I tried to program a function which creates the linear span of a list of independent vectors, but it seems that the last calculated vector overwrites all other elements. I'd be nice if someone could help me fixing it.
def span_generator(liste,n):
"""function to generate the span of a list of linear independent
vectors(in liste) in the n-dimensional vectorspace of a finite
field with characteristic 2, returns a list of all elements which
lie inside the span"""
results=[]
blank=[]
for i in range(n):
blank.append(0)
a=blank
if len(liste)>1:
listenwert=liste[-1]
liste.pop(-1)
values=span_generator(liste,n)
for i in range(2):
for j in range(len(values)):
for k in range(n):
a[k]=(i*listenwert[k]+values[j][k])%2
results.append(a)
else:
for i in range(2):
for j in range(n):
a[j]=(i*liste[0][j])
results.append(a)
print(results)
return results
print(span_generator([[1,0],[0,1]],2)) gives following results
[[1, 0], [1, 0]]
[[1, 1], [1, 1], [1, 1], [1, 1]]
[[1, 1], [1, 1], [1, 1], [1, 1]]
instead of the expected: [[0,0],[1,0],[0,1],[1,1]]
Edit: I tried to simplify the program with itertools.product, but it didn't solve the problem.
def span_generator(liste):
n=len(liste[0])
results=[]
coeff=list(itertools.product(range(2), repeat=n))
blank=[]
for i in range(n):
blank.append(0)
for i in range(len(coeff)):
a=blank
for j in range(len(coeff[0])):
for k in range(n):
a[k]=(a[k]+coeff[i][j]*liste[j][k])%2
results.append(a)
return results
Output: span_generator([[0,1],[1,0]])
[[0, 0], [0, 0], [0, 0], [0, 0]]
But it should give [[0,0],[0,1],[1,0],[1,1]]
Another example: span_generator([[0,1,1],[1,1,0]]) should give [[0,0,0],[0,1,1],[1,1,0],[1,0,1]] (2=0 since i'm calculating modulo 2)
Coefficients
You can use itertools.product to generate the coefficients:
n = len(liste[0])
coefficients = itertools.product(range(2), repeat=len(liste))
yields an iterator with this content:
[(0, 0), (0, 1), (1, 0), (1, 1)]
Linear combinations
You can then selectively multiply the results with the transpose of your liste (list(zip(*liste)))
for coeff in coefficients:
yield [sum((a * c) for a, c in zip(transpose[i], coeff)) for i in range(n)]
which take for each dimensionality (for i in range(n)) the sum of the products
def span_generator3(liste):
n = len(liste[0])
transpose = list(zip(*liste))
coefficients = itertools.product(range(2), repeat=len(liste))
for coeff in coefficients:
yield [sum((a * c) for a, c in zip(transpose[i], coeff)) % 2 for i in range(n)]
this produces an iterator. If you want the result in a list-form, just can list() on the iterator
Result
list(span_generator3([[1,2],[4,8]]))
output:
[[0, 0], [4, 8], [1, 2], [5, 10]]
Higher dimensions
list(sorted(span_generator3([[1,2, 4],[8, 16, 32], [64, 128, 256]])))
output:
[[0, 0, 0],
[1, 2, 4],
[8, 16, 32],
[9, 18, 36],
[64, 128, 256],
[65, 130, 260],
[72, 144, 288],
[73, 146, 292]]
Modulo 2
If you want the result modulo 2, that's just adding 2 characters in the right place
def span_generator3_mod2(liste):
n = len(liste[0])
transpose = list(zip(*liste))
coefficients = itertools.product(range(2), repeat=len(liste))
# print(list(itertools.product(range(2), repeat=len(liste))))
for coeff in coefficients:
yield [sum((a * c) for a, c in zip(transpose[i], coeff)) % 2 for i in range(n)]
list(span_generator3_mod2([[0,1,1],[1,1,0]])) gives
[[0, 0, 0], [1, 1, 0], [0, 1, 1], [1, 0, 1]]
Related
I am now using lists to represent the graph, which would be similar to previous question. I found out that the dict approach would be very long and complex, so decided to go with the list approach. But I am still facing a few roadblocks.
So for example, the graph:
is now represented as:
nodes = ["1", "2", "3", "4", "5"]
edges = [
[0, 2, 1, 2, 0],
[1, 0, 1, 0, 0],
[0, 2, 0, 0, 0],
[1, 0, 1, 0, 2],
[1, 2, 0, 0, 0],
]
Here, edge weights can only be 1 or 2 and 0 represents no edge from one node to other. The edges are directed, so every list in the matrix represents the edges coming toward the node.
Similar to the last question, I want all possible two-edge modifications on the graph. So, for example, if we add an edge from node "4" to "5" with weight of 1, and remove the edge with weight 1 coming from node "1" to "4", the new graph will look like:
edges = [
[0, 2, 1, 2, 0],
[1, 0, 1, 0, 0],
[0, 2, 0, 0, 0],
[0, 0, 1, 0, 2],
[1, 2, 0, 1, 0],
]
and this is one of the possible modifications.
I want to build a generator that can create all such modifications sequentially and pass it to me so that I can use them to test.
My code so far is like this:
def all_modification_generation(graph: list[list], iter_count: int = 0):
possible_weights = {-1, 0, 1}
node_len = len(graph)
for i in range(node_len**2):
ix_x = i // node_len
ix_y = i % node_len
if i == ix_y:
continue
for possible_pertubs in possible_weights - {graph[ix_x][ix_y]}:
graph[ix_x][ix_y] = possible_pertubs
if iter_count == 0:
all_modification_generation(graph=graph, iter_count=iter_count + 1)
else:
yield all_modification_generation(graph=graph)
My logic is, once I do one change, I can then loop over all other elements that come after it in the matrix. So this problem could be solved recursively. And once a node is explored, we do not need to take it into consideration for next loops, because it will just give us a duplicate result that we have already found. And because I need to check for 2 modifications, I am increasing iter_count after first iteration and then yielding the next time. I am skipping ix_x == ix_y cases because a self-looping edge does not make any sense in this context, so that change is not required to be recorded.
But even then, this does not output any result. What am I doing wrong? Any help is appreciated, thanks!
Edit: I think I have figured out a way to do the double modification without repetitive generation of modified matrices. Now the only problem is that there is quite a bit of code repetition and a 4-level nested for-loop.
I'm not sure how to call a generator recursively, but I feel that should be the way to go! Thanks J_H for pointing me to the right direction.
The working code is:
def all_modification_generation(graph: list[list]):
possible_weights = {-1, 0, 1}
node_len = len(graph)
for i in range(node_len**2):
ix_x1 = i // node_len
ix_y1 = i % node_len
if ix_x1 == ix_y1:
continue
for possible_pertubs in possible_weights - {graph[ix_x1][ix_y1]}:
cc1_graph = deepcopy(graph)
cc1_graph[ix_x1][ix_y1] = possible_pertubs
for j in range(i + 1, node_len**2):
ix_x2 = j // node_len
ix_y2 = j % node_len
if ix_x2 == ix_y2:
continue
for possible_perturbs2 in possible_weights - {cc1_graph[ix_x2][ix_y2]}:
cc2_graph = deepcopy(cc1_graph)
cc2_graph[ix_x2][ix_y2] = possible_perturbs2
yield cc2_graph
The quadratic looping is an interesting technique.
We do wind up with quite a few repeated
division results, from // node_len, but that's fine.
I had a "base + edits" datastructure in mind for this problem.
Converting array to list-of-lists would be straightforward.
After overhead, a 5-node graph consumes 25 bytes -- pretty compact.
Numpy offers good support for several styles of sparse
graphs, should that become of interest.
from typing import Generator, Optional
import numpy as np
class GraphEdit:
"""A digraph with many base edge weights plus a handful of edited weights."""
def __init__(self, edge: np.ndarray, edit: Optional[dict] = None):
a, b = edge.shape
assert a == b, f"Expected square matrix, got {a}x{b}"
self.edge = edge # We treat these as immutable weights.
self.edit = edit or {}
#property
def num_nodes(self):
return len(self.edge)
def __getitem__(self, item):
return self.edit.get(item, self.edge[item])
def __setitem__(self, item, value):
self.edit[item] = value
def as_array(g: GraphEdit) -> np.ndarray:
return np.array([[g[i, j] for j in range(g.num_nodes)] for i in range(g.num_nodes)])
def all_single_mods(g: GraphEdit) -> Generator[GraphEdit, None, None]:
"""Generates all possible single-edge modifications to the graph."""
orig_edit = g.edit.copy()
for i in range(g.num_nodes):
for j in range(g.num_nodes):
if i == j: # not an edge -- we don't support self-loops
continue
valid_weights = {0, 1, 2} - {g[i, j]}
for w in sorted(valid_weights):
yield GraphEdit(g.edge, {**orig_edit, (i, j): w})
def all_mods(g: GraphEdit, depth: int) -> Generator[GraphEdit, None, None]:
assert depth >= 1
if depth == 1:
yield from all_single_mods(g)
else:
for gm in all_single_mods(g):
yield from all_mods(gm, depth - 1)
def all_double_mods(g: GraphEdit) -> Generator[GraphEdit, None, None]:
"""Generates all possible double-edge modifications to the graph."""
yield from all_mods(g, 2)
Here's the associated test suite.
import unittest
from numpy.testing import assert_array_equal
import numpy as np
from .graph_edit import GraphEdit, all_double_mods, all_single_mods, as_array
class GraphEditTest(unittest.TestCase):
def setUp(self):
self.g = GraphEdit(
np.array(
[
[0, 2, 1, 2, 0],
[1, 0, 1, 0, 0],
[0, 2, 0, 0, 0],
[1, 0, 1, 0, 2],
[1, 2, 0, 0, 0],
],
dtype=np.uint8,
)
)
def test_graph_edit(self):
g = self.g
self.assertEqual(5, self.g.num_nodes)
self.assertEqual(2, g[0, 1])
g[0, 1] = 3
self.assertEqual(3, g[0, 1])
del g.edit[(0, 1)]
self.assertEqual(2, g[0, 1])
def test_non_square(self):
with self.assertRaises(AssertionError):
GraphEdit(np.array([[0, 0], [1, 1], [2, 2]]))
def test_all_single_mods(self):
g = GraphEdit(np.array([[0, 0], [1, 0]]))
self.assertEqual(4, len(list(all_single_mods(g))))
expected = [
np.array([[0, 1], [1, 0]]),
np.array([[0, 2], [1, 0]]),
np.array([[0, 0], [0, 0]]),
np.array([[0, 0], [2, 0]]),
]
for ex, actual in zip(
expected,
map(as_array, all_single_mods(g)),
):
assert_array_equal(ex, actual)
# Now verify that original graph is untouched.
assert_array_equal(
np.array([[0, 0], [1, 0]]),
as_array(g),
)
def test_all_double_mods(self):
g = GraphEdit(np.array([[0, 0], [1, 0]]))
self.assertEqual(16, len(list(all_double_mods(g))))
expected = [
np.array([[0, 0], [1, 0]]),
np.array([[0, 2], [1, 0]]),
np.array([[0, 1], [0, 0]]),
np.array([[0, 1], [2, 0]]),
np.array([[0, 0], [1, 0]]), # note the duplicate
np.array([[0, 1], [1, 0]]),
np.array([[0, 2], [0, 0]]), # and it continues on in this vein
]
for ex, actual in zip(
expected,
map(as_array, all_double_mods(g)),
):
assert_array_equal(ex, actual)
def test_many_mods(self):
self.assertEqual(40, len(list(all_single_mods(self.g))))
self.assertEqual(1_600, len(list(all_double_mods(self.g))))
self.assertEqual(1_600, len(list(all_mods(self.g, 2))))
self.assertEqual(64_000, len(list(all_mods(self.g, 3))))
self.assertEqual(2_560_000, len(list(all_mods(self.g, 4))))
One could quibble about the fact that
it produces duplicates, since inner and outer loops
know nothing of one another.
It feels like this algorithm wants to use an
itertools.combinations
approach, generating all modifications in lexicographic order.
Given a 1d vecotr x of size n, how can we construct an n-by-n matrix X consisting of all the rolled vectors of x in PyTorch?
For example
x = torch.tensor([1,2,3,4])
The expected output is
tensor([[1, 2, 3, 4],
[2, 3, 4, 1],
[3, 4, 1, 2],
[4, 1, 2, 3]])
Is there any better way than this?
N = x.shape[0]
A = torch.zeros(N, N)
for i in range(N):
A[i] = torch.roll(x, -i)
I am trying to add 3D matrix but third loop is not starting from 0.
Here shape of matrix is (2,3,3).
Code:
for i in range(0,r):
for j in range(0,c):
for l in range(0,k):
sum[i][j][k]=A1[i][j][k]+A2[i][j][k]
Output:
IndexError: index 3 is out of bounds for axis 0 with size 3
For element-wise addition of two matrices, you can simply use the + operator between two numpy arrays:
#create two matrices of random integers
matrix1 = np.random.randint(10, size=(2,3,3))
matrix2 = np.random.randint(10, size=(2,3,3))
#add the two matrices element-wise
sum_matrix = matrix1 + matrix2
print(matrix1, matrix2, sum_matrix, sep='\n__________\n')
I don't get IndexError. Maybe you post your whole code?
This is my code:
arr1 = [[[2, 4, 8], [7, 7, 1], [4, 9, 0]], [[5, 0, 0], [3, 8, 6], [0, 5, 8]]]
arr2 = [[[3, 8, 0], [1, 5, 2], [0, 3, 9]], [[9, 7, 7], [1, 2, 5], [1, 1, 3]]]
sumArr = [[[0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0],[0, 0, 0]]]
for i in range(2): #can also use range(0,2)
for j in range(3):
for k in range(3):
sumArr[i][j][k]=arr1[i][j][k]+arr2[i][j][k]
print(sumArr)
By the way, is it necessary to use for loop?
If not, you can use numpy library.
import numpy as np
Convert your manual array to numpy matrix array, then do addition.
arr1 = [[[2, 4, 8], [7, 7, 1], [4, 9, 0]], [[5, 0, 0], [3, 8, 6], [0, 5, 8]]]
arr2 = [[[3, 8, 0], [1, 5, 2], [0, 3, 9]], [[9, 7, 7], [1, 2, 5], [1, 1, 3]]]
m1 = np.array(arr1)
m2 = np.array(arr2)
print("M1: \n", m1)
print("M2: \n", m2)
print("Sum: \n", m1 + m2)
You iterate with 'l' in the third loop but to access in list, you used k. As a result, your code is trying to access k-th index which doesn't exists, and you're getting an error.
Use this:
for i in range(0, r):
for j in range(0, c):
for l in range(0, k):
sum[i][j][l] = A1[i][j][l] + A2[i][j][l]
Here are two related SO questions 1 2 that helped me formulate my preliminary solution.
The reason for wanting to do this is to feed permutations by edit distance into a Damerau-Levenshtein NFA; the number of permutations grows fast, so it's a good idea to delay (N-C) cycle N permutations candidates until (N-C) iterations of the NFA.
I've only studied engineering math up to Differential Equations and Discrete Mathematics, so I lack the foundation to approach this task from a formal perspective. If anyone can provide reference materials to help me understand this problem properly, I would appreciate that!
Through brief empirical analysis, I've noticed that I can generate the swaps for all C cycle N permutations with this procedure:
Generate all 2-combinations of N elements (combs)
Subdivide combs into arrays where the smallest element of each 2-combination is the same (ncombs)
Generate the cartesian products of the (N-C)-combinations of ncombs (pcombs)
Sum pcombs to get a list of the swaps that will generate all C cycle N permutations (swaps)
The code is here.
My Python is a bit rusty, so helpful advice about the code is appreciated (I have the feeling that lines 17, 20, and 21 should be combined. I'm not sure if I should be making lists of the results of itertools.(combinations|product). I don't know why line 10 can't be ncombs += ... instead of ncombs.append(...)).
My primary question is how to solve this question properly. I did the rounds on my own due diligence by finding a solution, but I am sure there's a better way. I've also only verified my solution for N=3 and N=4, is it really correct?
The ideal solution would be functionally identical to heap's algorithm, except it would generate the permutations in decreasing cycle order (by the minimum number of swaps to generate the permutation, increasing).
This is far from Heap's efficiency, but it does produce only the necessary cycle combinations restricted by the desired number of cycles, k, in the permutation. We use the partitions of k to create all combinations of cycles for each partition. Enumerating the actual permutations is just a cartesian product of applying each cycle n-1 times, where n is the cycle length.
Recursive Python 3 code:
from math import ceil
def partitions(N, K, high=float('inf')):
if K == 1:
return [[N]]
result = []
low = ceil(N / K)
high = min(high, N-K+1)
for k in range(high, low - 1, -1):
for sfx in partitions(N-k, K - 1, k):
result.append([k] + sfx)
return result
print("partitions(10, 3):\n%s\n" % partitions(10, 3))
def combs(ns, subs):
def g(i, _subs):
if i == len(ns):
return [tuple(tuple(x) for x in _subs)]
res = []
cardinalities = set()
def h(j):
temp = [x[:] for x in _subs]
temp[j].append(ns[i])
res.extend(g(i + 1, temp))
for j in range(len(subs)):
if not _subs[j] and not subs[j] in cardinalities:
h(j)
cardinalities.add(subs[j])
elif _subs[j] and len(_subs[j]) < subs[j]:
h(j)
return res
_subs = [[] for x in subs]
return g(0, _subs)
A = [1,2,3,4]
ns = [2, 2]
print("combs(%s, %s):\n%s\n" % (A, ns, combs(A, ns)))
A = [0,1,2,3,4,5,6,7,8,9,10,11]
ns = [3, 3, 3, 3]
print("num combs(%s, %s):\n%s\n" % (A, ns, len(combs(A, ns))))
def apply_cycle(A, cycle):
n = len(cycle)
last = A[ cycle[n-1] ]
for i in range(n-1, 0, -1):
A[ cycle[i] ] = A[ cycle[i-1] ]
A[ cycle[0] ] = last
def permutations_by_cycle_count(n, num_cycles):
arr = [x for x in range(n)]
cycle_combs = []
for partition in partitions(n, num_cycles):
cycle_combs.extend(combs(arr, partition))
result = {}
def f(A, cycle_comb, i):
if i == len(cycle_comb):
result[cycle_comb].append(A)
return
if len(cycle_comb[i]) == 1:
f(A[:], cycle_comb, i+1)
for k in range(1, len(cycle_comb[i])):
apply_cycle(A, cycle_comb[i])
f(A[:], cycle_comb, i+1)
apply_cycle(A, cycle_comb[i])
for cycle_comb in cycle_combs:
result[cycle_comb] = []
f(arr, cycle_comb, 0)
return result
result = permutations_by_cycle_count(4, 2)
print("permutations_by_cycle_count(4, 2):\n")
for e in result:
print("%s: %s\n" % (e, result[e]))
Output:
partitions(10, 3):
[[8, 1, 1], [7, 2, 1], [6, 3, 1], [6, 2, 2], [5, 4, 1], [5, 3, 2], [4, 4, 2], [4, 3, 3]]
# These are the cycle combinations
combs([1, 2, 3, 4], [2, 2]):
[((1, 2), (3, 4)), ((1, 3), (2, 4)), ((1, 4), (2, 3))]
num combs([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [3, 3, 3, 3]):
15400
permutations_by_cycle_count(4, 2):
((0, 1, 2), (3,)): [[2, 0, 1, 3], [1, 2, 0, 3]]
((0, 1, 3), (2,)): [[3, 0, 2, 1], [1, 3, 2, 0]]
((0, 2, 3), (1,)): [[3, 1, 0, 2], [2, 1, 3, 0]]
((1, 2, 3), (0,)): [[0, 3, 1, 2], [0, 2, 3, 1]]
((0, 1), (2, 3)): [[1, 0, 3, 2]]
((0, 2), (1, 3)): [[2, 3, 0, 1]]
((0, 3), (1, 2)): [[3, 2, 1, 0]]
I have a A = 10x1000 tensor and a B = 10x1000 index tensor. The tensor B has values between 0-999 and it's used to gather values from A (B[0,:] gathers from A[0,:], B[1,:] from A[1,:], etc...).
However, if I use tf.gather(A, B) I get an array of shape (10, 1000, 1000) when I'm expecting a 10x1000 tensor back. Any ideas how I could fix this?
EDIT
Let's say A= [[1, 2, 3],[4,5,6]] and B = [[0, 1, 1],[2,1,0]] What I want is to be able to sample A using the corresponding B. This should result in C = [[1, 2, 2],[6,5,4]].
Dimensions of tensors are known in advance.
First we 'unstack' both the parameters and indices (A and B respectively) along the first dimension. Then we apply tf.gather() such that rows of A correspond to the rows of B. Finally, we stack together the result.
import tensorflow as tf
import numpy as np
def custom_gather(a, b):
unstacked_a = tf.unstack(a, axis=0)
unstacked_b = tf.unstack(b, axis=0)
gathered = [tf.gather(x, y) for x, y in zip(unstacked_a, unstacked_b)]
return tf.stack(gathered, axis=0)
a = tf.convert_to_tensor(np.array([[1, 2, 3], [4, 5, 6]]), tf.float32)
b = tf.convert_to_tensor(np.array([[0, 1, 1], [2, 1, 0]]), dtype=tf.int32)
gathered = custom_gather(a, b)
with tf.Session() as sess:
sess.run(tf.global_variables_initializer())
print(sess.run(gathered))
# [[1. 2. 2.]
# [6. 5. 4.]]
For you initial case with shapes 1000x10 we get:
a = tf.convert_to_tensor(np.random.normal(size=(10, 1000)), tf.float32)
b = tf.convert_to_tensor(np.random.randint(low=0, high=999, size=(10, 1000)), dtype=tf.int32)
gathered = custom_gather(a, b)
print(gathered.get_shape().as_list()) # [10, 1000]
Update
The first dimension is unknown (i.e. None)
The previous solution works only if the first dimension is known in advance. If the dimension is unknown we solve it as follows:
We stack together two tensors such that the rows of both tensors are stacked together:
# A = [[1, 2, 3], [4, 5, 6]] [[[1 2 3]
# ---> [0 1 1]]
# [[4 5 6]
# B = [[0, 1, 1], [2, 1, 0]] [2 1 0]]]
We iterate over the elements of this stacked tensor (which consists of stacked together rows of A and B) and using tf.map_fn() function we apply tf.gather().
We stack back the elements we get with tf.stack()
import tensorflow as tf
import numpy as np
def custom_gather_v2(a, b):
def apply_gather(x):
return tf.gather(x[0], tf.cast(x[1], tf.int32))
a = tf.cast(a, dtype=tf.float32)
b = tf.cast(b, dtype=tf.float32)
stacked = tf.stack([a, b], axis=1)
gathered = tf.map_fn(apply_gather, stacked)
return tf.stack(gathered, axis=0)
a = np.array([[1, 2, 3], [4, 5, 6]], dtype=np.float32)
b = np.array([[0, 1, 1], [2, 1, 0]], dtype=np.int32)
x = tf.placeholder(tf.float32, shape=(None, 3))
y = tf.placeholder(tf.int32, shape=(None, 3))
gathered = custom_gather_v2(x, y)
with tf.Session() as sess:
sess.run(tf.global_variables_initializer())
print(sess.run(gathered, feed_dict={x:a, y:b}))
# [[1. 2. 2.]
# [6. 5. 4.]]
Use tf.gather with batch_dims=-1:
import numpy as np
import tensorflow as tf
rois = np.array([[1, 2, 3],[3, 2, 1]])
ind = np.array([[0, 2, 1, 1, 2, 0, 0, 1, 1, 2],
[0, 1, 2, 0, 2, 0, 1, 2, 2, 2]])
tf.gather(rois, ind, batch_dims=-1)
# output:
# <tf.Tensor: shape=(2, 10), dtype=int64, numpy=
# array([[1, 3, 2, 2, 3, 1, 1, 2, 2, 3],
# [3, 2, 1, 3, 1, 3, 2, 1, 1, 1]])>