let's assume I have an array of form
a = [
[ [12, 4, 2, 5] , [10, 12, 4, 2], [2, 2, 10, 2], [7, 10, 3 ,2] ],
[ [13, 23, 4, 5] , [10, 12, 4, 2], [2, 2, 12, 1], [7, 10, 3 ,2] ],
[ [4, 4, 2, 5] , [10, 12, 4, 2], [2, 2, 3, 2], [7, 10, 3 ,2] ],
]
I want to sort by the first column of each element, so the first entry of a would read:
[ [2, 2, 10, 2], [7, 10, 3 ,2], [10, 12, 4, 2], [12, 4, 2, 5] ]
I've found a solution that looks like a[a[:,1].argsort()] and tried a for loop over the first index like:
for i in range(0,4):
a[i,...] = a[a[i,0,:].argsort()]
or
for i in range(0,4):
a[i,...] = a[i, a[i,0,:].argsort()]
This doesn't work tho. I really can't get my head around this problem.
You can use argsort of the array indexed along the first columns, and expanding dims appropriately, and then np.take_along_axis:
ixs = a[...,0].argsort(1)[...,None]
np.take_along_axis(a, ixs, 1)
array([[[ 2, 2, 10, 2],
[ 7, 10, 3, 2],
[10, 12, 4, 2],
[12, 4, 2, 5]],
[[ 2, 2, 12, 1],
[ 7, 10, 3, 2],
[10, 12, 4, 2],
[13, 23, 4, 5]],
[[ 2, 2, 3, 2],
[ 4, 4, 2, 5],
[ 7, 10, 3, 2],
[10, 12, 4, 2]]])
Try:
sorted_array = [sorted(x, key=lambda y: y[0]) for x in a]
How does this work?
first we use list comprehension to map the array a into a new one, performing an operation on each element (e.g. [x + 1 for x in my_numbers])
The operation we perform on each element of a is to use sorted. This returns a copy of an array, sorted by comparing whatever value is returned by key.
key must be a function that returns a 'sortable' value. Here, we're using a lambda to create a one-line function that returns the first item from the array passed to it. This ensures that we sort based on the first column of each element of a.
If you don’t want to use lambda method try this one:
def select(item):
return item[0]
s_array=[]
for i in a:
s_array.append(sorted(i, key=select))
I have a numpy array x (with (n,4) shape) of integers like:
[[0 1 2 3],
[1 2 7 9],
[2 1 5 2],
...]
I want to transform the array into an array of pairs:
[0,1]
[0,2]
[0,3]
[1,2]
...
so first element makes a pair with other elements in the same sub-array. I have already a for-loop solution:
y=np.array([[x[j,0],x[j,i]] for i in range(1,4) for j in range(0,n)],dtype=int)
but since looping over numpy array is not efficient, I tried slicing as the solution. I can do the slicing for every column as:
y[1]=np.array([x[:,0],x[:,1]]).T
# [[0,1],[1,2],[2,1],...]
I can repeat this for all columns. My questions are:
How can I append y[2] to y[1],... such that the shape is (N,2)?
If number of columns is not small (in this example 4), how can I find y[i] elegantly?
What are the alternative ways to achieve the final array?
The cleanest way of doing this I can think of would be:
>>> x = np.arange(12).reshape(3, 4)
>>> x
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
>>> n = x.shape[1] - 1
>>> y = np.repeat(x, (n,)+(1,)*n, axis=1)
>>> y
array([[ 0, 0, 0, 1, 2, 3],
[ 4, 4, 4, 5, 6, 7],
[ 8, 8, 8, 9, 10, 11]])
>>> y.reshape(-1, 2, n).transpose(0, 2, 1).reshape(-1, 2)
array([[ 0, 1],
[ 0, 2],
[ 0, 3],
[ 4, 5],
[ 4, 6],
[ 4, 7],
[ 8, 9],
[ 8, 10],
[ 8, 11]])
This will make two copies of the data, so it will not be the most efficient method. That would probably be something like:
>>> y = np.empty((x.shape[0], n, 2), dtype=x.dtype)
>>> y[..., 0] = x[:, 0, None]
>>> y[..., 1] = x[:, 1:]
>>> y.shape = (-1, 2)
>>> y
array([[ 0, 1],
[ 0, 2],
[ 0, 3],
[ 4, 5],
[ 4, 6],
[ 4, 7],
[ 8, 9],
[ 8, 10],
[ 8, 11]])
Like Jaimie, I first tried a repeat of the 1st column followed by reshaping, but then decided it was simpler to make 2 intermediary arrays, and hstack them:
x=np.array([[0,1,2,3],[1,2,7,9],[2,1,5,2]])
m,n=x.shape
x1=x[:,0].repeat(n-1)[:,None]
x2=x[:,1:].reshape(-1,1)
np.hstack([x1,x2])
producing
array([[0, 1],
[0, 2],
[0, 3],
[1, 2],
[1, 7],
[1, 9],
[2, 1],
[2, 5],
[2, 2]])
There probably are other ways of doing this sort of rearrangement. The result will copy the original data in one way or other. My guess is that as long as you are using compiled functions like reshape and repeat, the time differences won't be significant.
Suppose the numpy array is
arr = np.array([[0, 1, 2, 3],
[1, 2, 7, 9],
[2, 1, 5, 2]])
You can get the array of pairs as
import itertools
m, n = arr.shape
new_arr = np.array([x for i in range(m)
for x in itertools.product(a[i, 0 : 1], a[i, 1 : n])])
The output would be
array([[0, 1],
[0, 2],
[0, 3],
[1, 2],
[1, 7],
[1, 9],
[2, 1],
[2, 5],
[2, 2]])
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 106 (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.
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.
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
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.
You can use np.array(itertools.product(a, b)).
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
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]
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]]
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')
Pandas' merge() offers a naive, fast solution to the problem:
# Given the lists
x, y, z = [1, 2, 3], [4, 5], [6, 7]
# Get dataframes with the 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 dataframe
df = pd.merge(x, pd.merge(y, z, left_index=True, right_index=True),
left_index=True, right_index=True)