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.
Related
Here I set a numpy memmap and set a value to one of the rows
import numpy as np
numpyMemmap = np.memmap( 'test.npy', dtype=np.uint16, mode='w+', shape=(500 , 512))
numpyMemmap[34] = np.ones(512)
numpyMemmap[34]
output
memmap([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1], dtype=uint16)
Now I save the memmap to a new file, and load a different memmap from that file
np.save('LocalMemmap.npy', numpyMemmap)
numpyMemmap2 = np.memmap( 'LocalMemmap.npy', dtype=np.uint16, mode='r', shape=(500 , 512))
numpyMemmap2[34]
Output
memmap([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1], dtype=uint16)
Why have the values changed?
For convenience, here is a colab notebook which reproduces the error
https://colab.research.google.com/drive/1WiSkFYRI0l7mkhtQzxZZDXVFw-IzzGpS?usp=sharing
If you want to memmap an array saved with numpy.save, you need to use numpy.load, not numpy.memmap:
numpyMemmap2 = np.load('LocalMemmap.npy', mmap_mode='r+')
With numpy.memmap, you're interpreting the NPY file header as part of the array.
(Also, your original test.npy isn't actually in NPY file format. np.memmap doesn't make NPY files, just like it doesn't read them.)
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.
I'm trying to automate chi squared calculations. I'm using scipy.stats.pearsonr. However, that's giving me different answers than SPSS is. Like, factor of 10 difference. (.07 --> .8)
I'm pretty sure that the data is the same in both cases because I'm printing out the crosstab in both cases (using pandas.crosstab) and the numbers are identical.
d1 = [1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1]
d2 = [1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 1, 2, 0, 2, 1, 2, 0, 0, 1]
print scipy.stats.stats.pearsonr(d1,d2)
gives:
(-0.065191159985573108, 0.61172152831874682)
(the 1st is the coefficient, the 2nd is the p value)
However SPSS says that the Pearson Chi-Square is .057.
Is there something I should check other than the crosstab?
Apparently you are computing the chi-squared statistic and p-value for the contingency table (i.e. "cross tab") of the data. The scipy function pearsonr is not the correct function to use for this. To do the calculation with scipy, you'll need to form the contingency table and then use scipy.stats.chi2_contingency.
There are several ways you could convert d1 and d2 into a contingency table. Here I'll use the Pandas function pandas.crosstab. Then I'll use chi2_contingency for the chi-squared test.
First, here is your data. I have them in numpy arrays, but this is not necessary:
In [49]: d1
Out[49]:
array([1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0,
1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1])
In [50]: d2
Out[50]:
array([1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1,
1, 2, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0,
1, 1, 0, 1, 2, 1, 0, 1, 1, 2, 0, 2, 1, 2, 0, 0, 1])
Use pandas to form the contingency table:
In [51]: import pandas as pd
In [52]: table = pd.crosstab(d1, d2)
In [53]: table
Out[53]:
col_0 0 1 2
row_0
0 5 7 4
1 10 34 3
Then use chi2_contingency for the chi-squared test:
In [54]: from scipy.stats import chi2_contingency
In [55]: chi2, p, dof, expected = chi2_contingency(table.values)
In [56]: p
Out[56]: 0.057230732412525138
The p value matches the value computed by SPSS.
Update: In SciPy 1.7.0 (targeted for mid-2021), you'll be able to create the contingency table with scipy.stats.contingency.crosstab:
In [33]: from scipy.stats.contingency import crosstab # Will be in SciPy 1.7.0
In [34]: d1
Out[34]:
array([1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1,
0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1])
In [35]: d2
Out[35]:
array([1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1,
1, 1, 2, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1,
1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 1, 2, 0, 2, 1, 2, 0, 0, 1])
In [36]: (vals1, vals2), table = crosstab(d1, d2)
In [37]: vals1
Out[37]: array([0, 1])
In [38]: vals2
Out[38]: array([0, 1, 2])
In [39]: table
Out[39]:
array([[ 5, 7, 4],
[10, 34, 3]])
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)