Why is python printing weird results while using the nested loop here?

Question

I was given an array of strings and I want to split each of the individual strings into separate characters and store them in a separate 2-D array. I wrote the following code:-

# The given array
grid = ['1112', '1912', '1892', '1234']

# Creating a new 2D array
mat = [[None]*len(grid)]*len(grid)
for i in range(0,len(grid)):
    for j in range(0,len(grid)):
        mat[i][j] = grid[i][j]
print(mat)

But doing this gives weird values in my two-dimensional array, mat. Each of the row of the 2D array mat gets printed as ['1','2','3','4'] but this should be the value of the last row only.

I shall be thankful if someone could answer my query.

Gugu72 · Accepted Answer · 2021-01-12 10:38:42Z

3

mat = [[None]*len(grid)]*len(grid)

This statement is creating a relational 2D array (matrix), so whenever you are updating a line, it will update all the other lines.

You should be using this to create an empty 2D array:

mat = [[None for i in len(grid)] for j in range(len(grid))]

As @marc commented, you can also pass one list comprehension as width = height here

mat = [[None]*len(grid) for _ in range(len(grid))]

edited Jan 12, 2021 at 10:38

answered Jan 12, 2021 at 10:24

Gugu72

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2 Comments

Marc Over a year ago

You could safely skip one of the list comprehension steps and use mat = [[None]*len(grid) for i in range(len(grid)].

Gugu72 Over a year ago

@Marc Ah yeah, I will add it alongside, if they later need a matrix with a width different from the height, they would still need to do 2 list comprehensions

Dan D · Accepted Answer · 2021-01-12 10:25:08Z

1

The multiplication to list makes the same reference in memory, code fixed as following:

# The given array
grid = ['1112', '1912', '1892', '1234']

# Creating a new 2D array
# mat = [[None]*len(grid)]*len(grid)
mat = []
for i in range(len(grid)):
    t = []
    for j in range(len(grid)):
        t.append(None)
    mat.append(t)

for i in range(0,len(grid)):
    for j in range(0,len(grid)):
        mat[i][j] = grid[i][j]
print(mat)

answered Jan 12, 2021 at 10:25

Dan D

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1 Comment

Gugu72 Over a year ago

This would work but is longer to process than my answer, with list comprehension, which is also more readable

Yang Liu · Accepted Answer · 2021-01-12 10:41:42Z

1

I have to agree this is easy to get confused in this problem.

Let's say len(grid) = 2, so we get mat = [[None]*2]*2

Here mat[0] points to the same pobject of mat[1], because *2 is giving two pointers which pointing to the same array. You can use for _ in range(2) instead to create two different objects - [[None]*len(grid) for _ in range(len(grid))]

We can use id to print the object's address to verify:

>>> mat = [[None]*2]*2
>>> id(mat[0])
76447616
>>> id(mat[1])
76447616
>>> grid = [0, 0]
>>> x = [[None]*len(grid) for _ in range(len(grid))]
>>> id(x[0])
76457744
>>> id(x[1])
76391968

answered Jan 12, 2021 at 10:41

Yang Liu

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