import numpy as np

the array looks like so:

 array = np.zeros((10,10))array = [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]]

the dictionary is like this:

 dict = {72: (3, 4), 11: (1, 5), 10: (2, 4), 43: (2, 3), 22: (24,35), 11: (8, 9)}

I want to iterate over the array and replace any grid points that match the grid coordinates in the dictionary with the corresponding value in the dictionary

i am after an output like this:

 [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0. 11.  0.  0.  0.  0.][ 0.  0.  0. 43. 10.  0.  0.  0.  0.  0.][ 0.  0.  0.  0. 72.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.][ 0.  0.  0.  0.  0.  0.  0.  0.  0. 11.][ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]]

** i have edited the question the provide co-ordinates that sit within the array except for 1 exception. I also provided an example of the desired output

I hope I understood your question correctly

array = np.zeros((10,10))
data = {72: (3, 4), 11: (1, 5), 10: (2, 4), 43: (2, 3), 22: (24,35)}for i in data.keys():try:array[data[i][0],data[i][1]] = float(i)except IndexError:pass
print array

I changed the indices such that it fits into your 10 x 10 array (I assume you work with a bigger array in your real example)

I iterate over all keys in the dictionary (the values). The program then tries to set this value in the array at the given coordinates. I pass IndexErrors for the case some coordinates are outside the array (like the last in this example.

EDIT

This solution only works if your keys are unique. If they are not I would recommend the solution of @Osssan.