Zip uneven numpy arrays

You can transpose b and c and then create a product of the a with the transposed array using itertools.product:

>>> from itertools import product
>>> [np.insert(j,0,i) for i,j in product(a,np.array((b,c)).T)]
[array([ 1.,  4.,  6.]), array([ 1.,  5.,  7.]), array([ 2.,  4.,  6.]), array([ 2.,  5.,  7.]), array([ 3.,  4.,  6.]), array([ 3.,  5.,  7.])]
>>> 

Let's say you have:

a = np.array([4., 5.])
b = np.array([1., 2., 3.])
c = np.array([6., 7.])
d = np.array([5., 1])
e = np.array([3., 2.])

Now, if you know before-hand which one is the longest array, which is b in this case, you can use an approach based upon np.meshgrid -

# Concatenate elements from identical positions from the equal arrays 
others = np.vstack((a,c,d,e)).T # If you have more arrays, edit this line

# Get grided version of the longest array and 
# grided-indices for indexing into others array
X,Y = np.meshgrid(np.arange(others.shape[0]),b)

# Concatenate grided longest array and grided indexed others for final output
out = np.hstack((Y.ravel()[:,None],others[X.ravel()]))

Sample run -

In [47]: b
Out[47]: array([ 1.,  2.,  3.])

In [48]: a
Out[48]: array([ 4.,  5.])

In [49]: c
Out[49]: array([ 6.,  7.])

In [50]: d
Out[50]: array([ 5.,  1.])

In [51]: e
Out[51]: array([ 3.,  2.])

In [52]: out
Out[52]: 
array([[ 1.,  4.,  6.,  5.,  3.],
       [ 1.,  5.,  7.,  1.,  2.],
       [ 2.,  4.,  6.,  5.,  3.],
       [ 2.,  5.,  7.,  1.,  2.],
       [ 3.,  4.,  6.,  5.,  3.],
       [ 3.,  5.,  7.,  1.,  2.]])

If the length differences are not extreme (check inputs first) I'd be tempted to pad out the shorter lists to the length of the longest with None and generate all the permutations (27 of them for 3 lists of 3 elements). Then

 results = []
 for candidate in possibles:
    if not (None in candidate): results.append(candidate)

Reasons not to do this: if the cube of the length of the longest list is significant in terms of memory usage (space to store N cubed possibles) or CPU usage.