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Set values around a specific position in a 2D array using Python

So say I have a 2D array such as:
[
[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, 3, 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]
]
And I want to set all the values 2 levels out around the 3 to a specific number like:
[
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 1, 1, 3, 1, 1, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 1, 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]
]
Note that the 3 could be in any position in the list, I'm using a random generator to get it. So how could I achieve this? Maybe using a for loop?

Carrying on from the comment - I find numpy super useful for slicing like this;
import numpy as np
arr = np.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, 3, 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]
])
xs, ys = np.where(arr == 3)
arr[xs[0] - 2: xs[0] + 3, ys[0] - 2: ys[0] + 3] = 1
arr[xs[0], ys[0]] = 3
Obviously possible in pure python/list form as well but you will be knee deep in double iteration probably

Here's a pure Python approach (albeit rather clumsy):
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, 3, 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]
]
for j, a in enumerate(A):
try:
i = a.index(3)
P = A[j-1] if j > 0 else None
Q = A[j+1] if j < len(a) - 1 else None
for k in range(max(0, i-2), min(i+3, len(a))):
if P:
P[k] = 1
if Q:
Q[k] = 1
a[k] = 1
a[i] = 3
break
except ValueError:
pass
print(A)

pure python
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, 3, 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]
]
location = [(i, j) for i, row in enumerate(A) for j, item in enumerate(row) if item == 3][0]
for i, row in enumerate(A):
for j, item in enumerate(row):
if (abs(i - location[0]) <= 2) and (abs(j - location[1]) <= 2) and not ((i, j) == location):
A[i][j] = 1

display two dimensional coordinate map in pyplot

I need a way to convert a coordinate map detailing a maze of 1s in a field of 0s into a set of plotted lines in pyplot.
To illustrate, I need to convert this
maze = [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 1, 1, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 1, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
into
plt.plot([4,4,4,4,4,3,2],[0,1,2,3,4,4,4])
plt.plot([3,4,4,4],[6,6,7,8])
such that if i need to, i can change the coordinate map and it will reflect the changes in the plotting.

Looks like you'd be better off with an imshow:
plt.imshow(maze)
Output:

can someone please explain this to me? [duplicate]

This question already has answers here:
List of lists changes reflected across sublists unexpectedly
(17 answers)
Closed 4 years ago.
def create_octahedron(size):
x = []
y = []
z = []
if size % 2 == 0 or size <= 1:
return x
for i in range(size):
x.append(0)
for i in range(size):
y.append(x)
for i in range(size):
z.append(y)
for i in range(size):
for u in range(size):
for v in range(size):
if i == len(z)//2:
if u == len(y)//2:
if v == len(x)//2:
z[3][3][3] = 1
print(z)
create_octahedron(7)
[[[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, 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, 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, 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, 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, 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, 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, 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, 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, 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]]]
this is the output i keep getting but the output I'm expecting is to only have a 1 at the middle of the entire equation not at. i am much less interested in how to fix this as i already know how. What i want to know is why this is giving this output.

Because you append the same list. In z, each row points the same Y, and in Y each row points to the same X. If you try z[0][0][0] = 2, you could see that every row's first element changes to 2.
To avoid this, create a new x/y list before append.

place the mydata_array into the random location of Big_array of zeros

mydata is an numpy array of shape(10,100,100) of the form(z,y,x). And i have created the empty array of shape(10,800,800). Now i need to place the mydata_array into some random locations of empty_array such that if I would plot the output, it should look like mydata is placed randomly in the ouput plot of array(10,800,800).
I used the np.hstack() and np.vstack().
But it places the mydata_array side by side. I need to place my_data_array in random location.
How could i do this? Any Suggestions please..
Regards
Raj

Here's a demonstration of placing several copies of one array inside another, using slice indexing:
In [802]: out = np.zeros((10,10),int)
In [803]: src = np.arange(6).reshape(2,3)
In [804]: out
Out[804]:
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]])
One copy in the upper left:
In [805]: out[:2,:3] = src
In [806]: out
Out[806]:
array([[0, 1, 2, 0, 0, 0, 0, 0, 0, 0],
[3, 4, 5, 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]])
Several more copies:
In [808]: out[4:6, 6:9] = src
In [809]: out[1:3, 4:7] = src
In [810]: out
Out[810]:
array([[0, 1, 2, 0, 0, 0, 0, 0, 0, 0],
[3, 4, 5, 0, 0, 1, 2, 0, 0, 0],
[0, 0, 0, 0, 3, 4, 5, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 2, 0],
[0, 0, 0, 0, 0, 0, 3, 4, 5, 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]])
Just repeat that kind of action for a selection of random locations. Make sure that the slice ranges match the src shape, and that they lie within the dimensions of the target array.
While may be possible to insert many copies at once (the flattening of the answer may be needed), let's start with understanding how to insert one copy at a time.
=========
#alvis' answer places the src items in shuffled order on one row of the out (or wrapped rows):
array([[2, 4, 5, 3, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
...
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
===================
Looped placement of multiple blocks:
def foo1(src, idx, NM):
out = np.zeros(NM, dtype=src.dtype)
n,m = src.shape
for i,j in idx:
out[i:i+n, j:j+m] = src
return out
idx=np.array([[0,0],[1,4],[4,4],[8,7],[7,2]])
In [940]: out1 = foo1(src, idx, (10,10))
In [941]: out1
Out[941]:
array([[0, 1, 2, 0, 0, 0, 0, 0, 0, 0],
[3, 4, 5, 0, 0, 1, 2, 0, 0, 0],
[0, 0, 0, 0, 3, 4, 5, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 1, 2, 0, 0, 0],
[0, 0, 0, 0, 3, 4, 5, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 1, 2, 0, 0, 0, 0, 0],
[0, 0, 3, 4, 5, 0, 0, 0, 1, 2],
[0, 0, 0, 0, 0, 0, 0, 3, 4, 5]])
================
Placement of a block with advanced indexing (arrays instead of slices):
In [880]: I = np.array([1,1,1,2,2,2])
In [881]: J = np.array([3,4,5,3,4,5])
In [882]: out[I,J] = src.flat
In [883]: out
Out[883]:
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 2, 0, 0, 0, 0],
[0, 0, 0, 3, 4, 5, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
...
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
And for multiple blocks
def foo2(src, idx, NM):
out = np.zeros(NM, dtype=src.dtype)
n,m = src.shape
ni = len(idx)
IJ = [np.mgrid[i:i+n, j:j+m] for i,j in idx]
IJ = np.concatenate(IJ, axis=1).reshape(2,-1)
out[IJ[0,:], IJ[1,:]] = np.tile(src,(ni,1)).flat
return out
In this small example the alternate is considerably slower (14x). For (1000,1000) out it is still slow (6x). Most of the time is spent in generating IJ.
This handles the I,J index calculation much faster (it needs to be generalize), but it is still slower than the looped slicing:
def foo3(src, idx, NM):
out = np.zeros(NM, dtype=src.dtype)
n,m = src.shape
ni = len(idx)
I = np.repeat((idx[:,[0]]+np.arange(2)).flatten(),3)
J = np.repeat((idx[:,[1]]+np.arange(3)),2,axis=0).flatten()
out[I, J] = np.tile(src,(ni,1)).flat
return out
This reminds me of work I did years ago to speed up the creation of a finite element stiffness matrix in MATLAB. There it was per-element stiffness blocks that needed to be placed in a large sparse global stiffness matrix.
==================
Regular pattern with broadcasting (see edit history)

According to your question, you don't need to preserve elements relatively to the first dimension of your array. For example, if there is one non-zero element a in (100,100) matrix z=0, and two elements b and c in the matrix z=1, then in your output all a, b, c can appear in z=0. In this case I suggest the following solution:
import numpy as np
#replace this with your input data
mydata = np.ones((10,100,100))
mydata_large = np.zeros((10,800,800))
mydata_flatten = mydata.flatten()
ind = np.array([i for i in range(len(mydata_flatten))])
np.random.shuffle(ind)
mydata_large_f = mydata_large.flatten()
np.put(mydata_large_f,ind[:len(mydata_flatten)],mydata_flatten)
mydata_large = np.reshape(mydata_large_f, (10,800,800))

Combining list of lists in python3.x

I have a list of lists such as
[[0, 0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 2, 0, 0, 0, 0, 0], [0, 0, 3, 0, 0, 0, 0], [0, 0, 0, 2, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 3, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 2, 0], [0, 0, 0, 0, 0, 0, 3]]
I would like to combine it into one integer list as
[0,1,0,2,3,2,0,3,0,2,3]
I couldn't find a exact way how to achieve it.
The pattern is that if there is any number other than 0 in the list add it as actual number found or else enter it as 0.

3>> [functools.reduce(operator.or_, x) for x in L]
[0, 1, 0, 2, 3, 2, 0, 3, 0, 2, 3]

Is your goal to flatten the 2d list? If so, b in the following snippet is what you want:
a = [[1,2,3],[4,5,6],[7,8,9]]
b = [val for sublist in a for val in sublist]

Just iterate like so:
new = [0 for i in range(len(old[0]))]
for a in old:
for b, c in enumerate(a):
new[b] += c

presuming you only get either all zeros or a single non zero digit you can call next on filter after filtering all the 0's with 0 as the default value to next.
l = [[0, 0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 2, 0, 0, 0, 0, 0], [0, 0, 3, 0, 0, 0, 0], [0, 0, 0, 2, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 3, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 2, 0], [0, 0, 0, 0, 0, 0, 3]]
print([next(filter(None,sub),0) for sub in l] )
[0, 1, 0, 2, 3, 2, 0, 3, 0, 2, 3]