Flatten arrays into a list of lists - python

I have this data structure:
[array([[0, 1, 0, 1, 1, 1, 0, 5, 1, 0, 2, 1]]), array([[0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0],
[1, 3, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0]]), array([[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], etc....
I want to flatten this into a list of lists something like:
[[0 1 0 1 1 1 0 5 1 0 2 1]
[1 6 1 0 0 1 1 1 2 0 2 0]
[2 0 5 0 5 2 2 0 6 3 2 2]
[1 0 1 1 1 1 0 2 0 0 0 1]]
How do we do this in python?


I believe you're looking for vstack:
>>> np.vstack(l)
array([[0, 1, 0, 1, 1, 1, 0, 5, 1, 0, 2, 1],
[0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0],
[1, 3, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
Note that this is equivalent to:
>>> np.concatenate(x, axis=0)
array([[0, 1, 0, 1, 1, 1, 0, 5, 1, 0, 2, 1],
[0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0],
[1, 3, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
To convert to list, use tolist:
>>> np.vstack(l).tolist()
[[0, 1, 0, 1, 1, 1, 0, 5, 1, 0, 2, 1], [0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0], [1, 3, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
# or
>>> np.concatenate(x, axis=0).tolist()
[[0, 1, 0, 1, 1, 1, 0, 5, 1, 0, 2, 1], [0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0], [1, 3, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0], [0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]


Use flatten:
print([i.flatten() for i in l])
Or:
print(list(map(lambda x: x.flatten(),l)))
Both output:
[array([0, 1, 0, 1, 1, 1, 0, 5, 1, 0, 2, 1]), array([0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0, 1, 3, 1, 0, 0, 1, 1, 1, 1, 0, 1,
0]), array([0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0])]

Related

How to add the appropriate number of zeros in a list of lists? (python)

I have this list of integers:
[
[0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0],
[1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0],
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0],
[0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0]
]
And I need to transform it in this list:
[
[0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0],
[1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0],
[1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0],
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0],
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0]
]
How can I do that?

From what I see, the sublists you want to get to have a length of 21.
Hence, if your original list of lists is l:
for x in l:
x.extend((21-len(x)) * [0])
In case you don't know the final length of each sublist, you can find it out in advance as follows:
max_len = max([len(x) for x in l])
and then use it instead of the 21 proposed:
for x in l:
x.extend((max_len - len(x)) * [0])

Using itertools.zip_longest and a double zip:
from itertools import zip_longest
## input
# l = [[...], [...], ...]
list(map(list, zip(*zip_longest(*l, fillvalue=0))))
Output:
[[0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0],
[1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0],
[1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0],
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0],
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0]]

For this, you need to find the Max. length of a list, so that you can make all the lists of same size, by adding zeroes to them.
Here is the code:
lst=[
[0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0],
[1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0],
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0],
[0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0]
]
for i in lst:
print(i)
max_len=0
for i in lst:
if len(i)>max_len:
max_len=len(i)
for i in lst:
for j in range(max_len-len(i)):
i.append(0)
print("\n\n")
for i in lst:
print(i)
OUTPUT (Before and After) :
[0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0]
[1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0]
[1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0]
[1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0]
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0]
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0]
[0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0]
[0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0]
[1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0]
[1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0]
[1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0]
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0]
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0]
[0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0]

Start by finding the longest list, loop over all list and append zeros based on the "length delta"
lst = [
[0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0],
[1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0],
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0],
[0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0]
]
_max_len = max(lst,key = len)
for l in lst:
l.extend((_max_len - len(l)) * [0])
for l in lst:
print(l)
output
[0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0]
[1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0]
[1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0]
[1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0]
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0]
[1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0]
[0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0]

removing the first 0 bit value

i want to get the binary value from a string, but why if i print the binary value and save it into an array there is always 0 bit get inserted automatically, i know that its from the bitwise operation inside the loop, how can i delete it and get the actual binary value from a string?
this is the code
def padding():
a = "Testing"
byte_a = a.encode('utf-8')
print(byte_a)
print(bin(byte_a[0]))
print(bin(byte_a[1]))
print(bin(byte_a[2]))
i = 0
x = []
#byte_a.append(1)
while True:
x.insert(i, ((byte_a[i//8]>>(7-(i%8)))&1))
print(x)
i+=1
and here is the output:
b'Testing'
0b1010100
0b1100101
0b1110011
[0]
[0, 1]
[0, 1, 0]
[0, 1, 0, 1]
[0, 1, 0, 1, 0]
[0, 1, 0, 1, 0, 1]
[0, 1, 0, 1, 0, 1, 0]
[0, 1, 0, 1, 0, 1, 0, 0]
[0, 1, 0, 1, 0, 1, 0, 0, 0]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0]
[0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1]
...
as you can see the array had the first value of 0 for every binary.
thanks guys

Tensorflow Initialize list with 0s and 1s similar to tf.one_hot

I currently have a list of values over a time sequence say the values are [1, 3, 5, 7, 3]. Currently I am using tf.one_hot to get a one hot vector/tensor representative for each value within the list.
1 = [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0]
3 = [0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0]
5 = [0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0]
7 = [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0]
3 = [0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0]
In Tensorflow is there a way/function that would allow me to do something similar but initialize all values from 0 to the value with 1s?
Desired Result:
1 = [1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0]
3 = [1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0]
5 = [1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0]
7 = [1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0]
3 = [1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0]

You are looking for tf.sequence_mask.
TensorFlow API

Order of repetition per row and column in Python [closed]

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Want to improve this question? Add details and clarify the problem by editing this post.
Closed 6 years ago.
Improve this question
I have been trying to figure the order of repetition per-row and just couldn't do it. Ok. Lets consider a ndarray of size (2, 11, 10)
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],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
[1, 0, 0, 1, 0, 1, 1, 1, 0, 0],
[1, 1, 0, 1, 1, 0, 1, 1, 0, 0],
[0, 1, 1, 1, 0, 0, 1, 1, 0, 1],
[1, 1, 1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 1],
[0, 1, 1, 1, 0, 0, 1, 1, 0, 1]
],
[
[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],
[1, 1, 0, 1, 0, 0, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 1, 1, 0, 0],
[1, 0, 0, 0, 1, 1, 0, 0, 1, 1],
[1, 1, 1, 0, 0, 1, 1, 1, 0, 1],
[1, 0, 0, 1, 1, 0, 1, 0, 1, 0],
[1, 0, 0, 0, 0, 0, 1, 0, 0, 0],
[1, 1, 1, 0, 0, 1, 1, 1, 0, 1]
]
])
What I wanted to is to get the order of every 1's per row based on a column. Whenever the first 1 is found in a row the order starts would start at 0; then goes to the second row if 1 is found here then the order is 1, but if the 1 is already present at the column index in the previous row, then it is ignored. For example
Lets consider these lists:
0 1 2 3 4 5 6 7 8 9 -> column index
0 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], -> no 1's no order here
1 [1, 1, 0, 0, 0, 1, 1, 1, 0, 0], -> order starts at 0
2 [0, 1, 0, 0, 0, 1, 0, 0, 1, 0], -> order starts at 1
At row index 0 there are no 1 so nothing happens, at row index 1 there are ones in column index [0,1,5,6,7] this will be equal to 0; the output should be
column order
0 0
1 0
2 -
3 -
4 -
5 0
6 0
7 0
8 -
9 -
At row index 2 there are 1 at column index [1,5,8] whos order is 1; in there 1 and 5 are ignored because it already has an order 0 to it, but for the unknown order it should be 1; the final output should be
column order
0 0
1 0
2 -
3 -
4 -
5 0
6 0
7 0
8 1
9 -
I have tried using Numpy's np.where method to the index values; something like this
index = np.asarray(np.where(a == 1)).T
I have no idea what to do next. Can anyone please help me?

Apparently, the desired result--based on comments on the question and an earlier version of this answer--is to find the "dense ranking" of the row index of the first 1 in each column. (See the docstring of scipy.stats.rankdata for the meaning of "dense ranking".) The result can be found using a combination of the .argmax() method and scipy.stats.rankdata.
Here's a function that computes the order for a two-dimensional array. The question doesn't define what should happen when a column is all zeros; order assigns that column the value -1.
from scipy.stats import rankdata
def order(x):
result = x.argmax(axis=0)
result[(x == 0).all(axis=0)] = -1
rank = rankdata(result, method='dense') - 1 - np.any(result < 0)
return rank
For example, here is the array y:
In [71]: y
Out[71]:
array([[0, 1, 0, 0, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 0, 0],
[1, 1, 1, 0, 1, 1, 0, 0],
[1, 0, 1, 1, 1, 1, 0, 0],
[1, 0, 0, 0, 1, 1, 1, 0]])
In [72]: order(y)
Out[72]: array([ 1, 0, 1, 2, 0, 0, 3, -1])
Here's the array a from the question:
In [73]: a
Out[73]:
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],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 0],
[1, 0, 0, 1, 0, 1, 1, 1, 0, 0],
[1, 1, 0, 1, 1, 0, 1, 1, 0, 0],
[0, 1, 1, 1, 0, 0, 1, 1, 0, 1],
[1, 1, 1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 1, 0, 1, 0, 0, 1, 1],
[0, 1, 1, 1, 0, 0, 1, 1, 0, 1]],
[[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],
[1, 1, 0, 1, 0, 0, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 1, 1, 0, 0],
[1, 0, 0, 0, 1, 1, 0, 0, 1, 1],
[1, 1, 1, 0, 0, 1, 1, 1, 0, 1],
[1, 0, 0, 1, 1, 0, 1, 0, 1, 0],
[1, 0, 0, 0, 0, 0, 1, 0, 0, 0],
[1, 1, 1, 0, 0, 1, 1, 1, 0, 1]]])
The function order() expects a two-dimensional array, so we must use a loop to get the order for each subarray in a:
In [74]: np.array([order(m) for m in a])
Out[74]:
array([[0, 0, 3, 3, 2, 0, 0, 0, 1, 4],
[2, 0, 3, 1, 0, 2, 2, 1, 0, 0]])

Append rows in array

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)

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