I have two numpy arrays that define the x and y axes of a grid. For example:
x = numpy.array([1,2,3])
y = numpy.array([4,5])
I'd like to generate the Cartesian product of these arrays to generate:
array([[1,4],[2,4],[3,4],[1,5],[2,5],[3,5]])
In a way that's not terribly inefficient since I need to do this many times in a loop. I'm assuming that converting them to a Python list and using itertools.product and back to a numpy array is not the most efficient form.
A canonical cartesian_product (almost)
There are many approaches to this problem with different properties. Some are faster than others, and some are more general-purpose. After a lot of testing and tweaking, I've found that the following function, which calculates an n-dimensional cartesian_product, is faster than most others for many inputs. For a pair of approaches that are slightly more complex, but are even a bit faster in many cases, see the answer by Paul Panzer.
Given that answer, this is no longer the fastest implementation of the cartesian product in numpy that I'm aware of. However, I think its simplicity will continue to make it a useful benchmark for future improvement:
def cartesian_product(*arrays):
la = len(arrays)
dtype = numpy.result_type(*arrays)
arr = numpy.empty([len(a) for a in arrays] + [la], dtype=dtype)
for i, a in enumerate(numpy.ix_(*arrays)):
arr[...,i] = a
return arr.reshape(-1, la)
It's worth mentioning that this function uses ix_ in an unusual way; whereas the documented use of ix_ is to generate indices into an array, it just so happens that arrays with the same shape can be used for broadcasted assignment. Many thanks to mgilson, who inspired me to try using ix_ this way, and to unutbu, who provided some extremely helpful feedback on this answer, including the suggestion to use numpy.result_type.
Notable alternatives
It's sometimes faster to write contiguous blocks of memory in Fortran order. That's the basis of this alternative, cartesian_product_transpose, which has proven faster on some hardware than cartesian_product (see below). However, Paul Panzer's answer, which uses the same principle, is even faster. Still, I include this here for interested readers:
def cartesian_product_transpose(*arrays):
broadcastable = numpy.ix_(*arrays)
broadcasted = numpy.broadcast_arrays(*broadcastable)
rows, cols = numpy.prod(broadcasted[0].shape), len(broadcasted)
dtype = numpy.result_type(*arrays)
out = numpy.empty(rows * cols, dtype=dtype)
start, end = 0, rows
for a in broadcasted:
out[start:end] = a.reshape(-1)
start, end = end, end + rows
return out.reshape(cols, rows).T
After coming to understand Panzer's approach, I wrote a new version that's almost as fast as his, and is almost as simple as cartesian_product:
def cartesian_product_simple_transpose(arrays):
la = len(arrays)
dtype = numpy.result_type(*arrays)
arr = numpy.empty([la] + [len(a) for a in arrays], dtype=dtype)
for i, a in enumerate(numpy.ix_(*arrays)):
arr[i, ...] = a
return arr.reshape(la, -1).T
This appears to have some constant-time overhead that makes it run slower than Panzer's for small inputs. But for larger inputs, in all the tests I ran, it performs just as well as his fastest implementation (cartesian_product_transpose_pp).
In following sections, I include some tests of other alternatives. These are now somewhat out of date, but rather than duplicate effort, I've decided to leave them here out of historical interest. For up-to-date tests, see Panzer's answer, as well as Nico Schlömer's.
Tests against alternatives
Here is a battery of tests that show the performance boost that some of these functions provide relative to a number of alternatives. All the tests shown here were performed on a quad-core machine, running Mac OS 10.12.5, Python 3.6.1, and numpy 1.12.1. Variations on hardware and software are known to produce different results, so YMMV. Run these tests for yourself to be sure!
Definitions:
import numpy
import itertools
from functools import reduce
### Two-dimensional products ###
def repeat_product(x, y):
return numpy.transpose([numpy.tile(x, len(y)),
numpy.repeat(y, len(x))])
def dstack_product(x, y):
return numpy.dstack(numpy.meshgrid(x, y)).reshape(-1, 2)
### Generalized N-dimensional products ###
def cartesian_product(*arrays):
la = len(arrays)
dtype = numpy.result_type(*arrays)
arr = numpy.empty([len(a) for a in arrays] + [la], dtype=dtype)
for i, a in enumerate(numpy.ix_(*arrays)):
arr[...,i] = a
return arr.reshape(-1, la)
def cartesian_product_transpose(*arrays):
broadcastable = numpy.ix_(*arrays)
broadcasted = numpy.broadcast_arrays(*broadcastable)
rows, cols = numpy.prod(broadcasted[0].shape), len(broadcasted)
dtype = numpy.result_type(*arrays)
out = numpy.empty(rows * cols, dtype=dtype)
start, end = 0, rows
for a in broadcasted:
out[start:end] = a.reshape(-1)
start, end = end, end + rows
return out.reshape(cols, rows).T
# from https://stackoverflow.com/a/1235363/577088
def cartesian_product_recursive(*arrays, out=None):
arrays = [numpy.asarray(x) for x in arrays]
dtype = arrays[0].dtype
n = numpy.prod([x.size for x in arrays])
if out is None:
out = numpy.zeros([n, len(arrays)], dtype=dtype)
m = n // arrays[0].size
out[:,0] = numpy.repeat(arrays[0], m)
if arrays[1:]:
cartesian_product_recursive(arrays[1:], out=out[0:m,1:])
for j in range(1, arrays[0].size):
out[j*m:(j+1)*m,1:] = out[0:m,1:]
return out
def cartesian_product_itertools(*arrays):
return numpy.array(list(itertools.product(*arrays)))
### Test code ###
name_func = [('repeat_product',
repeat_product),
('dstack_product',
dstack_product),
('cartesian_product',
cartesian_product),
('cartesian_product_transpose',
cartesian_product_transpose),
('cartesian_product_recursive',
cartesian_product_recursive),
('cartesian_product_itertools',
cartesian_product_itertools)]
def test(in_arrays, test_funcs):
global func
global arrays
arrays = in_arrays
for name, func in test_funcs:
print('{}:'.format(name))
%timeit func(*arrays)
def test_all(*in_arrays):
test(in_arrays, name_func)
# `cartesian_product_recursive` throws an
# unexpected error when used on more than
# two input arrays, so for now I've removed
# it from these tests.
def test_cartesian(*in_arrays):
test(in_arrays, name_func[2:4] + name_func[-1:])
x10 = [numpy.arange(10)]
x50 = [numpy.arange(50)]
x100 = [numpy.arange(100)]
x500 = [numpy.arange(500)]
x1000 = [numpy.arange(1000)]
Test results:
In [2]: test_all(*(x100 * 2))
repeat_product:
67.5 µs ± 633 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
dstack_product:
67.7 µs ± 1.09 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
cartesian_product:
33.4 µs ± 558 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
cartesian_product_transpose:
67.7 µs ± 932 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
cartesian_product_recursive:
215 µs ± 6.01 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
cartesian_product_itertools:
3.65 ms ± 38.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [3]: test_all(*(x500 * 2))
repeat_product:
1.31 ms ± 9.28 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
dstack_product:
1.27 ms ± 7.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
cartesian_product:
375 µs ± 4.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
cartesian_product_transpose:
488 µs ± 8.88 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
cartesian_product_recursive:
2.21 ms ± 38.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
cartesian_product_itertools:
105 ms ± 1.17 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [4]: test_all(*(x1000 * 2))
repeat_product:
10.2 ms ± 132 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
dstack_product:
12 ms ± 120 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
cartesian_product:
4.75 ms ± 57.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
cartesian_product_transpose:
7.76 ms ± 52.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
cartesian_product_recursive:
13 ms ± 209 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
cartesian_product_itertools:
422 ms ± 7.77 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In all cases, cartesian_product as defined at the beginning of this answer is fastest.
For those functions that accept an arbitrary number of input arrays, it's worth checking performance when len(arrays) > 2 as well. (Until I can determine why cartesian_product_recursive throws an error in this case, I've removed it from these tests.)
In [5]: test_cartesian(*(x100 * 3))
cartesian_product:
8.8 ms ± 138 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
cartesian_product_transpose:
7.87 ms ± 91.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
cartesian_product_itertools:
518 ms ± 5.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [6]: test_cartesian(*(x50 * 4))
cartesian_product:
169 ms ± 5.1 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
cartesian_product_transpose:
184 ms ± 4.32 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
cartesian_product_itertools:
3.69 s ± 73.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [7]: test_cartesian(*(x10 * 6))
cartesian_product:
26.5 ms ± 449 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
cartesian_product_transpose:
16 ms ± 133 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
cartesian_product_itertools:
728 ms ± 16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [8]: test_cartesian(*(x10 * 7))
cartesian_product:
650 ms ± 8.14 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
cartesian_product_transpose:
518 ms ± 7.09 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
cartesian_product_itertools:
8.13 s ± 122 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
As these tests show, cartesian_product remains competitive until the number of input arrays rises above (roughly) four. After that, cartesian_product_transpose does have a slight edge.
It's worth reiterating that users with other hardware and operating systems may see different results. For example, unutbu reports seeing the following results for these tests using Ubuntu 14.04, Python 3.4.3, and numpy 1.14.0.dev0+b7050a9:
>>> %timeit cartesian_product_transpose(x500, y500)
1000 loops, best of 3: 682 µs per loop
>>> %timeit cartesian_product(x500, y500)
1000 loops, best of 3: 1.55 ms per loop
Below, I go into a few details about earlier tests I've run along these lines. The relative performance of these approaches has changed over time, for different hardware and different versions of Python and numpy. While it's not immediately useful for people using up-to-date versions of numpy, it illustrates how things have changed since the first version of this answer.
A simple alternative: meshgrid + dstack
The currently accepted answer uses tile and repeat to broadcast two arrays together. But the meshgrid function does practically the same thing. Here's the output of tile and repeat before being passed to transpose:
In [1]: import numpy
In [2]: x = numpy.array([1,2,3])
...: y = numpy.array([4,5])
...:
In [3]: [numpy.tile(x, len(y)), numpy.repeat(y, len(x))]
Out[3]: [array([1, 2, 3, 1, 2, 3]), array([4, 4, 4, 5, 5, 5])]
And here's the output of meshgrid:
In [4]: numpy.meshgrid(x, y)
Out[4]:
[array([[1, 2, 3],
[1, 2, 3]]), array([[4, 4, 4],
[5, 5, 5]])]
As you can see, it's almost identical. We need only reshape the result to get exactly the same result.
In [5]: xt, xr = numpy.meshgrid(x, y)
...: [xt.ravel(), xr.ravel()]
Out[5]: [array([1, 2, 3, 1, 2, 3]), array([4, 4, 4, 5, 5, 5])]
Rather than reshaping at this point, though, we could pass the output of meshgrid to dstack and reshape afterwards, which saves some work:
In [6]: numpy.dstack(numpy.meshgrid(x, y)).reshape(-1, 2)
Out[6]:
array([[1, 4],
[2, 4],
[3, 4],
[1, 5],
[2, 5],
[3, 5]])
Contrary to the claim in this comment, I've seen no evidence that different inputs will produce differently shaped outputs, and as the above demonstrates, they do very similar things, so it would be quite strange if they did. Please let me know if you find a counterexample.
Testing meshgrid + dstack vs. repeat + transpose
The relative performance of these two approaches has changed over time. In an earlier version of Python (2.7), the result using meshgrid + dstack was noticeably faster for small inputs. (Note that these tests are from an old version of this answer.) Definitions:
>>> def repeat_product(x, y):
... return numpy.transpose([numpy.tile(x, len(y)),
numpy.repeat(y, len(x))])
...
>>> def dstack_product(x, y):
... return numpy.dstack(numpy.meshgrid(x, y)).reshape(-1, 2)
...
For moderately-sized input, I saw a significant speedup. But I retried these tests with more recent versions of Python (3.6.1) and numpy (1.12.1), on a newer machine. The two approaches are almost identical now.
Old Test
>>> x, y = numpy.arange(500), numpy.arange(500)
>>> %timeit repeat_product(x, y)
10 loops, best of 3: 62 ms per loop
>>> %timeit dstack_product(x, y)
100 loops, best of 3: 12.2 ms per loop
New Test
In [7]: x, y = numpy.arange(500), numpy.arange(500)
In [8]: %timeit repeat_product(x, y)
1.32 ms ± 24.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [9]: %timeit dstack_product(x, y)
1.26 ms ± 8.47 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
As always, YMMV, but this suggests that in recent versions of Python and numpy, these are interchangeable.
Generalized product functions
In general, we might expect that using built-in functions will be faster for small inputs, while for large inputs, a purpose-built function might be faster. Furthermore for a generalized n-dimensional product, tile and repeat won't help, because they don't have clear higher-dimensional analogues. So it's worth investigating the behavior of purpose-built functions as well.
Most of the relevant tests appear at the beginning of this answer, but here are a few of the tests performed on earlier versions of Python and numpy for comparison.
The cartesian function defined in another answer used to perform pretty well for larger inputs. (It's the same as the function called cartesian_product_recursive above.) In order to compare cartesian to dstack_prodct, we use just two dimensions.
Here again, the old test showed a significant difference, while the new test shows almost none.
Old Test
>>> x, y = numpy.arange(1000), numpy.arange(1000)
>>> %timeit cartesian([x, y])
10 loops, best of 3: 25.4 ms per loop
>>> %timeit dstack_product(x, y)
10 loops, best of 3: 66.6 ms per loop
New Test
In [10]: x, y = numpy.arange(1000), numpy.arange(1000)
In [11]: %timeit cartesian([x, y])
12.1 ms ± 199 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [12]: %timeit dstack_product(x, y)
12.7 ms ± 334 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
As before, dstack_product still beats cartesian at smaller scales.
New Test (redundant old test not shown)
In [13]: x, y = numpy.arange(100), numpy.arange(100)
In [14]: %timeit cartesian([x, y])
215 µs ± 4.75 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [15]: %timeit dstack_product(x, y)
65.7 µs ± 1.15 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
These distinctions are, I think, interesting and worth recording; but they are academic in the end. As the tests at the beginning of this answer showed, all of these versions are almost always slower than cartesian_product, defined at the very beginning of this answer -- which is itself a bit slower than the fastest implementations among the answers to this question.
>>> numpy.transpose([numpy.tile(x, len(y)), numpy.repeat(y, len(x))])
array([[1, 4],
[2, 4],
[3, 4],
[1, 5],
[2, 5],
[3, 5]])
See Using numpy to build an array of all combinations of two arrays for a general solution for computing the Cartesian product of N arrays.
You can just do normal list comprehension in python
x = numpy.array([1,2,3])
y = numpy.array([4,5])
[[x0, y0] for x0 in x for y0 in y]
which should give you
[[1, 4], [1, 5], [2, 4], [2, 5], [3, 4], [3, 5]]
I was interested in this as well and did a little performance comparison, perhaps somewhat clearer than in #senderle's answer.
For two arrays (the classical case):
For four arrays:
(Note that the length the arrays is only a few dozen entries here.)
Code to reproduce the plots:
from functools import reduce
import itertools
import numpy
import perfplot
def dstack_product(arrays):
return numpy.dstack(numpy.meshgrid(*arrays, indexing="ij")).reshape(-1, len(arrays))
# Generalized N-dimensional products
def cartesian_product(arrays):
la = len(arrays)
dtype = numpy.find_common_type([a.dtype for a in arrays], [])
arr = numpy.empty([len(a) for a in arrays] + [la], dtype=dtype)
for i, a in enumerate(numpy.ix_(*arrays)):
arr[..., i] = a
return arr.reshape(-1, la)
def cartesian_product_transpose(arrays):
broadcastable = numpy.ix_(*arrays)
broadcasted = numpy.broadcast_arrays(*broadcastable)
rows, cols = reduce(numpy.multiply, broadcasted[0].shape), len(broadcasted)
dtype = numpy.find_common_type([a.dtype for a in arrays], [])
out = numpy.empty(rows * cols, dtype=dtype)
start, end = 0, rows
for a in broadcasted:
out[start:end] = a.reshape(-1)
start, end = end, end + rows
return out.reshape(cols, rows).T
# from https://stackoverflow.com/a/1235363/577088
def cartesian_product_recursive(arrays, out=None):
arrays = [numpy.asarray(x) for x in arrays]
dtype = arrays[0].dtype
n = numpy.prod([x.size for x in arrays])
if out is None:
out = numpy.zeros([n, len(arrays)], dtype=dtype)
m = n // arrays[0].size
out[:, 0] = numpy.repeat(arrays[0], m)
if arrays[1:]:
cartesian_product_recursive(arrays[1:], out=out[0:m, 1:])
for j in range(1, arrays[0].size):
out[j * m : (j + 1) * m, 1:] = out[0:m, 1:]
return out
def cartesian_product_itertools(arrays):
return numpy.array(list(itertools.product(*arrays)))
perfplot.show(
setup=lambda n: 2 * (numpy.arange(n, dtype=float),),
n_range=[2 ** k for k in range(13)],
# setup=lambda n: 4 * (numpy.arange(n, dtype=float),),
# n_range=[2 ** k for k in range(6)],
kernels=[
dstack_product,
cartesian_product,
cartesian_product_transpose,
cartesian_product_recursive,
cartesian_product_itertools,
],
logx=True,
logy=True,
xlabel="len(a), len(b)",
equality_check=None,
)
Building on #senderle's exemplary ground work I've come up with two versions - one for C and one for Fortran layouts - that are often a bit faster.
cartesian_product_transpose_pp is - unlike #senderle's cartesian_product_transpose which uses a different strategy altogether - a version of cartesion_product that uses the more favorable transpose memory layout + some very minor optimizations.
cartesian_product_pp sticks with the original memory layout. What makes it fast is its using contiguous copying. Contiguous copies turn out to be so much faster that copying a full block of memory even though only part of it contains valid data is preferable to only copying the valid bits.
Some perfplots. I made separate ones for C and Fortran layouts, because these are different tasks IMO.
Names ending in 'pp' are my approaches.
1) many tiny factors (2 elements each)
2) many small factors (4 elements each)
3) three factors of equal length
4) two factors of equal length
Code (need to do separate runs for each plot b/c I couldn't figure out how to reset; also need to edit / comment in / out appropriately):
import numpy
import numpy as np
from functools import reduce
import itertools
import timeit
import perfplot
def dstack_product(arrays):
return numpy.dstack(
numpy.meshgrid(*arrays, indexing='ij')
).reshape(-1, len(arrays))
def cartesian_product_transpose_pp(arrays):
la = len(arrays)
dtype = numpy.result_type(*arrays)
arr = numpy.empty((la, *map(len, arrays)), dtype=dtype)
idx = slice(None), *itertools.repeat(None, la)
for i, a in enumerate(arrays):
arr[i, ...] = a[idx[:la-i]]
return arr.reshape(la, -1).T
def cartesian_product(arrays):
la = len(arrays)
dtype = numpy.result_type(*arrays)
arr = numpy.empty([len(a) for a in arrays] + [la], dtype=dtype)
for i, a in enumerate(numpy.ix_(*arrays)):
arr[...,i] = a
return arr.reshape(-1, la)
def cartesian_product_transpose(arrays):
broadcastable = numpy.ix_(*arrays)
broadcasted = numpy.broadcast_arrays(*broadcastable)
rows, cols = numpy.prod(broadcasted[0].shape), len(broadcasted)
dtype = numpy.result_type(*arrays)
out = numpy.empty(rows * cols, dtype=dtype)
start, end = 0, rows
for a in broadcasted:
out[start:end] = a.reshape(-1)
start, end = end, end + rows
return out.reshape(cols, rows).T
from itertools import accumulate, repeat, chain
def cartesian_product_pp(arrays, out=None):
la = len(arrays)
L = *map(len, arrays), la
dtype = numpy.result_type(*arrays)
arr = numpy.empty(L, dtype=dtype)
arrs = *accumulate(chain((arr,), repeat(0, la-1)), np.ndarray.__getitem__),
idx = slice(None), *itertools.repeat(None, la-1)
for i in range(la-1, 0, -1):
arrs[i][..., i] = arrays[i][idx[:la-i]]
arrs[i-1][1:] = arrs[i]
arr[..., 0] = arrays[0][idx]
return arr.reshape(-1, la)
def cartesian_product_itertools(arrays):
return numpy.array(list(itertools.product(*arrays)))
# from https://stackoverflow.com/a/1235363/577088
def cartesian_product_recursive(arrays, out=None):
arrays = [numpy.asarray(x) for x in arrays]
dtype = arrays[0].dtype
n = numpy.prod([x.size for x in arrays])
if out is None:
out = numpy.zeros([n, len(arrays)], dtype=dtype)
m = n // arrays[0].size
out[:, 0] = numpy.repeat(arrays[0], m)
if arrays[1:]:
cartesian_product_recursive(arrays[1:], out=out[0:m, 1:])
for j in range(1, arrays[0].size):
out[j*m:(j+1)*m, 1:] = out[0:m, 1:]
return out
### Test code ###
if False:
perfplot.save('cp_4el_high.png',
setup=lambda n: n*(numpy.arange(4, dtype=float),),
n_range=list(range(6, 11)),
kernels=[
dstack_product,
cartesian_product_recursive,
cartesian_product,
# cartesian_product_transpose,
cartesian_product_pp,
# cartesian_product_transpose_pp,
],
logx=False,
logy=True,
xlabel='#factors',
equality_check=None
)
else:
perfplot.save('cp_2f_T.png',
setup=lambda n: 2*(numpy.arange(n, dtype=float),),
n_range=[2**k for k in range(5, 11)],
kernels=[
# dstack_product,
# cartesian_product_recursive,
# cartesian_product,
cartesian_product_transpose,
# cartesian_product_pp,
cartesian_product_transpose_pp,
],
logx=True,
logy=True,
xlabel='length of each factor',
equality_check=None
)
As of Oct. 2017, numpy now has a generic np.stack function that takes an axis parameter. Using it, we can have a "generalized cartesian product" using the "dstack and meshgrid" technique:
import numpy as np
def cartesian_product(*arrays):
ndim = len(arrays)
return (np.stack(np.meshgrid(*arrays), axis=-1)
.reshape(-1, ndim))
a = np.array([1,2])
b = np.array([10,20])
cartesian_product(a,b)
# output:
# array([[ 1, 10],
# [ 2, 10],
# [ 1, 20],
# [ 2, 20]])
Note on the axis=-1 parameter. This is the last (inner-most) axis in the result. It is equivalent to using axis=ndim.
One other comment, since Cartesian products blow up very quickly, unless we need to realize the array in memory for some reason, if the product is very large, we may want to make use of itertools and use the values on-the-fly.
The Scikit-learn package has a fast implementation of exactly this:
from sklearn.utils.extmath import cartesian
product = cartesian((x,y))
Note that the convention of this implementation is different from what you want, if you care about the order of the output. For your exact ordering, you can do
product = cartesian((y,x))[:, ::-1]
I used #kennytm answer for a while, but when trying to do the same in TensorFlow, but I found that TensorFlow has no equivalent of numpy.repeat(). After a little experimentation, I think I found a more general solution for arbitrary vectors of points.
For numpy:
import numpy as np
def cartesian_product(*args: np.ndarray) -> np.ndarray:
"""
Produce the cartesian product of arbitrary length vectors.
Parameters
----------
np.ndarray args
vector of points of interest in each dimension
Returns
-------
np.ndarray
the cartesian product of size [m x n] wherein:
m = prod([len(a) for a in args])
n = len(args)
"""
for i, a in enumerate(args):
assert a.ndim == 1, "arg {:d} is not rank 1".format(i)
return np.concatenate([np.reshape(xi, [-1, 1]) for xi in np.meshgrid(*args)], axis=1)
and for TensorFlow:
import tensorflow as tf
def cartesian_product(*args: tf.Tensor) -> tf.Tensor:
"""
Produce the cartesian product of arbitrary length vectors.
Parameters
----------
tf.Tensor args
vector of points of interest in each dimension
Returns
-------
tf.Tensor
the cartesian product of size [m x n] wherein:
m = prod([len(a) for a in args])
n = len(args)
"""
for i, a in enumerate(args):
tf.assert_rank(a, 1, message="arg {:d} is not rank 1".format(i))
return tf.concat([tf.reshape(xi, [-1, 1]) for xi in tf.meshgrid(*args)], axis=1)
More generally, if you have two 2d numpy arrays a and b, and you want to concatenate every row of a to every row of b (A cartesian product of rows, kind of like a join in a database), you can use this method:
import numpy
def join_2d(a, b):
assert a.dtype == b.dtype
a_part = numpy.tile(a, (len(b), 1))
b_part = numpy.repeat(b, len(a), axis=0)
return numpy.hstack((a_part, b_part))
The fastest you can get is either by combining a generator expression with the map function:
import numpy
import datetime
a = np.arange(1000)
b = np.arange(200)
start = datetime.datetime.now()
foo = (item for sublist in [list(map(lambda x: (x,i),a)) for i in b] for item in sublist)
print (list(foo))
print ('execution time: {} s'.format((datetime.datetime.now() - start).total_seconds()))
Outputs (actually the whole resulting list is printed):
[(0, 0), (1, 0), ...,(998, 199), (999, 199)]
execution time: 1.253567 s
or by using a double generator expression:
a = np.arange(1000)
b = np.arange(200)
start = datetime.datetime.now()
foo = ((x,y) for x in a for y in b)
print (list(foo))
print ('execution time: {} s'.format((datetime.datetime.now() - start).total_seconds()))
Outputs (whole list printed):
[(0, 0), (1, 0), ...,(998, 199), (999, 199)]
execution time: 1.187415 s
Take into account that most of the computation time goes into the printing command. The generator calculations are otherwise decently efficient. Without printing the calculation times are:
execution time: 0.079208 s
for generator expression + map function and:
execution time: 0.007093 s
for the double generator expression.
If what you actually want is to calculate the actual product of each of the coordinate pairs, the fastest is to solve it as a numpy matrix product:
a = np.arange(1000)
b = np.arange(200)
start = datetime.datetime.now()
foo = np.dot(np.asmatrix([[i,0] for i in a]), np.asmatrix([[i,0] for i in b]).T)
print (foo)
print ('execution time: {} s'.format((datetime.datetime.now() - start).total_seconds()))
Outputs:
[[ 0 0 0 ..., 0 0 0]
[ 0 1 2 ..., 197 198 199]
[ 0 2 4 ..., 394 396 398]
...,
[ 0 997 1994 ..., 196409 197406 198403]
[ 0 998 1996 ..., 196606 197604 198602]
[ 0 999 1998 ..., 196803 197802 198801]]
execution time: 0.003869 s
and without printing (in this case it doesn't save much since only a tiny piece of the matrix is actually printed out):
execution time: 0.003083 s
This can also be easily done by using itertools.product method
from itertools import product
import numpy as np
x = np.array([1, 2, 3])
y = np.array([4, 5])
cart_prod = np.array(list(product(*[x, y])),dtype='int32')
Result:
array([[1, 4],
[1, 5],
[2, 4],
[2, 5],
[3, 4],
[3, 5]], dtype=int32)
Execution time: 0.000155 s
In the specific case that you need to perform simple operations such as addition on each pair, you can introduce an extra dimension and let broadcasting do the job:
>>> a, b = np.array([1,2,3]), np.array([10,20,30])
>>> a[None,:] + b[:,None]
array([[11, 12, 13],
[21, 22, 23],
[31, 32, 33]])
I'm not sure if there is any similar way to actually get the pairs themselves.
I'm a bit late to the party, but I encoutered a tricky variant of that problem.
Let's say I want the cartesian product of several arrays, but that cartesian product ends up being much larger than the computers' memory (however, the computation done with that product are fast, or at least parallelizable).
The obvious solution is to divide this cartesian product in chunks, and treat these chunks one after the other (in sort of a "streaming" manner). You can do that easily with itertools.product, but it's horrendously slow. Also, none of the proposed solutions here (as fast as they are) give us this possibility. The solution I propose uses Numba, and is slightly faster than the "canonical" cartesian_product mentioned here. It's pretty long because I tried to optimize it everywhere I could.
import numba as nb
import numpy as np
from typing import List
#nb.njit(nb.types.Tuple((nb.int32[:, :],
nb.int32[:]))(nb.int32[:],
nb.int32[:],
nb.int64, nb.int64))
def cproduct(sizes: np.ndarray, current_tuple: np.ndarray, start_idx: int, end_idx: int):
"""Generates ids tuples from start_id to end_id"""
assert len(sizes) >= 2
assert start_idx < end_idx
tuples = np.zeros((end_idx - start_idx, len(sizes)), dtype=np.int32)
tuple_idx = 0
# stores the current combination
current_tuple = current_tuple.copy()
while tuple_idx < end_idx - start_idx:
tuples[tuple_idx] = current_tuple
current_tuple[0] += 1
# using a condition here instead of including this in the inner loop
# to gain a bit of speed: this is going to be tested each iteration,
# and starting a loop to have it end right away is a bit silly
if current_tuple[0] == sizes[0]:
# the reset to 0 and subsequent increment amount to carrying
# the number to the higher "power"
current_tuple[0] = 0
current_tuple[1] += 1
for i in range(1, len(sizes) - 1):
if current_tuple[i] == sizes[i]:
# same as before, but in a loop, since this is going
# to get called less often
current_tuple[i + 1] += 1
current_tuple[i] = 0
else:
break
tuple_idx += 1
return tuples, current_tuple
def chunked_cartesian_product_ids(sizes: List[int], chunk_size: int):
"""Just generates chunks of the cartesian product of the ids of each
input arrays (thus, we just need their sizes here, not the actual arrays)"""
prod = np.prod(sizes)
# putting the largest number at the front to more efficiently make use
# of the cproduct numba function
sizes = np.array(sizes, dtype=np.int32)
sorted_idx = np.argsort(sizes)[::-1]
sizes = sizes[sorted_idx]
if chunk_size > prod:
chunk_bounds = (np.array([0, prod])).astype(np.int64)
else:
num_chunks = np.maximum(np.ceil(prod / chunk_size), 2).astype(np.int32)
chunk_bounds = (np.arange(num_chunks + 1) * chunk_size).astype(np.int64)
chunk_bounds[-1] = prod
current_tuple = np.zeros(len(sizes), dtype=np.int32)
for start_idx, end_idx in zip(chunk_bounds[:-1], chunk_bounds[1:]):
tuples, current_tuple = cproduct(sizes, current_tuple, start_idx, end_idx)
# re-arrange columns to match the original order of the sizes list
# before yielding
yield tuples[:, np.argsort(sorted_idx)]
def chunked_cartesian_product(*arrays, chunk_size=2 ** 25):
"""Returns chunks of the full cartesian product, with arrays of shape
(chunk_size, n_arrays). The last chunk will obviously have the size of the
remainder"""
array_lengths = [len(array) for array in arrays]
for array_ids_chunk in chunked_cartesian_product_ids(array_lengths, chunk_size):
slices_lists = [arrays[i][array_ids_chunk[:, i]] for i in range(len(arrays))]
yield np.vstack(slices_lists).swapaxes(0,1)
def cartesian_product(*arrays):
"""Actual cartesian product, not chunked, still fast"""
total_prod = np.prod([len(array) for array in arrays])
return next(chunked_cartesian_product(*arrays, total_prod))
a = np.arange(0, 3)
b = np.arange(8, 10)
c = np.arange(13, 16)
for cartesian_tuples in chunked_cartesian_product(*[a, b, c], chunk_size=5):
print(cartesian_tuples)
This would output our cartesian product in chunks of 5 3-uples:
[[ 0 8 13]
[ 0 8 14]
[ 0 8 15]
[ 1 8 13]
[ 1 8 14]]
[[ 1 8 15]
[ 2 8 13]
[ 2 8 14]
[ 2 8 15]
[ 0 9 13]]
[[ 0 9 14]
[ 0 9 15]
[ 1 9 13]
[ 1 9 14]
[ 1 9 15]]
[[ 2 9 13]
[ 2 9 14]
[ 2 9 15]]
If you're willing to understand what is being done here, the intuition behind the njitted function is to enumerate each "number" in a weird numerical base whose elements would be composed of the sizes of the input arrays (instead of the same number in regular binary, decimal or hexadecimal bases).
Obviously, this solution is interesting for large products. For small ones, the overhead might be a bit costly.
NOTE: since numba is still under heavy development, i'm using numba 0.50 to run this, with python 3.6.
Yet another one:
>>>x1, y1 = np.meshgrid(x, y)
>>>np.c_[x1.ravel(), y1.ravel()]
array([[1, 4],
[2, 4],
[3, 4],
[1, 5],
[2, 5],
[3, 5]])
Inspired by Ashkan's answer, you can also try the following.
>>> x, y = np.meshgrid(x, y)
>>> np.concatenate([x.flatten().reshape(-1,1), y.flatten().reshape(-1,1)], axis=1)
This will give you the required cartesian product!
This is a generalized version of the accepted answer (Cartesian product of multiple arrays using numpy.tile and numpy.repeat functions).
from functors import reduce
from operator import mul
def cartesian_product(arrays):
return np.vstack(
np.tile(
np.repeat(arrays[j], reduce(mul, map(len, arrays[j+1:]), 1)),
reduce(mul, map(len, arrays[:j]), 1),
)
for j in range(len(arrays))
).T
If you are willing to use PyTorch, I should think it is highly efficient:
>>> import torch
>>> torch.cartesian_prod(torch.as_tensor(x), torch.as_tensor(y))
tensor([[1, 4],
[1, 5],
[2, 4],
[2, 5],
[3, 4],
[3, 5]])
and you can easily get a numpy array:
>>> torch.cartesian_prod(torch.as_tensor(x), torch.as_tensor(y)).numpy()
array([[1, 4],
[1, 5],
[2, 4],
[2, 5],
[3, 4],
[3, 5]])
Related
To be clear, below is what I am trying to do. And the question is, how can I change the function oper_AB() so that instead of the nested for loop, I utilize the vectorization/broadcasting in numpy and get to the ret_list much faster?
def oper(a_1D, b_1D):
return np.dot(a_1D, b_1D) / np.dot(b_1D, b_1D)
def oper_AB(A_2D, B_2D):
ret_list = []
for a_1D in A_2D:
for b_1D in B_2D:
ret_list.append(oper(a_1D, b_1D))
return ret_list
Strictly addressing the question (with the reservation that I suspect the OP wants the norm, not the norm squared, as divisor below):
r = a # b.T / np.linalg.norm(b, axis=1)**2
Example:
np.random.seed(0)
a = np.random.randint(0, 10, size=(2,2))
b = np.random.randint(0, 10, size=(2,2))
Then:
>>> a
array([[5, 0],
[3, 3]])
>>> b
array([[7, 9],
[3, 5]])
>>> oper_AB(a, b)
[0.2692307692307692,
0.4411764705882353,
0.36923076923076925,
0.7058823529411765]
>>> a # b.T / np.linalg.norm(b, axis=1)**2
array([[0.26923077, 0.44117647],
[0.36923077, 0.70588235]])
>>> np.ravel(a # b.T / np.linalg.norm(b, axis=1)**2)
array([0.26923077, 0.44117647, 0.36923077, 0.70588235])
Speed:
n, m = 1000, 100
a = np.random.uniform(size=(n, m))
b = np.random.uniform(size=(n, m))
orig = %timeit -o oper_AB(a, b)
# 2.73 s ± 11 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
new = %timeit -o np.ravel(a # b.T / np.linalg.norm(b, axis=1)**2)
# 2.22 ms ± 33.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
orig.average / new.average
# 1228.78 (speedup)
Our solution is 1200x faster than the original.
Correctness:
>>> np.allclose(np.ravel(a # b.T / np.linalg.norm(b, axis=1)**2), oper_AB(a, b))
True
Speed on large array, comparison to #Ahmed AEK's solution:
n, m = 2000, 2000
a = np.random.uniform(size=(n, m))
b = np.random.uniform(size=(n, m))
new = %timeit -o np.ravel(a # b.T / np.linalg.norm(b, axis=1)**2)
# 86.5 ms ± 484 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
other = %timeit -o AEK(a, b) # Ahmed AEK's answer
# 102 ms ± 379 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
Our solution is 15% faster :-)
this should work.
result = (np.matmul(A_2D, B_2D.transpose())/np.sum(B_2D*B_2D,axis=1)).flatten()
but this second implementation will be faster because of cache utilization.
def oper_AB(A_2D, B_2D):
b_squared = np.sum(B_2D*B_2D,axis=1).reshape([-1,1])
b_normalized = B_2D/b_squared
del b_squared
returned_val = np.matmul(A_2D,b_normalized.transpose())
return returned_val.flatten()
the del is there just if the memory allocated by B_2D is too big, (or it's just me being used to working with multiple GB arrays)
Edit: as requested for A_1D - B_1D
def oper2_AB(A_2D, B_2D):
output = np.zeros([A_2D.shape[0]*B_2D.shape[0],A_2D.shape[1]],dtype=A_2D.dtype)
for i in range(len(A_2D)):
output[i*len(B_2D):(i+1)*len(B_2D)] = A_2D[i]-B_2D
return output
I have a 1D numpy array X with the shape (1000,). I want to inject in random (uniform) places 10 random (normal) values and thus obtain the numpy array of shape (1010,). How to do it efficiently in numpy?
You can use np.insert together with np.random.choice:
n = 10
np.insert(a, np.random.choice(len(a), size=n), np.random.normal(size=n))
Here's one based on masking -
def addrand(a, N):
n = len(a)
m = np.concatenate((np.ones(n, dtype=bool), np.zeros(N, dtype=bool)))
np.random.shuffle(m)
out = np.empty(len(a)+N, dtype=a.dtype)
out[m] = a
out[~m] = np.random.uniform(N)
return out
Sample run -
In [22]: a = 10+np.random.rand(20)
In [23]: a
Out[23]:
array([10.65458302, 10.18034826, 10.08652451, 10.03342622, 10.63930492,
10.48439184, 10.2859206 , 10.91419282, 10.56905636, 10.01595702,
10.21063965, 10.23080433, 10.90546147, 10.02823502, 10.67987108,
10.00583747, 10.24664158, 10.78030108, 10.33638157, 10.32471524])
In [24]: addrand(a, N=3) # adding 3 rand numbers
Out[24]:
array([10.65458302, 10.18034826, 10.08652451, 10.03342622, 0.79989563,
10.63930492, 10.48439184, 10.2859206 , 10.91419282, 10.56905636,
10.01595702, 0.23873077, 10.21063965, 10.23080433, 10.90546147,
10.02823502, 0.66857723, 10.67987108, 10.00583747, 10.24664158,
10.78030108, 10.33638157, 10.32471524])
Timings :
In [71]: a = np.random.rand(1000)
In [72]: %timeit addrand(a, N=10)
37.3 µs ± 273 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
# #a_guest's soln
In [73]: %timeit np.insert(a, np.random.choice(len(a), size=10), np.random.normal(size=10))
63.3 µs ± 2.18 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Note: If you are working with bigger arrays, it seems np.insert one is doing better.
You could use numpy.insert(arr, obj, values, axis=None).
import numpy as np
a = np.arange(1000)
a = np.insert(a, np.random.randint(low = 1, high = 999, size=10), np.random.normal(loc=0.0, scale=1.0, size=10))
Keep in mind that insert doesn't automatically change your original array, but it returns a modified copy.
Not sure if this is the most efficient way, but it works, at least.
A = np.arange(1000)
for i in np.random.randint(low = 0, high = 1000, size = 10):
A = np.concatenate((A[:i], [np.random.normal(),], A[i:]))
Edit, checking performance:
def insert_random(A):
for i in np.random.randint(low = 0, high = len(A), size = 10):
A = np.concatenate((A[:i], [np.random.normal(),], A[i:]))
return A
A = np.arange(1000)
%timeit test(A)
83.2 µs ± 2.47 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
So definitely not the most efficient. np.insert seems to be the way to go.
Say I have an array of distances x=[1,2,1,3,3,2,1,5,1,1].
I want to get the indices from x where cumsum reaches 10, in this case, idx=[4,9].
So the cumsum restarts after the condition are met.
I can do it with a loop, but loops are slow for large arrays and I was wondering if I could do it in a vectorized way.
A fun method
sumlm = np.frompyfunc(lambda a,b:a+b if a < 10 else b,2,1)
newx=sumlm.accumulate(x, dtype=np.object)
newx
array([1, 3, 4, 7, 10, 2, 3, 8, 9, 10], dtype=object)
np.nonzero(newx==10)
(array([4, 9]),)
Here's one with numba and array-initialization -
from numba import njit
#njit
def cumsum_breach_numba2(x, target, result):
total = 0
iterID = 0
for i,x_i in enumerate(x):
total += x_i
if total >= target:
result[iterID] = i
iterID += 1
total = 0
return iterID
def cumsum_breach_array_init(x, target):
x = np.asarray(x)
result = np.empty(len(x),dtype=np.uint64)
idx = cumsum_breach_numba2(x, target, result)
return result[:idx]
Timings
Including #piRSquared's solutions and using the benchmarking setup from the same post -
In [58]: np.random.seed([3, 1415])
...: x = np.random.randint(100, size=1000000).tolist()
# #piRSquared soln1
In [59]: %timeit list(cumsum_breach(x, 10))
10 loops, best of 3: 73.2 ms per loop
# #piRSquared soln2
In [60]: %timeit cumsum_breach_numba(np.asarray(x), 10)
10 loops, best of 3: 69.2 ms per loop
# From this post
In [61]: %timeit cumsum_breach_array_init(x, 10)
10 loops, best of 3: 39.1 ms per loop
Numba : Appending vs. array-initialization
For a closer look at how the array-initialization helps, which seems be the big difference between the two numba implementations, let's time these on the array data, as the array data creation was in itself heavy on runtime and they both depend on it -
In [62]: x = np.array(x)
In [63]: %timeit cumsum_breach_numba(x, 10)# with appending
10 loops, best of 3: 31.5 ms per loop
In [64]: %timeit cumsum_breach_array_init(x, 10)
1000 loops, best of 3: 1.8 ms per loop
To force the output to have it own memory space, we can make a copy. Won't change the things in a big way though -
In [65]: %timeit cumsum_breach_array_init(x, 10).copy()
100 loops, best of 3: 2.67 ms per loop
Loops are not always bad (especially when you need one). Also, There is no tool or algorithm that will make this quicker than O(n). So let's just make a good loop.
Generator Function
def cumsum_breach(x, target):
total = 0
for i, y in enumerate(x):
total += y
if total >= target:
yield i
total = 0
list(cumsum_breach(x, 10))
[4, 9]
Just In Time compiling with Numba
Numba is a third party library that needs to be installed.
Numba can be persnickety about what features are supported. But this works.
Also, as pointed out by Divakar, Numba performs better with arrays
from numba import njit
#njit
def cumsum_breach_numba(x, target):
total = 0
result = []
for i, y in enumerate(x):
total += y
if total >= target:
result.append(i)
total = 0
return result
cumsum_breach_numba(x, 10)
Testing the Two
Because I felt like it ¯\_(ツ)_/¯
Setup
np.random.seed([3, 1415])
x0 = np.random.randint(100, size=1_000_000)
x1 = x0.tolist()
Accuracy
i0 = cumsum_breach_numba(x0, 200_000)
i1 = list(cumsum_breach(x1, 200_000))
assert i0 == i1
Time
%timeit cumsum_breach_numba(x0, 200_000)
%timeit list(cumsum_breach(x1, 200_000))
582 µs ± 40.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
64.3 ms ± 5.66 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
Numba was on the order of 100 times faster.
For a more true apples to apples test, I convert a list to a Numpy array
%timeit cumsum_breach_numba(np.array(x1), 200_000)
%timeit list(cumsum_breach(x1, 200_000))
43.1 ms ± 202 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
62.8 ms ± 327 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
Which brings them to about even.
I have the following np.ndarray:
>>> arr
array([[1, 2],
[3, 4]])
I would like to split it to X and y where each array is coordinates and values, respectively. So far I managed to solve this using np.ndenumerate:
>>> X, y = zip(*np.ndenumerate(arr))
>>> X
((0, 0), (0, 1), (1, 0), (1, 1))
>>> y
(1, 2, 3, 4)
I'm wondering if there's a more idiomatic and faster way to achieve it, since the arrays I'm actually dealing with have millions of values.
I need the X and y array to pass them to a sklearn classifier later. The formats above seemed the most natural for me, but perhaps there's a better way I can pass them to the fit function.
Reshaping arr to y is easy, you can achieve it by y = arr.flatten(). I suggest treating generating X as a separate task.
Let's assume that your dataset is of shape NxM. In our benchmark we set N to 500 and M to 1000.
N = 500
M = 1000
arr = np.random.randn(N, M)
Then by using np.mgrid and transforming indices you can get the result as:
np.mgrid[:N, :M].transpose(1, 2, 0).reshape(-1, 2)
Benchmarks:
%timeit np.mgrid[:N, :M].transpose(1, 2, 0).reshape(-1, 2)
# 3.11 ms ± 35.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit zip(*np.ndenumerate(arr))
# 235 ms ± 1.57 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In your case you can unpack and get N and M by:
N, M = arr.shape
and then:
X = np.mgrid[:N, :M].transpose(1, 2, 0).reshape(-1, 2)
Use numpy.where with numpy.ravel():
import numpy as np
def ndenumerate(np_array):
return list(zip(*np.where(np_array+1))), np_array.ravel()
arr = np.random.randint(0, 100, (1000,1000))
X_new, y_new = ndenumerate(arr)
X,y = zip(*np.ndenumerate(arr))
Output (validation):
all(i1 == i2 for i1, i2 in zip(X, X_new))
# True
all(y == y_new)
# True
Benchmark (about 3x faster):
%timeit ndenumerate(arr)
# 234 ms ± 20.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit zip(*np.ndenumerate(arr))
# 877 ms ± 91.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
I want to sum a 2 dimensional array in python:
Here is what I have:
def sum1(input):
sum = 0
for row in range (len(input)-1):
for col in range(len(input[0])-1):
sum = sum + input[row][col]
return sum
print sum1([[1, 2],[3, 4],[5, 6]])
It displays 4 instead of 21 (1+2+3+4+5+6 = 21). Where is my mistake?
I think this is better:
>>> x=[[1, 2],[3, 4],[5, 6]]
>>> sum(sum(x,[]))
21
You could rewrite that function as,
def sum1(input):
return sum(map(sum, input))
Basically, map(sum, input) will return a list with the sums across all your rows, then, the outer most sum will add up that list.
Example:
>>> a=[[1,2],[3,4]]
>>> sum(map(sum, a))
10
This is yet another alternate Solution
In [1]: a=[[1, 2],[3, 4],[5, 6]]
In [2]: sum([sum(i) for i in a])
Out[2]: 21
And numpy solution is just:
import numpy as np
x = np.array([[1, 2],[3, 4],[5, 6]])
Result:
>>> b=np.sum(x)
print(b)
21
Better still, forget the index counters and just iterate over the items themselves:
def sum1(input):
my_sum = 0
for row in input:
my_sum += sum(row)
return my_sum
print sum1([[1, 2],[3, 4],[5, 6]])
One of the nice (and idiomatic) features of Python is letting it do the counting for you. sum() is a built-in and you should not use names of built-ins for your own identifiers.
This is the issue
for row in range (len(input)-1):
for col in range(len(input[0])-1):
try
for row in range (len(input)):
for col in range(len(input[0])):
Python's range(x) goes from 0..x-1 already
range(...)
range([start,] stop[, step]) -> list of integers
Return a list containing an arithmetic progression of integers.
range(i, j) returns [i, i+1, i+2, ..., j-1]; start (!) defaults to 0.
When step is given, it specifies the increment (or decrement).
For example, range(4) returns [0, 1, 2, 3]. The end point is omitted!
These are exactly the valid indices for a list of 4 elements.
range() in python excludes the last element. In other words, range(1, 5) is [1, 5) or [1, 4]. So you should just use len(input) to iterate over the rows/columns.
def sum1(input):
sum = 0
for row in range (len(input)):
for col in range(len(input[0])):
sum = sum + input[row][col]
return sum
Don't put -1 in range(len(input)-1) instead use:
range(len(input))
range automatically returns a list one less than the argument value so no need of explicitly giving -1
def sum1(input):
return sum([sum(x) for x in input])
Quick answer, use...
total = sum(map(sum,[array]))
where [array] is your array title.
In Python 3.7
import numpy as np
x = np.array([ [1,2], [3,4] ])
sum(sum(x))
outputs
10
It seems like a general consensus is that numpy is a complicated solution. In comparison to simpler algorithms. But for the sake of the answer being present:
import numpy as np
def addarrays(arr):
b = np.sum(arr)
return sum(b)
array_1 = [
[1, 2],
[3, 4],
[5, 6]
]
print(addarrays(array_1))
This appears to be the preferred solution:
x=[[1, 2],[3, 4],[5, 6]]
sum(sum(x,[]))
def sum1(input):
sum = 0
for row in input:
for col in row:
sum += col
return sum
print(sum1([[1, 2],[3, 4],[5, 6]]))
Speed comparison
import random
import timeit
import numpy
x = [[random.random() for i in range(100)] for j in range(100)]
xnp = np.array(x)
Methods
print("Sum python array:")
%timeit sum(map(sum,x))
%timeit sum([sum(i) for i in x])
%timeit sum(sum(x,[]))
%timeit sum([x[i][j] for i in range(100) for j in range(100)])
print("Convert to numpy, then sum:")
%timeit np.sum(np.array(x))
%timeit sum(sum(np.array(x)))
print("Sum numpy array:")
%timeit np.sum(xnp)
%timeit sum(sum(xnp))
Results
Sum python array:
130 µs ± 3.24 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
149 µs ± 4.16 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
3.05 ms ± 44.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.58 ms ± 107 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Convert to numpy, then sum:
1.36 ms ± 90.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
1.63 ms ± 26.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Sum numpy array:
24.6 µs ± 1.95 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
301 µs ± 4.78 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
def sum1(input):
sum = 0
for row in range (len(input)-1):
for col in range(len(input[0])-1):
sum = sum + input[row][col]
return sum
print (sum1([[1, 2],[3, 4],[5, 6]]))
You had a problem with parenthesis at the print command....
This solution will be good now
The correct solution in Visual Studio Code