python - Simple Numpy 2-D matrix and meshgrid - ValueError: The truth value of an array with more than one element is ambiguous -

i'm new python, , task simple, i'm not getting it. i'm trying model light transmission through 2-d aperture using grid: if light transmitted, grid-element 1; if not, 0.

i'm having trouble when iterating through 2-d matrix. think there issue when comparing element of numpy array scalar value, because 2 conditions must met. tried a.any() , a.all(), &, , np.logical_and(), can't seem them work.

def make_object(x,y,a,b):     f = np.zeros((len(x),len(y)))      if np.abs(x) < (a/2.0):         if np.abs(y) < (b/2.0):             f[x][y] = 1.0     return f   = 6.0  # width of slit b = 6.0  # height of slit n = 20.0 x = np.linspace(-10.0,10.0,n) y = np.linspace(-10.0,10.0,n) x,y = np.meshgrid(x,y)  z = make_object(x,y,a,b) print z 

i error message:

--------------------------------------------------------------------------- valueerror                                traceback (most recent call last) <ipython-input-13-fef282d4a308> in <module>()      29 x,y = np.meshgrid(x,y)      30  ---> 31 z = make_object(x,y,a,b)      32       33   <ipython-input-13-fef282d4a308> in make_object(x, y, a, b)       8     f = np.zeros((len(x),len(y)))       9  ---> 10     if np.abs(x) < (a/2.0):      11         if np.abs(y) < (b/2.0):      12             f[x][y] = 1.0  valueerror: truth value of array more 1 element ambiguous. use a.any() or a.all() 

any ideas i'm doing wrong?

edit: code used without vectorization

for i, xi in enumerate(x):     j, yj in enumerate(y):         if np.abs(xi) < a/2:             if np.abs(yj) < b/2:                 f[i][j] = 1.0 

the problem x , y arrays, not individual elements. heres code evaluates before reaches error:

if array([true, false, ... ]): 

to python, evaluating whether entire array true or false makes no sense. if want use if statements, have iterate through array , check if individual elements less cutoff rather entire arrays.

for in range(f.shape[0]):     j in range(f.shape[1]):         if x[i][j] < a/2:            if y[i][j] < b/2:                 f[i][j] = 1 

however, since using numpy, not need if statements , can take advantage of vectorization , solve problem in single line.

f[ np.logical_and(np.abs(x) < a/2.0,  np.abs(y) < b/2.0) ] = 1