CODEWITHSUNDEEP
python
yakxxx
yakxxx

Reputation: 2941

2d array of zeros

There is no array type in python, but to emulate it we can use lists. I want to have 2d array-like structure filled in with zeros. My question is: what is the difference, if any, in this two expressions:

zeros = [[0 for i in xrange(M)] for j in xrange(M)]

and

zeros = [[0]*M]*N

Will zeros be same? which one is better to use by means of speed and readability?

Upvotes: 68

Views: 149955

Answers (5)

Somen Chanda
Somen Chanda

Reputation: 1

time = np.arange(0,12,2)
N = [np.zeros_like(time) for _ in range(2)]
print(N)

Result:

[array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]), array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])]

N[0][1] = 2
print(N)

Result:

[array([0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]), array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])]

If we want the length of the array N to be same as of another array S, say, then:

N = [np.zeros_like(time) for _ in range(len(S))]

In this way we can have two arrays S and N where each element in S will have a corresponding array-element in N. For ex., S=[1-30:days in a month] and N=[hourly temperature for each day: 30 such arrays]. This N can be initialized just like above.

Upvotes: -3

mgilson
mgilson

Reputation: 309831

You should use numpy.zeros. If that isn't an option, you want the first version. In the second version, if you change one value, it will be changed elsewhere in the list -- e.g.:

>>> a = [[0]*3]*3
>>> a
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> a[0][0] = 1
>>> a
[[1, 0, 0], [1, 0, 0], [1, 0, 0]]

This is because (as you read the expression from the inside out), you create a list of 10 zeros. You then create a list of 10 references to that initial list of 10 zeros.

Note that this will also work and it avoids the nested list comprehension:

zeros = [ [0]*M for _ in range(N) ]

If numpy isn't on the table, this is the form I would use.

(Use xrange if you're still stuck in the python2.x dark ages :).)

Upvotes: 112

Crabigator360
Crabigator360

Reputation: 1032

Zhe Hu's answer is the safer one and should have been the best answer. This is because if we use the accepted answer method

a = [[0] * 2] * 2
a[0][0] = 1
print(a)

will give the answer

[[1,0],[1,0]]

So even though you just want to update the first row first column value, all the values in the same column get updated. However

a = [[0] * 2 for _ in range(2)]
a[0][0] = 1
print(a)

gives the correct answer

[[1,0],[0,0]]

Upvotes: 5

Ivan Mushketyk
Ivan Mushketyk

Reputation: 8285

In second case you create a list of references to the same list. If you have code like:

[lst] * N

where the lst is a reference to a list, you will have the following list:

[lst, lst, lst, lst, ..., lst]

But because the result list contains references to the same object, if you change a value in one row it will be changed in all other rows.

Upvotes: 25

Zhe Hu
Zhe Hu

Reputation: 4007

for Python 3 (no more xrange), the preferred answer

zeros = [ [0] * N for _ in range(M)]

for M x N array of zeros

Upvotes: 53

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