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I'm looking for an efficient algorithm to solve the following problem. Say you have set S = {1,2,...n}, set T of non-empty subsets of S, and integer x > 0. You want to know if it's possible to select one member from each subset in T, so that each member of S is selected at most x times.
Example 1:
S = {1,2,3,...,10}
T = {{1},{4,5},{1,2},{6,7,8},{4}}
x = 1
one solution would be {1,5,2,6,4}
Example 2:
S = {1,2,3,...,10}
T = {{1},{4,5},{1},{6,7,8},{4}}
x = 1
no solution since 1 would have to be selected twice
Here I have a python program that solves the problem, but can be very slow. (I'm not a computer scientist, but think the upper bound on the cost would be n^m where m is the size of T). I represent T as a 2d list of 0's and 1's, where each row in the list corresponds to a subset of T, each position in a row corresponds to a member of S, and the value in the position (either 0 or 1) corresponds to whether or not that member is present in the subset. For example, the row [0,1,1,1,0,0,1,0,1] would correspond to the subset {2,3,4,7,9}.
def f():
t = [[1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0],
[0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1],
[0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
solution_len = len(t)
x = 3
elem_counts = [0]*solution_len
solution = [0]*solution_len # positions correspond to members of t, values correspond to members of s
input_rows_len = len(t[0])
a = 0
t = sorted(t)
while a <= solution_len:
if a == solution_len:
return solution[:solution_len]
if solution[a] == input_rows_len:
solution[a] = 0
a -= 1
if a == -1:
return False
elem_counts[solution[a]] -= 1
solution[a] += 1
continue
bin_val = t[a][solution[a]]
if elem_counts[solution[a]] == x or bin_val == 0:
solution[a] += 1
else:
elem_counts[solution[a]] += 1
a += 1
return False
there is a library that I used once for a similar combinatorial problem. I had the same issue of performance and this library is fast. it's called python-constraint. here is the link: https://pypi.org/project/python-constraint/
you simply create a problem object, add variables, and then add certain constraints to the object and finally get the solutions.
I have a list that creates a square image but I want to create it in a round shape. Can I create using any loop? I have tried in many ways but my code can not generate image in round shape. Can anyone help me with this?
def home(request):
abcd=abc([[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1],
[1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1],
[1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1],
[1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1],
[0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0],
[1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1],
[1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1],
[1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1],
[1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]])
qr_file = os.path.join("media/prashant.jpg")
img_file = open(qr_file, 'wb')
abcd.save(img_file, 'JPEG')
img_file.close()
If an array would suffice, you could use skimage.morphology
There are a number of shapes available here. Disk will create a 2D array with a circle.
import skimage
radius = 3
circle = skimage.morphology.disk(radius)
I have the following binary dataset:
X = [
[1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1],
[1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1],
[1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0],
[1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0],
[1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0]
]
I performed Kmeans clustering on this data:
kmeans = KMeans(n_clusters=4)
kmeans.fit(X)
centers = kmeans.cluster_centers_
Now I want to display the resulting clustering on a scatter graph
import matplotlib.pyplot as plt
plt.figure()
plt.scatter(X, X, s=50, cmap='viridis')
plt.scatter(centers, centers, c='black', s=200, alpha=0.5);
plt.show()
But no scatter graph is showing up. Does anyone have an idea where I am going wrong? Any suggestion will be highly appreciated.
after a lot of searching I haven't been able to find the answer to what seems like a simple question.
I have some code that is doing a Monte Carlo simulation and storing the results in a nested list. Here are the results I generate from a 10-trial simulation:
[[1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1], [1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1], [1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1], [0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1], [1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0], [1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1], [1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0], [1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1]]
Where I'm stuck is I'd like to find the mean of the 0th item in each list, the 1st item, and so on. I generally use numpy.mean for this, but how do I instruct it to only average the nth item?
You can use np.mean with axis=0:
lst = [[1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1], [1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1], [1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1], [0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1], [1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0], [1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1], [1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0], [1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1]]
np.mean(lst, axis=0)
# array([ 0.9, 1. , 0.8, 0.9, 0.6, 0.8, 0.5, 0.7, 0.8, 0.5, 0.7, 0.5, 0.6])
If I understood the question well, the answer is the same as #Psidom proposed but over axis=1. Also, you may need to convert it to a numpy array beforehand:
lst = np.array([[1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1],
[1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1], # and so on...)
np.mean(lst, axis=1)
I've been using scipy.cluster.vq.kmeans for doing some k-means clustering, but was wondering if there's a way to determine which centroid each of your data points is (putativly) associated with.
Clearly you could do this manually, but as far as I can tell the kmeans function doesn't return this?
There is a function kmeans2 in scipy.cluster.vq that returns the labels, too.
In [8]: X = scipy.randn(100, 2)
In [9]: centroids, labels = kmeans2(X, 3)
In [10]: labels
Out[10]:
array([2, 1, 2, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 2, 2, 1, 2, 1, 2, 1, 2, 0,
1, 0, 2, 0, 1, 2, 0, 1, 0, 1, 1, 2, 2, 2, 2, 1, 2, 1, 1, 1, 2, 0, 0,
2, 2, 0, 1, 0, 0, 0, 2, 2, 2, 0, 0, 1, 2, 1, 0, 0, 0, 2, 1, 1, 1, 1,
1, 0, 0, 1, 0, 1, 2, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 2, 0, 2, 2, 0,
1, 1, 0, 1, 0, 0, 0, 2])
Otherwise, if you must use kmeans, you can also use vq to get labels:
In [17]: from scipy.cluster.vq import kmeans, vq
In [18]: codebook, distortion = kmeans(X, 3)
In [21]: code, dist = vq(X, codebook)
In [22]: code
Out[22]:
array([1, 0, 1, 0, 2, 2, 2, 0, 1, 1, 0, 2, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1,
2, 2, 1, 2, 0, 1, 1, 0, 2, 2, 0, 1, 0, 1, 0, 2, 1, 2, 0, 2, 1, 1, 1,
0, 1, 2, 0, 1, 2, 2, 1, 1, 1, 2, 2, 0, 0, 2, 2, 2, 2, 1, 0, 2, 2, 2,
0, 1, 1, 2, 1, 0, 0, 0, 0, 1, 2, 1, 2, 0, 2, 0, 2, 2, 1, 1, 1, 1, 1,
2, 0, 2, 0, 2, 1, 1, 1])
Documentation: scipy.cluster.vq