Python: Remove sublist from a nested list following a criterion - python

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)

Related

pattern restriction in substring- Python

I found a question in glassdoor. I do not have additional clarification
Input : an int array [1,0,0,1,1,0,0,1,0,1,0,0,0,1]
you have to come up with a program that will give all possible subsets of the array based on the pattern.
Pattern restrictions were the string array should start with 1 and end with 1. So there will be many sub arrays like from index 0 to 3 and 0 to 4 and index 7 to 9
To solve this I was thinking of using 2 for loops and if both cases the values are equal to 1 then print them.
v=[1,0,0,1,1,0,0,1,0,1,0,0,0,1]
resultList=[]
for i in range(0,len(v)-1):
for j in range(i+1, len(v)):
if v[i]==1 and v[j]==1:
r=v[i:j]
resultList.append(r)
print(resultList)
Output:[[1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 1, 0, 0], [1, 0, 0, 1, 1, 0, 0, 1, 0], [1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0], [1], [1, 1, 0, 0],
I only see 1 correct value so far in output [1, 0, 0, 1]. Should I have used set instead of list? I tried that but that approach did not work either. Can someone kindly give some directions on how to solve this problem?
Thanks for your time.

You can use itertools.combinations to pick 2 indices where the values are non-zeroes in the list:
from itertools import combinations
a = [1,0,0,1,1,0,0,1,0,1,0,0,0,1]
[a[i: j + 1] for i, j in combinations((i for i, n in enumerate(a) if n), 2)]
This returns:
[[1, 0, 0, 1], [1, 0, 0, 1, 1], [1, 0, 0, 1, 1, 0, 0, 1], [1, 0, 0, 1, 1, 0, 0, 1, 0, 1], [1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1], [1, 1], [1, 1, 0, 0, 1], [1, 1, 0, 0, 1, 0, 1], [1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1], [1, 0, 0, 1], [1, 0, 0, 1, 0, 1], [1, 0, 0, 1, 0, 1, 0, 0, 0, 1], [1, 0, 1], [1, 0, 1, 0, 0, 0, 1], [1, 0, 0, 0, 1]]

The probelm is in v[i:j]. Change v[i:j] to v[i:j+1]

Creating a list of bits?

Currently, I have two functions: char2bin and segmentString.
segmentString takes a string and a fill character and returns lists of 8 character strings. For example, if there is a 13 character string, it splits it into a list of two strings where the second string has 3 fill characters to make it a complete 8.
>>>segmentString("Hello, World!", "-")
['Hello, W', 'orld!---']
char2bin takes individual string characters (single character) and turns them into a list of 8 bits. It does not work for multiple character strings. For example,
>>>char2bin('a')
[0,1,1,0,0,0,0,1]
>>>char2bin('abc')
(ERROR)
I need to create a function (in this example, let's call it framer) that takes the result from segmentString and convert it into a list of bits, where each list of bits are contained in a separate list within a list.
For example, from the segmentString function, this would create a list of two strings. Each letter of each separate string is converted into a list of bits, and each list of bits is contained as a list for each string.
>>>F=framer("Hello, World!", "-")
>>>F
[[[0, 1, 0, 0, 1, 0, 0, 0], [0, 1, 1, 0, 0, 1, 0, 1], [0,1, 1, 0, 1, 1, 0, 0], [0, 1, 1, 0, 1, 1, 0, 0], [0, 1, 1, 0, 1, 1,1,1], [0, 0, 1, 0, 1, 1, 0, 0], [0, 0, 1, 0, 0, 0, 0, 0], [0, 1,1, 1,0, 1, 1, 1]], [[0, 1, 1, 0, 1, 1, 1, 1], [0, 1, 1, 1, 0, 0,1, 0], [0,1, 1, 0, 1, 1, 0, 0], [0, 1, 1, 0, 0, 1, 0, 0], [0, 0,1, 0, 0, 0, 0,1], [0, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1,1, 0], [0, 1, 1, 1,1, 1, 1, 0]]]
As you can see, there is one general list that contains two lists that contain 8 lists of bits, which were converted from a string character by char2bin.
How would I do this?

You can use a list comprehension for this:
def char2bin(byte):
return list(map(int, format(byte, '08b')))
def segmentString(text, padding, chunksize):
for index in range(0, len(text), chunksize):
yield text[index:index + chunksize].ljust(chunksize, padding)
def framer(text, padding='-', chunksize=8, encoding='utf8'):
return [[char2bin(byte) for byte in segment] for segment in
segmentString(text.encode(encoding), padding.encode(encoding), chunksize)]
This uses utf8 encoding, but since your input text is all ascii characters, there's one byte per character.
>>> framer('Hello, World!')
[[[0, 1, 0, 0, 1, 0, 0, 0],
[0, 1, 1, 0, 0, 1, 0, 1],
[0, 1, 1, 0, 1, 1, 0, 0],
[0, 1, 1, 0, 1, 1, 0, 0],
[0, 1, 1, 0, 1, 1, 1, 1],
[0, 0, 1, 0, 1, 1, 0, 0],
[0, 0, 1, 0, 0, 0, 0, 0],
[0, 1, 0, 1, 0, 1, 1, 1]],
[[0, 1, 1, 0, 1, 1, 1, 1],
[0, 1, 1, 1, 0, 0, 1, 0],
[0, 1, 1, 0, 1, 1, 0, 0],
[0, 1, 1, 0, 0, 1, 0, 0],
[0, 0, 1, 0, 0, 0, 0, 1],
[0, 0, 1, 0, 1, 1, 0, 1],
[0, 0, 1, 0, 1, 1, 0, 1],
[0, 0, 1, 0, 1, 1, 0, 1]]]
Non-ascii characters require multiple bits to encode.
>>> framer('💩', padding='\x00')
[[[1, 1, 1, 1, 0, 0, 0, 0],
[1, 0, 0, 1, 1, 1, 1, 1],
[1, 0, 0, 1, 0, 0, 1, 0],
[1, 0, 1, 0, 1, 0, 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],
[0, 0, 0, 0, 0, 0, 0, 0]]]

You could either use list comprehensions or make use of the itertools module.
You can learn more about list comprehensions here, and more about itertootls here.

You can use below code to achieve your goal.
def segment_string(s, fill_by):
l = []
while s:
if len(s) < 8:
s = s + (fill_by) * (8 - len(s))
l.append(s[0:8])
s = s[8:]
return l # ['Hello, W', 'orld!---']
def char2bin(ch):
a = bin(ord(ch))[2:]
l = [int(c) for c in a]
if len(l) < 8:
l = ([0] * (8 - len(l))) + l # Adding extra 0s to front (if len(l) < 8)
return l # [0, 1, 0, 0, 1, 0, 0, 0]
def framer(s, fill_by='-'):
segments = segment_string(s, fill_by) # Calling segment_string()
print(segments)
arr = []
for segment in segments:
arr2 = []
for ch in segment:
arr3 = char2bin(ch); # Calling char2bin()
arr2.append(arr3)
arr.append(arr2)
return arr # final list to be returned
if __name__ == "__main__":
f = framer('Hello, World!', '~')
print(f)
Output »
[[[0, 1, 0, 0, 1, 0, 0, 0], [0, 1, 1, 0, 0, 1, 0, 1], [0, 1, 1, 0, 1, 1, 0, 0], [0, 1, 1, 0, 1, 1, 0, 0], [0, 1, 1, 0, 1, 1, 1, 1], [0, 0, 1, 0, 1, 1, 0, 0], [0, 0, 1, 0, 0, 0, 0, 0], [0, 1, 0, 1, 0, 1, 1, 1]], [[0, 1, 1, 0, 1, 1, 1, 1], [0, 1, 1, 1, 0, 0, 1, 0], [0, 1, 1, 0, 1, 1, 0, 0], [0, 1, 1, 0, 0, 1, 0, 0], [0, 0, 1, 0, 0, 0, 0, 1], [0, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 0]]]
# >>> bin(126)
# '0b1111110'
# >>>
# >>> chr(126)
# '~'
# >>>

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

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

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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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