I want to replace an item in a numpy array composed of 0 and 1. I want to replace 1 with the list [0,1] and the 0 with the list [1,0].
I found the function numpy.where(), but it doesn't work with lists.
Is there a function or something similar to do the equivalent for numpy.where(vector==1,[0,1], [1,0]) ?
Simple indexing should do the job:
In [156]: x = np.array([0,1,0,0,1,1,0])
In [157]: y = np.array([[1,0],[0,1]])
In [158]: y[x]
Out[158]:
array([[1, 0],
[0, 1],
[1, 0],
[1, 0],
[0, 1],
[0, 1],
[1, 0]])
And just to be sure this is a general solution, not some 'boolean' fluke
In [162]: x = np.array([0,1,0,0,1,2,2])
In [163]: y = np.array([[1,0],[0,1],[1,1]])
In [164]: y[x]
Out[164]:
array([[1, 0],
[0, 1],
[1, 0],
[1, 0],
[0, 1],
[1, 1],
[1, 1]])
You can actually use where for that if you wish:
>>> import numpy as np
>>> vector = np.random.randint(0, 2, (8, 8))
>>> vector
array([[1, 0, 0, 0, 0, 0, 1, 1],
[1, 1, 0, 1, 0, 0, 0, 1],
[1, 0, 0, 1, 1, 1, 0, 0],
[1, 0, 1, 1, 0, 0, 0, 0],
[0, 1, 0, 1, 1, 1, 1, 1],
[1, 0, 1, 1, 0, 0, 0, 0],
[0, 1, 1, 0, 1, 1, 0, 0],
[1, 1, 1, 1, 1, 1, 0, 0]])
>>> np.where(vector[..., None] == 1, [0,1], [1,0])
# btw. if vector has only 0 and 1 entries you can leave out the " == 1 "
array([[[0, 1],
[1, 0],
[1, 0],
[1, 0],
[1, 0],
[1, 0],
[0, 1],
[0, 1]],
[[0, 1],
[0, 1],
[1, 0],
[0, 1],
[1, 0],
[1, 0],
[1, 0],
etc.
Looks like you want to make one-hot vector encoding:
a = np.array([1, 0, 1, 1, 0]) # initial array
b = np.zeros((len(a), 2)) # where 2 is number of classes
b[np.arange(len(a)), a] = 1
print(b)
result:
array([[ 0., 1.],
[ 1., 0.],
[ 0., 1.],
[ 0., 1.],
[ 1., 0.]])
Since it's not possible to grow the array dynamically in NumPy, you've to convert the array to list, do the substitution as desired, and then convert the result back to NumPy array like:
In [21]: vector = np.array([0, 0, 1, 0, 1])
In [22]: vlist = vector.tolist()
In [23]: vlist
Out[23]: [0, 0, 1, 0, 1]
In [24]: new_list = [[1, 0] if el == 0 else [0, 1] for idx, el in enumerate(vlist)]
In [25]: result = np.array(new_list)
In [26]: result
Out[26]:
array([[1, 0],
[1, 0],
[0, 1],
[1, 0],
[0, 1]])
If your 1D array is not that large enough, then list comprehension is not that bad of an approach.
hpaulj's answer would be the ideal way of doing this. An alternative is to reshape your original numpy array, and apply your standard np.where() operation.
import numpy as np
x = np.array([0,1,0,0,1,1,0])
x = np.reshape(x,(len(x),1))
# Equivalently, if you want to be explicit:
# x = np.array([[e] for e in x])
print np.where(x==1,[1,0],[0,1])
# Result:
array([[0, 1],
[1, 0],
[0, 1],
[0, 1],
[1, 0],
[1, 0],
[0, 1]])