You can try this:

import numpy as np
listvals = []
for i in range(10):
    listvals.append((xc, yc))
mat_vals = np.vstack(listvals)

This will output an ndarray like this:

[[ 2  0]
 [ 3  1]
 [ 4  2]
 [ 5  3]
 [ 6  4]
 [ 7  5]
 [ 8  6]
 [ 9  7]
 [10  8]
 [11  9]]

Or this maybe is better:

import numpy as np
list_xc = []
list_yc = []
for i in np.arange(10): 
    list_xc.append(xc)
    list_yc.append(yc)
mat_xc = np.asarray(list_xc)
mat_yc = np.asarray(list_yc)
mat_f = np.column_stack((mat_xc, mat_yc))

The numpy array you are defining is not the right shape for the loop you are using. b = np.array([]) gives you an array of shape (0,)

You can use something like np.zeros to define your 2D array.

import numpy as np
b = np.zeros((3,6))
for i in range(2):
    for j in range(5):
        b[i][j]=i+j
print(b)

The output will be

[[0. 1. 2. 3. 4. 0.]
 [1. 2. 3. 4. 5. 0.]
 [0. 0. 0. 0. 0. 0.]]

Another option is to make a 2D list, fill it up in for loops, and convert to numpy array later

import numpy as np

#2D array
b = [ [0 for _ in range(6)] for _ in range(3)]

#Fill up array with for loops
for i in range(2):
    for j in range(5):
        b[i][j]=i+j

#Convert to numpy array
print(np.array(b))

The output will be

[[0 1 2 3 4 0]
 [1 2 3 4 5 0]
 [0 0 0 0 0 0]]

First lets focus on the append operation,

import numpy as np
a = np.array( [1,2,3] )
b = np.array( [4,5,6] )

np.append( a, b )

produces

array( [1, 2, 3, 4, 5, 6] )

What you might want is

np.append( [a], [b], 0 )

which produces

array([ [1, 2, 3], [4, 5, 6] ])

Note that here a 0 appears in the third parameter to specify the axis for the append operation.

Regarding the error, the index h, tells numpy to append to the 'h'-th element. That's probably not what you intended. Also, the index is not correct since 'h' is always one more than the number of elements in the array at the point where the call occurs.

Note this:

>>> matrix = [[0] * 3] * 3
>>> [x for x in matrix]
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> [id(x) for x in matrix]
[32484168, 32484168, 32484168]
>>>

Three rows but only one object.

See the docs especially Note 2 about the s * n operation.

Fix:

>>> m2= [[0] * 3 for i in xrange(5)]
>>> [id(x) for x in m2]
[32498152, 32484808, 32498192, 32499952, 32499872]
>>>

Update: Here are some samples of code that gets an answer simply (i.e. without iter()):

>>> nrows = 2; ncols = 4
>>> zeroes = [[0 for j in xrange(ncols)] for i in xrange(nrows)]
>>> zeroes
[[0, 0, 0, 0], [0, 0, 0, 0]]
>>> ap = [[ncols * i + j for j in xrange(ncols)] for i in xrange(nrows)]
>>> ap
[[0, 1, 2, 3], [4, 5, 6, 7]]
>>>

It is just a shortcut for this:

x = []
for j in range(column):
    _ = []
    for i in range(row):
        _.append(i)
    x.append(_)

you can put the inner for loop into one by saying that the list is composed of a whole bunch of 0's. We do that for each i in range(row), so we can say _ = [0 for i in range(row)]. This is how that looks:

x = []
for j in range(column):
    x.append([0 for i in range(row)])

We can now see that x is composed of a bunch of [0 for i in range(row)]'s and we do that once for each of j in range(column), so we can simplify it to this:

x = [[0 for i in range(row)] for j in range(column)]