Compare multidimensional array in python

Question

I have a multidimensional array called old_arr which is like this [[8,8,8,8,0,0,0,0,6,6,5,5],[...]] then I have an updated multidimensional array new_arr like this [[9,9,6,7,3,6,5,0,6,4,3,4],[...]] What I want to do is to update the new_arr so that if the value in it corresponds to a 0 in old_arr then the value should be 0 otherwise keep the new value. So in the above example the new_arr would look like this [[9,9,6,7,0,0,0,0,6,4,3,4],[...]] where the 3,6,5 where replaced by 0. Any advice?

Also I want to know if it is possible to update the cell to 0 only if 4 out of it's 8 surrounding neighbour cells have the value 0 as well? Like new_arr and old_arr are multidimensional array (lists) which represents rows and cols so they are like a big table as shown in the below image where the blue cell in the new_arr will only be updated to zero if the corresponding cell in the old_arr is 0 and 4 of its neighbour cells are 0 (white cells in the photo)

So I need to check all 8neighbour cells (sometimes 6 or 7 depending on the cell position where it's in the middle (8) or edges(7) or corners (6) ) if they are zeros or not and count them, if the count is 4 or more then set the cell value to 0.

So if old_arr is

[[8,8,8,8,0,0,0,0,6,6,5,5],
 [8,8,8,8,0,x,0,0,6,6,5,5],
 [8,8,8,8,0,0,0,0,6,6,5,5],
 [8,8,8,8,0,0,0,0,6,6,5,5],....]

Where x is a zero

And new_arr is

[[9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],....]

For the highlighted cell, the corresponding cell in the new_arr will be zero because the highlighted cell in old_arr is 0 and more than 4 of its neighbor cells are zeros as well.

Updated new_arr is

[[9,9,6,7,3,0,0,0,6,4,3,4],
 [9,9,6,7,0,0,0,0,6,4,3,4],
 [9,9,6,7,0,0,0,0,6,4,3,4],
 [9,9,6,7,0,0,0,0,6,4,3,4],....]

Their are no array lists in python? Array lists are only in java. — RoadRunner
– RoadRunner, Commented Jan 11, 2018 at 8:50
By "array list", do you mean "list"? or "list of lists"? because you don't mean "array list". — khelwood
– khelwood, Commented Jan 11, 2018 at 8:58
@khelwood I mean multidimensional array as shown in the example — Tak
– Tak, Commented Jan 11, 2018 at 8:59
For your highlighted cell [8,8,8,8,0,**0**,0,0,6,6,5,5], only 3 of the neighbouring cells are 0? Unless I'm not looking at it right. — RoadRunner
– RoadRunner, Commented Jan 12, 2018 at 10:23

RoadRunner · Accepted Answer · 2018-01-12 11:46:15Z

3

Assuming old_arr and new_arr are the same length, you could do something like this:

old_arr = [[8,8,8,8,0,0,0,0,6,6,5,5], [8,8,7,0,0,0,0,0,6,6,5,5]]
new_arr = [[9,9,6,7,3,6,5,0,6,4,3,4], [9,0,6,7,4,6,5,0,6,4,3,4]]

new_arr = [[x if old[i] else 0 for i, x in enumerate(new)] for old, new in zip(old_arr, new_arr)]

print(new_arr)

which outputs:

[[9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4], [9, 0, 6, 0, 0, 0, 0, 0, 6, 4, 3, 4]]

UPDATE:

Here is a brute force solution that deals with neighbouring cells:

old_arr = [[8,8,8,8,0,0,0,0,6,6,5,5],
           [8,8,8,8,0,0,0,0,6,6,5,5],
           [8,8,8,8,0,0,0,0,6,6,5,5],
           [8,8,8,8,0,0,0,0,6,6,5,5]]

new_arr = [[9,9,6,7,3,6,5,0,6,4,3,4],
           [9,9,6,7,3,6,5,0,6,4,3,4],
           [9,9,6,7,3,6,5,0,6,4,3,4],
           [9,9,6,7,3,6,5,0,6,4,3,4]]

def first_last(row, next_row, old, new):
    for i in range(len(new[row])):
        count = 0
        if old[row][i] == 0:
            if old[row][i-1] == 0:
                count += 1
            if old[row][i+1] == 0:
                count += 1
            if old[next_row][i] == 0:
                count += 1
            if old[next_row][i-1] == 0:
                count += 1
            if old[next_row][i+1] == 0:
                count += 1  

        if count > 4:
            new[row][i] = 0

def middle(old, new):
    for i, l in enumerate(new[1:-1]):
        for j in range(len(l)):
            count = 0
            if old[i][j] == 0:
                if old[i][j-1] == 0:
                    count += 1
                if old[i][j+1] == 0:
                    count += 1
                if old[i-1][j] == 0:
                    count += 1
                if old[i-1][j-1] == 0:
                    count += 1
                if old[i-1][j+1] == 0:
                    count += 1
                if old[i+1][j] == 0:
                    count += 1
                if old[i+1][j-1] == 0:
                    count += 1
                if old[i+1][j+1] == 0:
                    count += 1

            if count > 4:
                l[j] = 0

# first row
first_last(0, 1, old_arr, new_arr)

# middle rows
middle(old_arr, new_arr)

# last row
first_last(-1, -2, old_arr, new_arr)

print(new_arr)

Which Outputs:

[[9, 9, 6, 7, 3, 0, 0, 0, 6, 4, 3, 4], 
 [9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4], 
 [9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4], 
 [9, 9, 6, 7, 3, 0, 0, 0, 6, 4, 3, 4]]

Note: This could be made better, but you can optimize it to your liking.

edited Jan 12, 2018 at 11:46

answered Jan 11, 2018 at 8:59

RoadRunner

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9 Comments

Tak Over a year ago

Is it possible to update it to 0 only if 4 out of it's surrounding neighbour cells have the value 0 as well? Like they are multidimensional array (lists) which represents rows and cols so they are like a big table. So in this image i.sstatic.net/vKthv.jpg the blue cell in the new_arr will only be updated to zero if the corresponding cell in the old_arr is 0 and 4 of it's neighbour cells are 0 (white cells in the photo)

RoadRunner Over a year ago

@Tak definetly possible, I'll update my answer. Can you elaborate "4 out of its surrounding neighbour cells"? Is that 2 on each side of the zero? Or is that 4 on each side of the zero? I don't understand this.

Tak Over a year ago

I check all 8neighbour cells (sometimes 6 or 7 depending on the cell position) if the are zeros or not and count them, if the count is 4 or more then set the cell value to 0

RoadRunner Over a year ago

@Tak Sorry I was a bit busy, forgot to respond. Could you perhaps post a example input and expected output for this behaviour? This would make it alot easier.

Tak Over a year ago

Question updated, please let me know if I need to clarify anything more. Thanks

|

user2314737 · Accepted Answer · 2018-01-11 09:02:42Z

2

A simple solution with a list comprehension:

>>> old_arr=[9,9,6,7,3,6,5,0,6,4,3,4]
>>> new_arr=[8,8,8,8,0,0,0,0,6,6,5,5]
>>> [new_arr[i] if old_arr[i] else 0 for i in range(len(new_arr))]
[9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4]

answered Jan 11, 2018 at 9:02

user2314737

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Comments

Serge Ballesta · Accepted Answer · 2018-01-11 09:49:29Z

2

Lists are modifiable sequences in Python so you can change them in place, which can make sense for large data sets. You can do that simply with 2 nested loops:

for i, l in enumerate(new_arr):
    for j in range(len(l)):
        if old_arr[i][j] == 0:
            l[j] = 0

edited Jan 11, 2018 at 9:49

answered Jan 11, 2018 at 9:05

Serge Ballesta

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3 Comments

Tak Over a year ago

Gives a Type Error range() integer end argument expected, got list

Tak Over a year ago

@SergeBallesta but how to save the updates back to new_arr?

Serge Ballesta Over a year ago

@Tak: the updates are done directly in new_arr. What is your problem?

Reshma Parab · Accepted Answer · 2018-01-11 09:56:21Z

0

def custom_compare(list_old, list_new):
a=[]
for num, each_list_old in enumerate(list_old):
    for num1,each_list_new_entry in enumerate(list_new[num]):
        if each_list_old[num1] != 0:
            a.append(each_list_new_entry)
        else:
            a.append(0)
    list_new[num] = a
    a=[]
print "Finally new_arr = ",list_new

Eg: old_arr = [[8,8,8,8,0,0,0,0,6,6,5,5],[1,2,3,4,5,6,7,8]] new_arr = [[9,9,6,7,3,6,5,0,6,4,3,4],[8,3,4,5,6,78,8,9]] custom_compare(old_arr, new_arr) Finally new_arr = [[9, 9, 6, 7, 0, 0, 0, 0, 6, 4, 3, 4], [8, 3, 4, 5, 6, 78, 8, 9]]

answered Jan 11, 2018 at 9:56

Reshma Parab

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Comments

user9158931 · Accepted Answer · 2018-01-12 16:11:42Z

A simple approach would be something like this:

first=[[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5],[8,8,8,8,0,0,0,0,6,6,5,5]]
second = [[9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4],
 [9,9,6,7,3,6,5,0,6,4,3,4]]

for first_1,second_1 in zip(first,second):
    for index,value in enumerate(first_1):
        if value==0:
            try:
                if first_1[index-1]==0 and first_1[index+1]==0:
                    second_1[index]=0
                    second_1[index-1]=0
                    second_1[index+1]=0
            except IndexError:
                pass
print(second)

output:

[[9 9 6 7 0 0 0 0 6 4 3 4]
 [9 9 6 7 0 0 0 0 6 4 3 4]
 [9 9 6 7 0 0 0 0 6 4 3 4]
 [9 9 6 7 0 0 0 0 6 4 3 4]]

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