Python Numpy 2D Array Number of Rows For Empty Matrix is Still 1

Question

Say I have a matrices A, B, and C. When I initialize the matrices to

A = np.array([[]])
B = np.array([[1,2,3]])
C = np.array([[1,2,3],[4,5,6]])

Then

B.shape[0]
C.shape[0]

give 1 and 2, respectively (as expected), but

A.shape[0]

gives 1, just like B.shape[0].

What is the simplest way to get the number of rows of a given matrix, but still ensure that an empty matrix like A gives a value of zero.

After searching stack overflow for awhile, I couldn't find an answer, so I'm posting my own below, but if you can come up with a cleaner, more general answer, I'll accept your answer instead. Thanks!

user2357112 · Accepted Answer · 2016-04-04 22:10:17Z

2

A = np.array([[]])

That's a 1-by-0 array. You seem to want a 0-by-3 array. Such an array is almost completely useless, but if you really want one, you can make one:

A = np.zeros([0, 3])

Then you'll have A.shape[0] == 0.

answered Apr 4, 2016 at 22:10

user2357112

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cameronroytaylor Over a year ago

Thanks. That doesn't solve my problem, since I'm pulling in the array from a csv, which may potentially be blank. Your answer might be helpful to someone else though.

hpaulj · Accepted Answer · 2016-04-04 22:59:16Z

1

You could qualify the shape[0] by the test of whether size is 0 or not.

In [121]: A.shape[0]*(A.size>0)
Out[121]: 0
In [122]: B.shape[0]*(B.size>0)
Out[122]: 1
In [123]: C.shape[0]*(C.size>0)
Out[123]: 2

or test the number of columns

In [125]: A.shape[0]*(A.shape[1]>0)
Out[125]: 0

What's distinctive about A is the number of columns, the 2nd dimension.

answered Apr 4, 2016 at 22:59

hpaulj

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Comments

cameronroytaylor · Accepted Answer · 2016-04-04 23:09:44Z

0

Using

A.size/(len(A[0]) or 1)
B.size/(len(B[0]) or 1)
C.size/(len(C[0]) or 1)

yields 0, 1, and 2, respectively.

answered Apr 4, 2016 at 23:09

cameronroytaylor

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