7])).T.reshape(-1,3)
Out[116]:
array([[1, 4, 6],
[1, 5, 6],
[2, 4, 6],
[2, 5, 6],
[3, 4, 6],
[3, 5, 6],
[1, 4, 7],
[1, 5, 7],
[2, 4, 7],
[2, 5, 7],
[3, 4, 7],
[3, 5, 7]])
```

Note that the order of the final resultant is slightly different.

For a pure NumPy implementation of the Cartesian product of one-dimensional arrays (or flat Python lists), just use meshgrid(), roll the axes with transpose(), and reshape to the desired output:

```
def cartprod(*arrays):
N = len(arrays)
return transpose(meshgrid(*arrays, indexing='ij'),
roll(arange(N + 1), -1)).reshape(-1, N)
```

Note this has the convention of the last axis changing the fastest ("C style" or "row-major").

```
In [88]: cartprod([1,2,3], [4,8], [100, 200, 300, 400], [-5, -4])
Out[88]:
array([[ 1, 4, 100, -5],
[ 1, 4, 100, -4],
[ 1, 4, 200, -5],
[ 1, 4, 200, -4],
[ 1, 4, 300, -5],
[ 1, 4, 300, -4],
[ 1, 4, 400, -5],
[ 1, 4, 400, -4],
[ 1, 8, 100, -5],
[ 1, 8, 100, -4],
[ 1, 8, 200, -5],
[ 1, 8, 200, -4],
[ 1, 8, 300, -5],
[ 1, 8, 300, -4],
[ 1, 8, 400, -5],
[ 1, 8, 400, -4],
[ 2, 4, 100, -5],
[ 2, 4, 100, -4],
[ 2, 4, 200, -5],
[ 2, 4, 200, -4],
[ 2, 4, 300, -5],
[ 2, 4, 300, -4],
[ 2, 4, 400, -5],
[ 2, 4, 400, -4],
[ 2, 8, 100, -5],
[ 2, 8, 100, -4],
[ 2, 8, 200, -5],
[ 2, 8, 200, -4],
[ 2, 8, 300, -5],
[ 2, 8, 300, -4],
[ 2, 8, 400, -5],
[ 2, 8, 400, -4],
[ 3, 4, 100, -5],
[ 3, 4, 100, -4],
[ 3, 4, 200, -5],
[ 3, 4, 200, -4],
[ 3, 4, 300, -5],
[ 3, 4, 300, -4],
[ 3, 4, 400, -5],
[ 3, 4, 400, -4],
[ 3, 8, 100, -5],
[ 3, 8, 100, -4],
[ 3, 8, 200, -5],
[ 3, 8, 200, -4],
[ 3, 8, 300, -5],
[ 3, 8, 300, -4],
[ 3, 8, 400, -5],
[ 3, 8, 400, -4]])
```

If you want to change the *first* axis fastest ("Fortran style" or "column-major"), just change the `order`

parameter of `reshape()`

like this: `reshape((-1, N), order='F')`

You can use `np.array(itertools.product(a, b))`

.

Pandas `merge`

offers a naive, fast solution to the problem:

```
# given the lists
x, y, z = [1, 2, 3], [4, 5], [6, 7]
# get dfs with same, constant index
x = pd.DataFrame({'x': x}, index=np.repeat(0, len(x)))
y = pd.DataFrame({'y': y}, index=np.repeat(0, len(y)))
z = pd.DataFrame({'z': z}, index=np.repeat(0, len(z)))
# get all permutations stored in a new df
df = pd.merge(x, pd.merge(y, z, left_index=True, right_index=True),
left_index=True, right_index=True)
```