Numpy assignment behaviour

By using expression a = b you actually copy the reference to b. If you want to copy its value, you should rather walk through all b's items and copy their values to a.

In numpy you should use copy function.

>>> import numpy
>>> b = numpy.array([-0.06106568, -0.10843541, -0.0694688 ,  0.02464023, -0.03686665,
...        -0.0582096 , -0.13476669, -0.08505708,  0.00391955, -0.12300518,
...        -0.01183732, -0.05374973, -0.12300518, -0.05312849,  0.01963862,
...         0.00155719, -0.10843541, -0.08490177, -0.08505708, -0.02026149,
...        -0.01777489,  0.01183732, -0.11575136,  0.04278603, -0.0694688 ,
...        -0.06106568, -0.08755022, -0.01660802, -0.06087603, -0.06582411])
>>> a = numpy.copy(b)
>>> a
array([-0.06106568, -0.10843541, -0.0694688 ,  0.02464023, -0.03686665,
       -0.0582096 , -0.13476669, -0.08505708,  0.00391955, -0.12300518,
       -0.01183732, -0.05374973, -0.12300518, -0.05312849,  0.01963862,
        0.00155719, -0.10843541, -0.08490177, -0.08505708, -0.02026149,
       -0.01777489,  0.01183732, -0.11575136,  0.04278603, -0.0694688 ,
       -0.06106568, -0.08755022, -0.01660802, -0.06087603, -0.06582411])
>>> a[a<= 0] = 0
>>> a[a> 0] = 1
>>> a
array([ 0.,  0.,  0.,  1.,  0.,  0.,  0.,  0.,  1.,  0.,  0.,  0.,  0.,
        0.,  1.,  1.,  0.,  0.,  0.,  0.,  0.,  1.,  0.,  1.,  0.,  0.,
        0.,  0.,  0.,  0.])
>>> b
array([-0.06106568, -0.10843541, -0.0694688 ,  0.02464023, -0.03686665,
       -0.0582096 , -0.13476669, -0.08505708,  0.00391955, -0.12300518,
       -0.01183732, -0.05374973, -0.12300518, -0.05312849,  0.01963862,
        0.00155719, -0.10843541, -0.08490177, -0.08505708, -0.02026149,
       -0.01777489,  0.01183732, -0.11575136,  0.04278603, -0.0694688 ,
       -0.06106568, -0.08755022, -0.01660802, -0.06087603, -0.06582411])

There are some native ways to do it, but if you work with mathematics, using only numpy is strongly recommended.

UPDATE

I don't have in my mind any non-numpy way which will cause entire compatibility with your case.

When you did a=b then you made a named ref to b so a and b were views on the same object, if you wanted a copy then use np.copy:

In [35]:
a = np.copy(b)
a[a<= 0] = 0
a[a> 0] = 1
a

Out[35]:
array([ 0.,  0.,  0.,  1.,  0.,  0.,  0.,  0.,  1.,  0.,  0.,  0.,  0.,
        0.,  1.,  1.,  0.,  0.,  0.,  0.,  0.,  1.,  0.,  1.,  0.,  0.,
        0.,  0.,  0.,  0.])

In [36]:
b

Out[36]:
array([-0.06106568, -0.10843541, -0.0694688 ,  0.02464023, -0.03686665,
       -0.0582096 , -0.13476669, -0.08505708,  0.00391955, -0.12300518,
       -0.01183732, -0.05374973, -0.12300518, -0.05312849,  0.01963862,
        0.00155719, -0.10843541, -0.08490177, -0.08505708, -0.02026149,
       -0.01777489,  0.01183732, -0.11575136,  0.04278603, -0.0694688 ,
       -0.06106568, -0.08755022, -0.01660802, -0.06087603, -0.06582411])

You are actually just pointing a to b, that both contain the same list. Essentially python variables are names that are given to objects. You want to create a new object containing the new list here. You have many options:

You could use the built-in list() function:

a = list(b)

Or you could import copy and use that.

import copy
a = copy.copy(b)

or in the case of numpy:

import numpy
a = np.copy(b)