Convert string array to numpy array

Question

I'm attempting to convert the string '[ 0. 0. 1.]' to a numpy array.

This is the code I've written but is more complicated that needs be ?

arr = []
s = '[ 0. 0. 1.]'
arr.append(int(s.split(" ")[1].replace("." , '')))
arr.append(int(s.split(" ")[3].replace("." , '')))
arr.append(int(s.split(" ")[5].replace("]" , '').replace("." , '')))
arr = np.array(arr)
print(arr)
print(type(arr))
print(type(arr[0]))

Above code prints :

[0 0 1]
<class 'numpy.ndarray'>
<class 'numpy.int64'>

Is there a cleaner method to convert string '[ 0. 0. 1.]' to numpy int array type ?

• Paul Panzer: depends on how flexible you want to be. np.array(s[1:-1].split(), float).astype(int)
• Paul Panzer: or even np.matrix(s).A1.astype(int)
• blue-sky: @PaulPanzer s = '[ 0. 0. 1.]' np.matrix(s) returns 'TypeError: Invalid data string supplied: ['
• Paul Panzer: Strange, it works for me. Which versions are you on?
• Paul Panzer: Looks like it changed sometime during 1.13.x. So if you are 1.12 or less you'll have to do np.matrix(s[1:-1]).A1.astype(int)

Answer 1

Numpy as can handle it much easier than all the answers:

s = '[ 0. 0. 1.]'
np.fromstring(s[1:-1],sep=' ').astype(int)

Answer 2

IN:

import numpy as np
s = '[ 0. 0. 1.]'
out = np.array([int(i.replace('.','')) for i in s[s.find('[')+1:s.find(']')].split()])
print(type(out))

OUT:

<class 'numpy.ndarray'>