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I'm not sure how to explain why I want to do this but...why does this code

x = np.array([1,2,3,4]) d = np.empty((0, 4)) d = np.append(d,[x],axis=0) x[0]=8 d = np.append(d,[x],axis=0)

give me this

array([[ 1., 2., 3., 4.], [ 8., 2., 3., 4.]])

while this code

x = np.array([1,2,3,4]) d = np.empty((0, 4)) d = np.append(d,[x],axis=0) x = d[0,] x[0]=8 d = np.append(d,[x],axis=0)

gives me this?

array([[ 8., 2., 3., 4.], [ 8., 2., 3., 4.]])

Thanks in advance for any help here!

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1 Answer

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The issue is not with append but rather the assignment (bindings) of numpy arrays as pointers to the same location in memory.

Consider the following example

import numpy as np x = np.array([1, 2, 3, 4]) y = x y[0] = 8 print(x)

One would expect that since one did not modify the array x, that the print statement would yield the output [1,2,3,4] but in fact the real output is [8,2,3,4]. The arrays x and y are indistinguishable because they are pointers and via the assignment y=x they point to the same address of memory.

I assume that in your example, pointers became shared with the command x=d[0,].

See for a more thorough discussion.

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