I´m using pybrain in order to train a simple neural network in which the input is going to be a 7x5 Matrix.
The following are the inputs:
A = [[0, 0, 1, 0, 0],
[0, 1, 1, 0, 0],
[0, 1, 0, 1, 0],
[0, 1, 0, 1, 0],
[1, 1, 1, 1, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1]]
E = [[1, 1, 1, 1, 1],
[1, 0, 0, 0, 0],
[1, 0, 0, 0, 0],
[1, 1, 1, 1, 0],
[1, 0, 0, 0, 0],
[1, 0, 0, 0, 0],
[1, 1, 1, 1, 1]]
I = [[0, 0, 1, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 1, 0, 0]]
O = [[1, 1, 1, 1, 0],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 1, 1, 1, 0]]
U = [[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[0, 1, 0, 0, 1],
[0, 0, 1, 1, 0]]
I thought writing something like:
ds = SupervisedDataSet(1, 1)
ds.addSample((A), ("A",))
might work, but I´m getting:
ValueError: cannot copy sequence with size 7 to array axis with dimension 1
Is there any way I can give this datasets to pyBrain?
First you have to know that SupervisedDataSet works with list, so you will need to convert the 2D arrays into a list. You can do it with something like this:
def convertToList (matrix):
list = [ y for x in matrix for y in x]
return list
Then you will need to give the new list to the method SupervisedDataSet.
Also if you would like to use that info to make the network you should use some number to identify the letter like A = 1, E = 2, I = 3, O = 4, U = 5. So to do this, the second parameter for SupervisedDataSet should be just a number 1. In this way you are saying something like "For a list with 35 elements use these numbers to identify a single number".
Finally your code should look like this:
ds = SupervisedDataSet(35, 1)
A2 = convertToList(A)
ds.addSample(A2, (1,))
E2 = convertToList(E)
ds.addSample(E2, (2,))
I2 = convertToList(I)
ds.addSample(I2, (3,))
O2 = convertToList(O)
ds.addSample(O2, (4,))
U2 = convertToList(U)
ds.addSample(U2, (5,))
Hope this could help.
Related
I have a 10x10 array with zeros and ones.
I would like to:
find the position of each cell with a value of 1.
replace all the neighbors with 1. neighbors= any cell to a n=1 distance (also diagonal).
Example:
array([[0, 1, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 1, 0, 0],
[1, 0, 0, 0, 0],
[0, 0, 0, 1, 1]])
output:
array([[1, 1, 1, 1, 0],
[1, 1, 1, 1, 0],
[1, 1, 1, 1, 0],
[1, 1, 1, 1, 1],
[1, 1, 1, 1, 1]])
I am trying finding indexes but It does not work:
a=np.where(a==1)+1
From other post I also try getting the neighbors with this function:
def n_closest(x,n,d=1):
return x[n[0]-d:n[0]+d+1,n[1]-d:n[1]+d+1]
But this does not work for the edges
Thanks
If you don't mind using scipy, a 2D convolution will solve the problem quickly:
import numpy as np
from scipy import signal
# Input array
X = np.array([[0, 1, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 1, 0, 0],
[1, 0, 0, 0, 0],
[0, 0, 0, 1, 1]])
# We apply a 2D convolution with a 3x3 kernel and we check which value are bigger than 0.
R = (signal.convolve2d(X,np.ones((3,3)),mode='same')>0).astype(int)
# R = array([[1, 1, 1, 0, 0],
# [1, 1, 1, 1, 0],
# [1, 1, 1, 1, 0],
# [1, 1, 1, 1, 1],
# [1, 1, 1, 1, 1]])
# Finally we extract the index
x,y = np.where(R)
I would like to generate an array as follows:
[[0,0,0],
[0,0,1],
[0,1,0],
[0,1,1],
[1,0,0],
[1,0,1],
[1,1,0]
[1,1,1]]
I tried to achieve this by setting 3 for loops, but I wish to go further to 4, 5, and higher bit-numbers, so the last method would not scale easly to these numbers.
Is there any simple way for doing this?
I can't figure out why you want this, but here goes:
For 3:
>>> [[int(x) for x in "{0:03b}".format(y)] for y in range(8)]
[[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 1, 1], [1, 0, 0], [1, 0, 1], [1, 1, 0], [1, 1, 1]]
>>>
For 5:
>>> [[int(x) for x in "{0:05b}".format(y)] for y in range(32)]
[[0, 0, 0, 0, 0], [0, 0, 0, 0, 1], [0, 0, 0, 1, 0], [0, 0, 0, 1, 1], [0, 0, 1, 0, 0], [0, 0, 1, 0, 1], [0, 0, 1, 1, 0], [0, 0, 1, 1, 1], [0, 1, 0, 0, 0], [0, 1, 0, 0, 1], [0, 1, 0, 1, 0], [0, 1, 0, 1, 1], [0, 1, 1, 0, 0], [0, 1, 1, 0, 1], [0, 1, 1, 1, 0], [0, 1, 1, 1, 1], [1, 0, 0, 0, 0], [1, 0, 0, 0, 1], [1, 0, 0, 1, 0], [1, 0, 0, 1, 1], [1, 0, 1, 0, 0], [1, 0, 1, 0, 1], [1, 0, 1, 1, 0], [1, 0, 1, 1, 1], [1, 1, 0, 0, 0], [1, 1, 0, 0, 1], [1, 1, 0, 1, 0], [1, 1, 0, 1, 1], [1, 1, 1, 0, 0], [1, 1, 1, 0, 1], [1, 1, 1, 1, 0], [1, 1, 1, 1, 1]]
>>>
Matching your formatting is harder.
You can use itertools.product to do this.
>>> import itertools
>>> list(itertools.product([0,1], repeat=3))
[(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)]
https://docs.python.org/3/library/itertools.html#itertools.product
You can use a recursive function like the following:
def generate_binary_entries(n, t=[[]]): # n: length of bit number
if n == 0:
return t
new_t = []
for entry in t:
new_t.append(entry + [0])
new_t.append(entry + [1])
return generate_binary_entries(n - 1, new_t)
Then
generate_binary_entries(4)
generates
[[0, 0, 0, 0],
[0, 0, 0, 1],
[0, 0, 1, 0],
[0, 0, 1, 1],
[0, 1, 0, 0],
[0, 1, 0, 1],
[0, 1, 1, 0],
[0, 1, 1, 1],
[1, 0, 0, 0],
[1, 0, 0, 1],
[1, 0, 1, 0],
[1, 0, 1, 1],
[1, 1, 0, 0],
[1, 1, 0, 1],
[1, 1, 1, 0],
[1, 1, 1, 1]]
I want to define an array with a given number of columns (let's say n=5) and in each cell of the array, the value can be either 0 or 1. And I would like to create all possibilities of ones and zeros, which means, that each row would represent one possible vector with n elements.
In other words, I want the table to look like this:
I know that create the vector of ones and zeros is quite easy but how can I ensure that the vectors would not repeat in the table and that there will be all possible combinations included (If my math is correct the table should have 2**5 = 32 rows)
How can I do it in Python? Thank you very much
Easy with itertools:
itertools.product(*[[0, 1]] * 3)
results in
[(0, 0, 0),
(0, 0, 1),
(0, 1, 0),
(0, 1, 1),
(1, 0, 0),
(1, 0, 1),
(1, 1, 0),
(1, 1, 1)]
You could generate all the numbers up to 32, and convert each to binary representation using bit shifts.
combs = [[(n >> p) & 1 for p in range(4, -1, -1)] for n in range(32)]
which gives combs as:
[
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 1],
[0, 0, 0, 1, 0],
[0, 0, 0, 1, 1],
[0, 0, 1, 0, 0],
[0, 0, 1, 0, 1],
[0, 0, 1, 1, 0],
[0, 0, 1, 1, 1],
[0, 1, 0, 0, 0],
[0, 1, 0, 0, 1],
[0, 1, 0, 1, 0],
[0, 1, 0, 1, 1],
[0, 1, 1, 0, 0],
[0, 1, 1, 0, 1],
[0, 1, 1, 1, 0],
[0, 1, 1, 1, 1],
[1, 0, 0, 0, 0],
[1, 0, 0, 0, 1],
[1, 0, 0, 1, 0],
[1, 0, 0, 1, 1],
[1, 0, 1, 0, 0],
[1, 0, 1, 0, 1],
[1, 0, 1, 1, 0],
[1, 0, 1, 1, 1],
[1, 1, 0, 0, 0],
[1, 1, 0, 0, 1],
[1, 1, 0, 1, 0],
[1, 1, 0, 1, 1],
[1, 1, 1, 0, 0],
[1, 1, 1, 0, 1],
[1, 1, 1, 1, 0],
[1, 1, 1, 1, 1]
]
Alternatively, you could use a recursive generation function:
def gimme_combs(n):
if n == 1: return [[0], [1]]
lower_combs = gimme_combs(n - 1)
return [[0] + c for c in lower_combs] + \
[[1] + c for c in lower_combs]
which would give the same result when called with:
combs = gimme_combs(5)
After reading an interesting topic on scipy.ndimage.label (Variable area threshold for identifying objects - python), I'd like to include an 'error margin' in the labelling.
In the above linked discussion:
How can the blue dot on top be included, too (let's say it is wrongly disconnected from the orange, biggest, object)?
I found the structure attribute, which should be able to include that dot by changing the array (from np.ones(3,3,3) to anything more than that (I'd like it to be 3D). However, adjusting the 'structure' attribute to a larger array does not seem to work, unfortunately. It either gives an error of dimensions (RuntimeError: structure and input must have equal rank
) or it does not change anything..
Thanks!
this is the code:
labels, nshapes = ndimage.label(a, structure=np.ones((3,3,3)))
in which a is a 3D array.
Here's a possible approach that uses scipy.ndimage.binary_dilation. It is easier to see what is going on in a 2D example, but I'll show how to generalize to 3D at the end.
In [103]: a
Out[103]:
array([[0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[1, 1, 0, 0, 1, 0, 0],
[1, 1, 0, 0, 0, 1, 1],
[0, 0, 0, 0, 0, 1, 1],
[1, 1, 1, 0, 0, 0, 0]])
In [104]: from scipy.ndimage import label, binary_dilation
Extend each "shape" by one pixel down and to the right:
In [105]: b = binary_dilation(a, structure=np.array([[0, 0, 0], [0, 1, 1], [0, 1, 1]])).astype(int)
In [106]: b
Out[106]:
array([[0, 0, 0, 1, 1, 0, 0],
[0, 0, 0, 1, 1, 0, 0],
[1, 1, 1, 0, 1, 1, 0],
[1, 1, 1, 0, 1, 1, 1],
[1, 1, 1, 0, 0, 1, 1],
[1, 1, 1, 1, 0, 1, 1]])
Apply label to the padded array:
In [107]: labels, numlabels = label(b)
In [108]: numlabels
Out[108]: 2
In [109]: labels
Out[109]:
array([[0, 0, 0, 1, 1, 0, 0],
[0, 0, 0, 1, 1, 0, 0],
[2, 2, 2, 0, 1, 1, 0],
[2, 2, 2, 0, 1, 1, 1],
[2, 2, 2, 0, 0, 1, 1],
[2, 2, 2, 2, 0, 1, 1]], dtype=int32)
By multiplying a by labels, we get the desired array of labels of a:
In [110]: alab = labels*a
In [111]: alab
Out[111]:
array([[0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[2, 2, 0, 0, 1, 0, 0],
[2, 2, 0, 0, 0, 1, 1],
[0, 0, 0, 0, 0, 1, 1],
[2, 2, 2, 0, 0, 0, 0]])
(This assumes that the values in a are 0 or 1. If they are not, you can use alab = labels * (a > 0).)
For a 3D input, you have to change the structure argument to binary_dilation:
struct = np.zeros((3, 3, 3), dtype=int)
struct[1:, 1:, 1:] = 1
b = binary_dilation(a, structure=struct).astype(int)
I have N positions, and each position can be either 0 or 1. I have fixed number of 1s, and I want to permutate these fixed number of 1s in these N positions.
from itertools import permutations
p = [0 for k in xrange(6)]
for k in xrange(0,3):
p[k] = 1
print(list(permutations(p)))
But above result contains four [0,0,0,1,1,1] in the list. I only want one of them. How can I get rid of these duplicates?
You could grab the positions of the 1s instead:
from itertools import combinations
def place_ones(size, count):
for positions in combinations(range(size), count):
p = [0] * size
for i in positions:
p[i] = 1
yield p
In action:
>>> list(place_ones(6, 3))
[
[1, 1, 1, 0, 0, 0],
[1, 1, 0, 1, 0, 0],
[1, 1, 0, 0, 1, 0],
[1, 1, 0, 0, 0, 1],
[1, 0, 1, 1, 0, 0],
[1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 0, 1],
[1, 0, 0, 1, 1, 0],
[1, 0, 0, 1, 0, 1],
[1, 0, 0, 0, 1, 1],
[0, 1, 1, 1, 0, 0],
[0, 1, 1, 0, 1, 0],
[0, 1, 1, 0, 0, 1],
[0, 1, 0, 1, 1, 0],
[0, 1, 0, 1, 0, 1],
[0, 1, 0, 0, 1, 1],
[0, 0, 1, 1, 1, 0],
[0, 0, 1, 1, 0, 1],
[0, 0, 1, 0, 1, 1],
[0, 0, 0, 1, 1, 1],
]
Set is perfect for this, as set does not not contain any duplicated element:
set(permutations(p))