Given this 2D Array - [[0, 0, -1], [1, 0, -2], [2, 0, -3], [3, 0, 0]]
How can I code something in Python that moves everything to the end? (If the end is empty).
So I get something like this - [[0, 0, 0], [1, 0, -1], [2, 0, -2], [3, 0, -3]]
I tried something like this:
count = 0
while count < row: # row is 4 in this case.
if my_array[row-1][2] == 0:
tp = my_array[count][2]
my_array[count][2] = 0
my_array[count+1][2] = tp
else:
break
count += 1
return my_array
But this is what I get instead:
[[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, -1]]
Update:
Thanks #furas, I spent hours trying to figure this out
How can I also handle a gap in the middle? So if I get something like this:
data = [[0, 0, -1], [0, 0, 0], [0, 0, -3], [0, 0, -4]]
expected = [[0, 0, 0], [0, 0, -1], [0, 0, -3], [0, 0, -4]]
rows = len(data)
if data[-1][-1] == 0:
count = rows-1
while count > 0:
data[count][-1] = data[count-1][-1]
count -= 1
data[0][-1] = 0
print(data)
print(expected)
print(data == expected) # False
Thanks
UPDATE Nov.29 -
I have certain scenarios like these that don't work? Can you help?
DATA: [[0, 0, 0, 0], [1, 2, 1, 0], [2, 1, 2, 0], [0, 0, 0, 0]]
EXPECT: [[0, 0, 0, 0], [0, 0, 0, 0], [1, 2, 1, 0], [2, 1, 2, 0]]
Or if it's negative:
DATA: [[0, 0, 0, 0], [-101, -101, -101, 0], [2, 3, 2, 0], [3, 2, 3, 0]]
EXPECT: [[0, 0, 0, 0], [0, 0, 0, 0], [2, 3, 2, 0], [3, 2, 3, 0]]
Thanks A lot!!
I got something like this
def move(data):
row = len(data)-1
prev = row-1
while row > 0 and prev >= 0:
if data[row][-1] == 0:
while prev >= 0 and data[prev][-1] == 0:
prev -= 1
data[row][-1] = data[prev][-1]
data[prev][-1] = 0
row -= 1
prev -= 1
return data
# --- test ---
examples = [
{
'data': [[0, 0, -1], [0, 0, 0], [0, 0, -3], [0, 0, -4]],
'expected': [[0, 0, 0], [0, 0, -1], [0, 0, -3], [0, 0, -4]],
},
{
'data': [[0, 0, -1], [1, 0, -2], [2, 0, -3], [3, 0, 0]],
'expected': [[0, 0, 0], [1, 0, -1], [2, 0, -2], [3, 0, -3]],
},
]
for ex in examples:
print(ex['data'])
result = move(ex['data'])
print(ex['expected'])
print(result == ex['expected']) # True
I use rows = len(data) and [-1] so it can have any size.
External while check if place row is empty and start internal while which is looking for not-empty elements prev before row. After that it move non-empty element from prev to row and put zero in prev. And it starts again with next row place.
It looks more complicated then other solutions.
Here's a reasonably efficient way to do this. We grab the last column of the grid into a list named lastcol, and then sort lastcol with a key function that tests if each item is not equal to zero. That will move all the zeros to the start of lastcol without disturbing the order of the other elements. And then we copy lastcol back to the grid.
I've modified your test data slightly so that we can see that the code handles gaps and positive & negative values correctly.
grid = [[0, 0, -1], [1, 0, 2], [2, 0, 0], [3, 0, -3], [4, 0, 0]]
print(grid)
lastcol = [u[-1] for u in grid]
lastcol.sort(key=(0).__ne__)
for row, u in zip(grid, lastcol):
row[-1] = u
print(grid)
output
[[0, 0, -1], [1, 0, 2], [2, 0, 0], [3, 0, -3], [4, 0, 0]]
[[0, 0, 0], [1, 0, 0], [2, 0, -1], [3, 0, 2], [4, 0, -3]]
Assuming I'm understanding the goal correctly -- that you'd want a column [0, 1, 0, 2, 3] to become [0, 0, 1, 2, 3] -- then with the use of a helper function to transpose the array, all we need to do is sort the columns using bool as the key function, because non-zero numbers are false, and False < True.
def transpose(seq):
return [list(part) for part in zip(*seq)]
def push(seq):
return transpose(sorted(part, key=bool) for part in transpose(seq))
which gives me
In [29]: push([[0, 0, -1], [1, 0, -2], [2, 0, -3], [3, 0, 0]])
Out[29]: [[0, 0, 0], [1, 0, -1], [2, 0, -2], [3, 0, -3]]
In [30]: push([[0, 0, -1], [0, 0, 0], [0, 0, -3], [0, 0, -4]])
Out[30]: [[0, 0, 0], [0, 0, -1], [0, 0, -3], [0, 0, -4]]
Related
Assuming the sudoku puzzle is a 9 by 9 and is filled with 0-9. I want is 9 lists, with each list containing a 3 by 3 Sudoku box.
This is what I have:
grid = [[5, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 1, 0, 0, 8, 7, 0, 6, 0],
[0, 0, 0, 0, 0, 3, 0, 0, 0],
[0, 5, 0, 0, 6, 1, 0, 7, 0],
[0, 0, 2, 0, 0, 0, 9, 0, 0],
[0, 0, 0, 4, 0, 0, 0, 0, 0],
[0, 0, 0, 5, 0, 0, 0, 4, 0],
[9, 0, 0, 0, 4, 8, 7, 0, 0],
[0, 8, 0, 3, 0, 0, 0, 0, 0]]
list1 = []
for i in range(0,3):
for j in range(0,3):
list1.append(grid[i][j])
list2 = []
for i in range(0,3):
for j in range(4,7):
list2.append(grid[i][j])
and it goes on and on.......
How do I simplify this for loop for getting a list of 3 by 3 boxes in Sudoku puzzle?
You need to add a step parameter to your range()s so that you start reading in each box at the appropriate row / column index:
from itertools import product
boxes = []
for row_start, col_start in product(range(0, 9, 3), repeat=2):
boxes.append([
[grid[row][col] for col in range(col_start, col_start + 3)]
for row in range(row_start, row_start + 3)
])
print(boxes)
This outputs:
[
[[5, 0, 0], [0, 1, 0], [0, 0, 0]],
[[0, 0, 0], [0, 8, 7], [0, 0, 3]],
[[0, 0, 1], [0, 6, 0], [0, 0, 0]],
[[0, 5, 0], [0, 0, 2], [0, 0, 0]],
[[0, 6, 1], [0, 0, 0], [4, 0, 0]],
[[0, 7, 0], [9, 0, 0], [0, 0, 0]],
[[0, 0, 0], [9, 0, 0], [0, 8, 0]],
[[5, 0, 0], [0, 4, 8], [3, 0, 0]],
[[0, 4, 0], [7, 0, 0], [0, 0, 0]]
]
from pprint import pprint as pp
SIZE = 3
sudoku = [[[0] * SIZE for _ in range(SIZE)] for _ in range(SIZE * SIZE)]
# example of populating it
for i in range(SIZE):
for j in range(SIZE):
counter = i * SIZE + j
sudoku[counter][i][j] = counter + 1
pp(sudoku)
output
[[[1, 0, 0], [0, 0, 0], [0, 0, 0]],
[[0, 2, 0], [0, 0, 0], [0, 0, 0]],
[[0, 0, 3], [0, 0, 0], [0, 0, 0]],
[[0, 0, 0], [4, 0, 0], [0, 0, 0]],
[[0, 0, 0], [0, 5, 0], [0, 0, 0]],
[[0, 0, 0], [0, 0, 6], [0, 0, 0]],
[[0, 0, 0], [0, 0, 0], [7, 0, 0]],
[[0, 0, 0], [0, 0, 0], [0, 8, 0]],
[[0, 0, 0], [0, 0, 0], [0, 0, 9]]]
I wrote the function below and it works if the len(list_of_list) is even. I'm having trouble when it's odd. When I run it, it ends up with an Assertion Error. How do I reverse only the elements in the first and last list index and not the ones in between when the list of lists is odd?
def flip_diag(list_of_list):
for i in range(int(len(list_of_list) % 2 != 0)):
list_of_list[0][::-1] = reversed(list_of_list[0][::-1])
list_of_list[-1][::-1] = reversed(list_of_list[-1][::-1])
for i in range(int(len(list_of_list) % 2 == 0)):
reversed_list = [elem[::-1] for elem in list_of_list]
return reversed_list
if __name__ == '__main__':
assert flip_diag([[1, 0, 0, 0], [0, 2, 0, 0], [0, 0, 3, 0], [0, 0, 0, 4]]) == [[0, 0, 0, 1],[0, 0, 2, 0], [0, 3, 0, 0], [4, 0, 0, 0]]
assert flip_diag([[0, 0, 0], [0, 1, 0], [0, 0, 1]]) == [[0, 0, 0], [0, 1, 0], [1, 0, 0]]
assert flip_diag([[1, 0, 0], [0, 1, 0], [0, 0, 1]]) == [[0, 0, 1], [0, 1, 0], [1, 0, 0]]
assert flip_diag([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) == [[3, 2, 1], [4, 5, 6], [9, 8, 7]]
one way to do it via list comprehension:
def flip_diag(list_of_list):
if len(list_of_list) % 2 != 0:
return [item[::-1] if index in [0, len(list_of_list)-1] else item for index, item in enumerate(list_of_list)]
return [item[::-1] for index, item in enumerate(list_of_list)]
If the length of list_of_list is odd, you can have the below for loop in place to only reverse the first and last index elements each list inside list_of_list:
>> list_of_list = [[1, 0, 0, 0], [0, 2, 0, 0], [0, 0, 3, 0]]
>> for i in range(int(len(list_of_list) % 2 != 0)):
>> reversed_list = [elem[-1:-2:-1]+elem[1:-1]+elem[0:1] for elem in list_of_list]
>> print(reversed_list)
[[0, 0, 0, 1], [0, 2, 0, 0], [0, 0, 3, 0]]
If you want to use map and lambda you can do it like below:
>> list_of_list = [[1, 0, 0, 0], [0, 2, 0, 0], [0, 0, 3, 0]]
>> for i in range(int(len(list_of_list) % 2 != 0)):
>> reversed_list = list(map(lambda elem: elem[-1:-2:-1]+elem[1:-1]+elem[0:1] ,list_of_list))
>> print(reversed_list)
[[0, 0, 0, 1], [0, 2, 0, 0], [0, 0, 3, 0]]
I have an array of numbers between 0 and 3 and I want to create a 2D array of their binary digits.
in the future may be I need to have array of numbers between 0 and 7 or 0 to 15.
Currently my array is defined like this:
a = np.array([[0], [1], [2], [3]], dtype=np.uint8)
I used numpy unpackbits function:
b = np.unpackbits(a, axis=1)
and the result is this :
array([[0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0, 1, 0],
[0, 0, 0, 0, 0, 0, 1, 1]], dtype=uint8)
As you can see it created a 2d array with 8 items in column while I'm looking for 2 columns 2d array.
here is my desired array:
array([[0, 0],
[0, 1],
[1, 0],
[1, 1]])
Is this related to data type uint8 ?
what is your idea?
One way of approaching the problem is to just adapt your b to match your desired output via a simple slicing, similarly to what suggested in #GrzegorzSkibinski answer:
import numpy as np
def gen_bits_by_val(values):
n = int(max(values)).bit_length()
return np.unpackbits(values, axis=1)[:, -n:].copy()
print(gen_bits_by_val(a))
# [[0 0]
# [0 1]
# [1 0]
# [1 1]]
Alternatively, you could create a look-up table, similarly to what suggested in #WarrenWeckesser answer, using the following:
import numpy as np
def gen_bits_by_num(n):
values = np.arange(2 ** n, dtype=np.uint8).reshape(-1, 1)
return np.unpackbits(values, axis=1)[:, -n:].copy()
bits2 = gen_bits_by_num(2)
print(bits2)
# [[0 0]
# [0 1]
# [1 0]
# [1 1]]
which allows for all kind of uses thereby indicated, e.g.:
bits4 = gen_bits_by_num(4)
print(bits4[[1, 3, 12]])
# [[0 0 0 1]
# [0 0 1 1]
# [1 1 0 0]]
EDIT
Considering #PaulPanzer answer the line:
return np.unpackbits(values, axis=1)[:, -n:]
has been replaced with:
return np.unpackbits(values, axis=1)[:, -n:].copy()
which is more memory efficient.
It could have been replaced with:
return np.unpackbits(values << (8 - n), axis=1, count=n)
with similar effects.
You can use the count keyword. It cuts from the right so you also have to shift bits before applying unpackbits.
b = np.unpackbits(a<<6, axis=1, count=2)
b
# array([[0, 0],
# [0, 1],
# [1, 0],
# [1, 1]], dtype=uint8)
This produces a "clean" array:
b.flags
# C_CONTIGUOUS : True
# F_CONTIGUOUS : False
# OWNDATA : True
# WRITEABLE : True
# ALIGNED : True
# WRITEBACKIFCOPY : False
# UPDATEIFCOPY : False
In contrast, slicing the full 8-column output of unpackbits is in a sense a memory leak because the discarded columns will stay in memory as long as the slice lives.
You can truncate b to keep just the columns since the first column with 1:
b=b[:, int(np.argwhere(b.max(axis=0)==1)[0]):]
For such a small number of bits, you can use a lookup table.
For example, here bits2 is an array with shape (4, 2) that holds the bits of the integers 0, 1, 2, and 3. Index bits2 with the values from a to get the bits:
In [43]: bits2 = np.array([[0, 0], [0, 1], [1, 0], [1, 1]])
In [44]: a = np.array([[0], [1], [2], [3]], dtype=np.uint8)
In [45]: bits2[a[:, 0]]
Out[45]:
array([[0, 0],
[0, 1],
[1, 0],
[1, 1]])
This works fine for 3 or 4 bits, too:
In [46]: bits4 = np.array([[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]])
In [47]: bits4
Out[47]:
array([[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]])
In [48]: x = np.array([0, 1, 5, 14, 9, 8, 15])
In [49]: bits4[x]
Out[49]:
array([[0, 0, 0, 0],
[0, 0, 0, 1],
[0, 1, 0, 1],
[1, 1, 1, 0],
[1, 0, 0, 1],
[1, 0, 0, 0],
[1, 1, 1, 1]])
Given this 2D Array -
[[0, 0, -1], [0, 0, -2], [0, 0, 0], [0, 0, 3], [0, 0, 0]]
How can I code something in Python that shifts once to the end each time it's called? However, it has to stop if it reaches a positive value.
So I get something like this
[[0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, 3], [0, 0, 0]]
This is what I have. It works for condition 1 but it does not work for condition 2. What can I tweak so condition 2 also works?
# Condition 1 - Works
data = [[0, 0, -1], [0, 0, -2], [0, 0, -3], [0, 0, 0], [0, 0, 0]]
expected = [[0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, -3], [0, 0, 0]] # My Results.
# Condition 2 - Does Not Work
data = [[0, 0, -1], [0, 0, -2], [0, 0, 0], [0, 0, 3], [0, 0, 0]]
expected = [[0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, 3], [0, 0, 0]] # This is what I want to get. (But it doesn't work)
rows = len(data)
if data[-1][-1] == 0:
count = rows-1
while count > 0:
data[count][-1] = data[count-1][-1]
count -= 1
data[0][-1] = 0
print(data)
print(expected)
This is what I am getting currently. I want to get the expected for Condition 2 listed in the code snipped above.:
Condition 2 Result:
[[0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, 0], [0, 0, 3]]
Thanks
Update #Furas:
Something like this to find the positive value location?
possitive_value = []
for i in range(len(data)):
if data[i][-1] > 0:
possitive_value.append(i, -1)
I think you have to find first positive value and use its position as rows
EDIT: I changed name count to last to make it more readable
# --- function ---
def move(data):
rows = len(data)
# - find positive -
for x in range(rows):
if data[x][-1] > 0:
rows = x # use its position as `rows`
break # don't seach other positiove values
# - star moving -
# set "last" checked row
last = rows-1
# check "last" row
if data[last][-1] == 0:
# move previous values
while last > 0:
data[last][-1] = data[last-1][-1]
last -= 1
# put 0 in first place
data[0][-1] = 0
# --- tests ---
examples = [
{
# Condition 1 - Works
'data': [[0, 0, -1], [0, 0, -2], [0, 0, -3], [0, 0, 0], [0, 0, 0]],
'expected': [[0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, -3], [0, 0, 0]], # My Results.
},
{
# Condition 2 - Works
'data': [[0, 0, -1], [0, 0, -2], [0, 0, 0], [0, 0, 3], [0, 0, 0]],
'expected': [[0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, 3], [0, 0, 0]], # This is what I want to get. (But it doesn't work)
}
]
for example in examples:
data = example['data']
expected = example['expected']
print(' before:', data)
move(data)
print(' after:', data)
print('expected:', expected)
print(' correct:', data == expected)
print('---')
Result:
before: [[0, 0, -1], [0, 0, -2], [0, 0, -3], [0, 0, 0], [0, 0, 0]]
after: [[0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, -3], [0, 0, 0]]
expected: [[0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, -3], [0, 0, 0]]
correct: True
---
before: [[0, 0, -1], [0, 0, -2], [0, 0, 0], [0, 0, 3], [0, 0, 0]]
after: [[0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, 3], [0, 0, 0]]
expected: [[0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, 3], [0, 0, 0]]
correct: True
---
EDIT:
BTW: instead of while you can use for with reversed range() if it makes it more readable
def move(data):
rows = len(data)
# - find positive -
for x in range(rows):
if data[x][-1] > 0:
rows = x # use its position as `rows`
break # don't seach other positiove values
# - star moving -
# set "last" checked row
last = rows-1
# check "last" row
if data[last][-1] == 0:
# move previous values
for pos in range(last, 0, -1): # range with reversed order
data[pos][-1] = data[pos-1][-1]
# put 0 in first place
data[0][-1] = 0
BTW: both versions move items in original data ("in-place") so they don't need return data.
Here's a variation, a function that can be called and will shift data by one every time. I wasn't entirely clear from your question if this was the behavior you wanted.
import numpy as np
def shift_data(data):
ele_idx = [i for i,x in enumerate(data) if x != [0,0,0]]
zero_idx = [i for i,x in enumerate(data) if x == [0,0,0] and i != 0]
if max(np.diff(ele_idx)) == 1 or max(np.diff([i for i,x in enumerate(data) if x == [0,0,0]])) == 1:#things are consecutive
data.insert(0,data.pop(-1))
else:
tt = [l for l in data[ele_idx[0]:zero_idx[0]+1]]
tt.insert(0,tt.pop(-1))
data = data[:ele_idx[0]] + tt + data[zero_idx[0]+1:]
return data
data = [[0, 0, -1], [0, 0, 0], [0, 0, -2], [0, 0, 0], [0, 0, 3]]
print(data)
for _ in range(0,6):
data = shift_data(data)
print(data)
This code outputs:
[[0, 0, -1], [0, 0, 0], [0, 0, -2], [0, 0, 0], [0, 0, 3]]
[[0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, 0], [0, 0, 3]]
[[0, 0, 0], [0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, 3]]
[[0, 0, 3], [0, 0, 0], [0, 0, 0], [0, 0, -1], [0, 0, -2]]
[[0, 0, -2], [0, 0, 3], [0, 0, 0], [0, 0, 0], [0, 0, -1]]
[[0, 0, -1], [0, 0, -2], [0, 0, 3], [0, 0, 0], [0, 0, 0]]
[[0, 0, 0], [0, 0, -1], [0, 0, -2], [0, 0, 3], [0, 0, 0]]
So I have several 3D arrays that I need to add together. Each array consists of entries with either 0 or 1. All arrays also have the same dimension. Now, when I add these arrays together some of the values overlap (which they do). However, I just need to know how the structure of the total combined array is, which means that I don't need the values 1, 2 or 3 when 2 or 3 arrays have overlapped. This also just need to be one, and of course, wherever there is a zero, the value zero just need to remain zero.
So basically what I have is:
array1 =
[[[1, 0, 0], [0, 0, 0], [0, 0, 0]],
[[0, 1, 0], [0, 0, 0], [0, 0, 0]],
[[0, 0, 1], [1, 1, 1], [0, 0, 0]]]
array2 =
[[[1, 0, 0], [0, 1, 0], [0, 0, 0]],
[[0, 0, 0], [1, 1, 0], [0, 0, 0]],
[[0, 0, 1], [0, 1, 0], [0, 0, 0]]]
So when adding them together I get:
array_total = array1 + array2 =
[[[2, 0, 0], [0, 1, 0], [0, 0, 0]],
[[0, 1, 0], [1, 1, 0], [0, 0, 0]],
[[0, 0, 2], [1, 2, 1], [0, 0, 0]]]
Where I actually want it to give me:
array_total = array1 + array2 =
[[[1, 0, 0], [0, 1, 0], [0, 0, 0]],
[[0, 1, 0], [1, 1, 0], [0, 0, 0]],
[[0, 0, 1], [1, 1, 1], [0, 0, 0]]]
So can anyone give me a hint to how this is done ?
(Assuming those are numpy arrays, or array1 + array2 would behave differently).
If you want to "change all positive values to 1", you can do this
array_total[array_total > 0] = 1
But what you actually want is an array that has a 1 where array1 or array2 has a 1, so just write it directly like that:
array_total = array1 | array2
Example:
>>> array1 = np.array([[[1, 0, 0], [0, 0, 0], [0, 0, 0]],
... [[0, 1, 0], [0, 0, 0], [0, 0, 0]],
... [[0, 0, 1], [1, 1, 1], [0, 0, 0]]])
>>> array2 = np.array([[[1, 0, 0], [0, 1, 0], [0, 0, 0]],
... [[0, 0, 0], [1, 1, 0], [0, 0, 0]],
... [[0, 0, 1], [0, 1, 0], [0, 0, 0]]])
>>> array1 | array2
array([[[1, 0, 0], [0, 1, 0], [0, 0, 0]],
[[0, 1, 0], [1, 1, 0], [0, 0, 0]],
[[0, 0, 1], [1, 1, 1], [0, 0, 0]]])