Append rows in array - python

I am making a Draughts game in python, I made an array 10 by 10 and I need to append values within the entire row so that is eventually looks like this;
(
[0, 1, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 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, 2, 0, 2, 0, 2, 0, 2, 0, 2],
[2, 0, 2, 0, 2, 0, 2, 0, 2, 0],
[0, 2, 0, 2, 0, 2, 0, 2, 0, 2],
[2, 0, 2, 0, 2, 0, 2, 0, 2, 0],
)
Here is my attempt at it so far, I know it's incorrect;
__author__ = 'Matt'
import array
Board_Array = array(10, 10)
pieces = ['Empty', 'White_Piece', 'Black_Piece', 'Upgraded_White_Piece', 'Upgraded_Black_Piece']
list(enumerate(pieces))
if Board_Array.array_equals == [1, 0]:
for i in range(10):
if (i%2) == 0:
array.pop([i])
array.insert(i,1)


You could use a nested list comprehension:
In [173]: [[((i+j) % 2)*k for i in range(10)] for k in (1,1,0,2,2)
for j in (0,1)]
Out[173]:
[[0, 1, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 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, 2, 0, 2, 0, 2, 0, 2, 0, 2],
[2, 0, 2, 0, 2, 0, 2, 0, 2, 0],
[0, 2, 0, 2, 0, 2, 0, 2, 0, 2],
[2, 0, 2, 0, 2, 0, 2, 0, 2, 0]]
This is equivalent to
result = []
for k in (1,1,0,2,2):
for j in (0,1):
row = []
for i in range(10):
row.append(((i+j) % 2)*k)
result.append(row)

Related

Convert array of indices to array of zeros where the index is one

I have the following numpy array representing indexes:
array([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, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3,
3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5,
5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5])
How can I convert this to an array of zeros where only the given index is a one?
Without numpy, this works:
a = [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, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3,
3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5,
5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5]
for i in range(len(a)):
idx = a[i]
a[i] = [0 for i in range(6)]
a[i][idx] = 1
print(a)
>>> [[1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1]]
However, I am looking for a numpy function (preferably no iterating).

Maybe you could use np.identity:
>>> import numpy as np
>>> np.identity(6, np.int64)
array([[1, 0, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0],
[0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 1, 0],
[0, 0, 0, 0, 0, 1]])

Use your "index" array to index into the identity matrix:
In [2]: indices = np.array([0, 0, 0, ...]) # provided by OP
In [3]: np.identity(6, int)[indices]
Out[3]:
array([[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
.
.
.
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 1]])

You can use advanced indexing:
b = np.zeros((len(a), 6), dtype=a.dtype)
b[range(len(a)), a.tolist()] = 1
output:
array([[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
...
[0, 0, 1, 0, 0, 0],
[0, 0, 0, 1, 0, 0],
[0, 0, 0, 1, 0, 0],
[0, 0, 0, 1, 0, 0],
[0, 0, 0, 1, 0, 0],
[0, 0, 0, 1, 0, 0],
[0, 0, 0, 1, 0, 0],
[0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 1, 0],
[0, 0, 0, 0, 1, 0],
[0, 0, 0, 0, 1, 0],
...
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 1]])

Search through a 2-dimensional list without numpy

I'm looking to define a function that accepts two parameters: an int and a list.
If the function finds the integer in the list it returns its coordinates.
For example how would I do that for the number 4 in the following list, without using numpy?
l = [
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 2, 1, 1, 0, 1, 1, 1, 0],
[0, 1, 0, 1, 0, 0, 0, 1, 0],
[0, 1, 0, 1, 1, 1, 0, 1, 0],
[0, 1, 0, 0, 0, 1, 0, 1, 0],
[0, 1, 1, 1, 0, 1, 0, 1, 0],
[0, 0, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 1, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 0, 0, 0, 0, 1, 0],
[0, 1, 0, 1, 1, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 1, 1, 0, 1, 1, 4, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0]
]
You can assume that the target will always show up only once and will always be contained in the list.

The target will always show up only once and will always be contained in the list
You can use enumerate to enumerate the outer lists and the elements of the inner lists.
def coords(lst, find):
return next((i, j) for i, sub in enumerate(lst)
for j, x in enumerate(sub)
if x == find)
Demo with your list l:
>>> coords(l, 2)
>>> (1, 1)
>>> coords(l, 1)
>>> (1, 2)
In case you later want to adapt the function to work properly if the target is not in the list, remember that next takes an optional default argument.

You can do something like this:
l = [
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 2, 1, 1, 0, 1, 1, 1, 0],
[0, 1, 0, 1, 0, 0, 0, 1, 0],
[0, 1, 0, 1, 1, 1, 0, 1, 0],
[0, 1, 0, 0, 0, 1, 0, 1, 0],
[0, 1, 1, 1, 0, 1, 0, 1, 0],
[0, 0, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 1, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 0, 0, 0, 0, 1, 0],
[0, 1, 0, 1, 1, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 1, 1, 0, 1, 1, 4, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0]
]
def findElement(element, l):
for i in range(len(l)):
for j in range(len(l[i])):
if element==l[i][j]:
return (i,j)
return None
print(findElement(4,l))
Output:
(11, 7)

I would used solution like this:
#!/usr/bin/env ipython
# ---------------------
l = [
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 2, 1, 1, 0, 1, 1, 1, 0],
[0, 1, 0, 1, 0, 0, 0, 1, 0],
[0, 1, 0, 1, 1, 1, 0, 1, 0],
[0, 1, 0, 0, 0, 1, 0, 1, 0],
[0, 1, 1, 1, 0, 1, 0, 1, 0],
[0, 0, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 1, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 0, 0, 0, 0, 1, 0],
[0, 1, 0, 1, 1, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 1, 1, 0, 1, 1, 4, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0]
]
# ----------------------------------
def search(value,listin):
coords = [[ival,kkval] for ival,dd in enumerate(listin) for kkval,val in enumerate(dd) if val==value]
return coords
# ----------------------------------
result = search(4,l)
print result
where I defined a function search, which can be used to search for certain value from an input list.

Here is my approach:
def matrix_search(target, matrix):
for row_index, row in enumerate(matrix):
try:
return (row_index, row.index(target))
except ValueError:
pass
raise ValueError('Target {} not found'.format(target))
Sample usage:
print(matrix_search(4, l))
Notes
To search a simple list, use the .index() method
The .index() method will either return the index of the element if found or throw a ValueError if not found. In our context, we just ignore this exception and move on to the next row.
At the end of the loop, we will throw an exception because the element is not found

alternating values in numpy

Trying to make my code more efficient and readable and i'm stuck. Assume I want to build something like a chess board, with alternating black and white colors on an 8x8 grid. So, using numpy, I have done this:
import numpy as np
board = np.zeros((8,8), np.int32)
for ri in range(8):
for ci in range(8):
if (ci + ri) % 2 == 0:
board[ri,ci] = 1
Which nicely outputs:
array([[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1]], dtype=int32)
That I can then parse as white squares or black squares. However, in practice my array is much larger, and this way is very inefficient and unreadable. I assumed numpy already has this figured out, so I tried this:
board = np.zeros(64, np.int32)
board[::2] = 1
board = board.reshape(8,8)
But that output is wrong, and looks like this:
array([[1, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 1, 0]], dtype=int32)
Is there a better way to achieve what I want that works efficiently (and preferably, is readable)?
Note: i'm not attached to 1's and 0's, this can easily be done with other types of values, even True/False or strings of 2 kinds, as long as it works

Here's one approach using slicing with proper starts and stepsize of 2 in two steps -
board = np.zeros((8,8), np.int32)
board[::2,::2] = 1
board[1::2,1::2] = 1
Sample run -
In [229]: board = np.zeros((8,8), np.int32)
...: board[::2,::2] = 1
...: board[1::2,1::2] = 1
...:
In [230]: board
Out[230]:
array([[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1]], dtype=int32)
Other tricky ways -
1) Broadcasted comparison :
In [254]: r = np.arange(8)%2
In [255]: (r[:,None] == r)*1
Out[255]:
array([[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1]])
2) Broadcasted addition :
In [279]: r = np.arange(8)
In [280]: 1-(r[:,None] + r)%2
Out[280]:
array([[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1]])

Just found out an alternative answer by myself, so posting it here for future reference to anyone who's interested:
a = np.array([[1,0],[0,1]])
b = np.tile(a, (4,4))
Results:
array([[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1]])

I think the following is also a good way of doing it for a variable input
import sys
lines = sys.stdin.readlines()
n = int(lines[0])
import numpy as np
a = np.array([[1,0], [0,1]],dtype=np.int)
outputData= np.tile(a,(n//2,n//2))
print(outputData)

You can achieve this for single even input number n
import numpy as np
i = np.eye(2)
i = i[::-1]
k = np.array(i, dtype = np.int)
print(np.tile(k,(n//2,n//2)))

I tried and found this to be shorter one for any giver number:
n = int(input())
import numpy as np
c = np.array([[0,1], [1, 0]])
print(np.tile(c, reps=(n//2, n//2)))

Python: Remove sublist from a nested list following a criterion

I'm trying to remove sublists from a nested list containing all the possible permutation of [1, 1, 1, 0, 0, 0, 0]
[[0, 1, 0, 1, 0, 0, 1], [0, 1, 0, 0, 1, 1, 0], [0, 0, 1, 1, 0, 0, 1], [0, 0, 1, 1, 1, 0, 0], [0, 1, 1, 0, 0, 1, 0], [1, 0, 0, 0, 1, 1, 0], [0, 0, 1, 0, 0, 1, 1], [0, 0, 1, 0, 1, 1, 0], [0, 0, 1, 0, 1, 0, 1], [0, 1, 0, 0, 0, 1, 1], [0, 1, 1, 1, 0, 0, 0], [1, 1, 0, 0, 1, 0, 0], [0, 1, 0, 0, 1, 0, 1], [1, 1, 0, 1, 0, 0, 0], [0, 1, 0, 1, 1, 0, 0], [1, 0, 1, 1, 0, 0, 0], [0, 1, 1, 0, 1, 0, 0], [0, 1, 0, 1, 0, 1, 0], [1, 0, 0, 1, 1, 0, 0], [0, 0, 0, 1, 0, 1, 1], [1, 0, 1, 0, 1, 0, 0], [1, 1, 0, 0, 0, 1, 0], [1, 0, 0, 0, 0, 1, 1], [1, 0, 1, 0, 0, 1, 0], [0, 0, 0, 0, 1, 1, 1], [0, 0, 0, 1, 1, 0, 1], [0, 0, 0, 1, 1, 1, 0], [1, 0, 0, 0, 1, 0, 1], [1, 1, 1, 0, 0, 0, 0], [0, 1, 1, 0, 0, 0, 1], [1, 0, 0, 1, 0, 1, 0], [1, 0, 0, 1, 0, 0, 1], [1, 1, 0, 0, 0, 0, 1], [1, 0, 1, 0, 0, 0, 1], [0, 0, 1, 1, 0, 1, 0]]
I want to remove all the sublists in which there are 3 consecutive 0 or two couples of consecutive 0 (eg. i want to remove [1, 0, 1, 0, 0, 0, 1] or [0, 0, 1, 1, 0, 0, 1]).
Can someone give me an advice on how to proceed? Thanks in advance!

You could define such a methode to find out if a given permutation p has those triple zeros or two double zeros:
def has_triple_zeros(p):
for i, e in enumerate(p[:-2]): # e are elements (0s and 1s) of the perm
if e == 0: # we encounter a 0
if p[i+1] == 0 and p[i+2] == 0: # the two following are also 0s
return True
return False # we made it to the end, no triple 0s
def has_two_double_zeros(p):
nb_doubles = 0
i = 0 # init loop
while i < len(p[:-1]):
if p[i] == 0: # we encounter a first 0
if p[i+1] == 0: # there is one next to it
nb_doubles += 1
i += 1 # skip the next element (already treated, cannot start new double)
i += 1 # increment the loop
return nb_doubles == 2
for p in lst: # here, lst is your list of permutations
print(p, has_two_double_zeros(p), has_triple_zeros(p))
Then just read your list of permutations and delete if it matches one of your criteria. This is an idea:
res = list() # result
for p in lst:
if not (has_two_double_zeros(p) or has_triple_zeros(p)):
res.append(p)
print(res)

How to get the length of repeated numbers column wise?

I am trying to get the length of repeated numbers in Python Numpy. For example, let's consider a simple ndarray
import numpy as np
a = np.array([
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[1, 0, 1, 0, 0, 1, 1, 1, 0, 1],
[0, 1, 0, 1, 0, 1, 0, 0, 1, 0],
[1, 1, 0, 0, 1, 1, 1, 1, 0, 0],
])
The first column has [0, 1, 0, 1], the position of 1 is 1, now start counting from there, we get ones = 2 and zeros = 1. So I have to start counting ones and zeros when 1 is encountered (starting position).
so the answer for a would be
ones = [2, 2, 1, 1, 1, 3, 2, 2, 1, 1]
zeros = [1, 0, 2, 1, 0, 0, 1, 1, 1, 2]
Can any one please help me out?
Update
3D array:
a = np.array([
[
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[1, 1, 0, 0, 0, 1, 1, 1, 0, 0],
[0, 1, 0, 0, 0, 1, 0, 0, 1, 0],
[1, 1, 0, 0, 1, 1, 1, 1, 0, 0],
],
[
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 1, 0, 0, 1, 0, 0, 0, 1, 1],
[0, 1, 0, 1, 0, 0, 0, 1, 0, 0],
[1, 1, 0, 1, 0, 1, 1, 1, 0, 0],
]
])
The expected output should be
ones = [
[2, 3, 0, 0, 1, 3, 2, 2, 1, 0],
[1, 3, 0, 2, 1, 1, 1, 2, 1, 1]
]
zeros = [
[1, 0, 0, 0, 0, 0, 1, 1, 1, 0],
[0, 0, 0, 0, 2, 0, 0, 0, 2, 2]
]

With focus on performance, here's one generic approach for ndarrays -
ones_count = a.sum(-2)
zeros_count = (a.shape[-2] - ones_count - a.argmax(-2))*a.any(-2)
One alternative to get zeros_count with selections using np.where, would be -
zeros_count = np.where(a.any(-2),a.shape[-2] - ones_count - a.argmax(-2),0)
Sample runs
2D case :
In [60]: a
Out[60]:
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[1, 0, 1, 0, 0, 1, 1, 1, 0, 1],
[0, 1, 0, 1, 0, 1, 0, 0, 1, 0],
[1, 1, 0, 0, 1, 1, 1, 1, 0, 0]])
In [61]: ones_count = a.sum(-2)
...: zeros_count = (a.shape[-2] - ones_count - a.argmax(-2))*a.any(-2)
...:
In [62]: ones_count
Out[62]: array([2, 2, 1, 1, 1, 3, 2, 2, 1, 1])
In [63]: zeros_count
Out[63]: array([1, 0, 2, 1, 0, 0, 1, 1, 1, 2])
3D case :
In [65]: a = np.array([
...: [
...: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
...: [1, 1, 0, 0, 0, 1, 1, 1, 0, 0],
...: [0, 1, 0, 0, 0, 1, 0, 0, 1, 0],
...: [1, 1, 0, 0, 1, 1, 1, 1, 0, 0],
...: ],
...: [
...: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
...: [0, 1, 0, 0, 1, 0, 0, 0, 1, 1],
...: [0, 1, 0, 1, 0, 0, 0, 1, 0, 0],
...: [1, 1, 0, 1, 0, 1, 1, 1, 0, 0],
...: ]
...: ])
In [66]: ones_count = a.sum(-2)
...: zeros_count = (a.shape[-2] - ones_count - a.argmax(-2))*a.any(-2)
...:
In [67]: ones_count
Out[67]:
array([[2, 3, 0, 0, 1, 3, 2, 2, 1, 0],
[1, 3, 0, 2, 1, 1, 1, 2, 1, 1]])
In [68]: zeros_count
Out[68]:
array([[1, 0, 0, 0, 0, 0, 1, 1, 1, 0],
[0, 0, 0, 0, 2, 0, 0, 0, 2, 2]])
and so on for higher dim arrays.

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