Transpose, then unpack:

>>> x, y, z = data.T
>>> x
array([1, 4, 7])

One way of doing it using hstack:

import numpy as np
a = np.array([['burger flipper',  'part time',  '12-5.00'],
             ['spam flipper',    'full time',  '98-10.00']])
a = np.hstack((a[:,:2], map(lambda x: x.split('-'), a[:,2])))
print a

Output:

[['burger flipper' 'part time' '12' '5.00']
 ['spam flipper' 'full time' '98' '10.00']]

A bit of explanation:

1. The function numpy.hstack allows you to horizontally stack multiple numpy arrays. For example,

   np.hstack((a[:,[0,1]], a[:,[2]]))
   

   produces the original array a with three columns. Note the use of brackets in a[:,[2]], [a:,2] will not work as it produces a single dimensional array (len(a[:,2].shape) equals 1).

2. The map statement apply a function lambda x: x.split('-') to the problematic column (i.e. the 3rd column) of the array. Each call to the lambda function returns a list containing the separated job codes and wage, such as ['12', '5.00']. Thus, the map statement produces a list of list which looks like [['12', '5.00'], ['98', '10.00']]. This can be converted to a numpy array with 2 columns when being fed to hstack.

The code hstack first two columns of the original array with the list of list obtained via map, resulting in an array similar to what you want in the end.

I think NumPy is good for this:

>>> import numpy as np
>>> my_list = [[1,2,3],[4,5,6],[7,8,9]]
>>> x = np.array(my_list)
>>> np.transpose(x).tolist()
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

If you're using Numpy, first find the rows where the third column has your desired value, then extract the rows using indexing.

Demo

>>> import numpy
>>> A = numpy.array([[1, 0, 1],
                     [2, 0, 1],
                     [3, 0, 0],
                     [4, 0, 0],
                     [5, 0, 0]])
>>> A1 = A[A[:, 2] == 1, :] # extract all rows with the third column 1
>>> A0 = A[A[:, 2] == 0, :] # extract all rows with the third column 0
>>> A0
array([[3, 0, 0],
       [4, 0, 0],
       [5, 0, 0]])
>>> A1
array([[1, 0, 1],
       [2, 0, 1]])