That is a pure python solution and it produces a list .

With simple python operations, you can map inner list with float. That will convert all string elements to float and assign it as the zero indexed item of your list.

a = [['-0.99' , '0.56' , '0.56' , '0.56', '-2.02' , '-0.96']]

a[0] = map(float, a[0])

print a
[[-0.99, 0.56, 0.56, 0.56, -2.02, -0.96]]

Update: Try the following

a = [['-0.99' , '0.56' , '0.56' , '0.56', '-2.02' , '-0.96', '', 'nan']]
for _position, _value in enumerate(a[0]):
    try:
        _new_value = float(_value)
    except ValueError:
        _new_value = 0.0
    a[0][_position] = _new_value

[[-0.99, 0.56, 0.56, 0.56, -2.02, -0.96, 0.0, nan]]

It enumerates the objects in the list and try to parse them to float, if it fails, then replace it with 0.0

import numpy as np
nums_str = ['1','23','345']
nums_str_np = np.asarray(nums_str)
nums_int_np = nums_str_np.astype('int')

nums_int_np - is now np array of integers.

You could use the following code:

X[X=='y'] = 1
X[X=='n'] = 0

This replaces the indexes of 'y' with 1 and of 'n' with 0. Generally the X=='y' returns a Boolean array which contains True where the 'y' and False everywhere else and so on.

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'>

Can just you convert the timestamp to a epoch time?