Numpy: Divide each row by a vector element

Here you go. You just need to use None (or alternatively np.newaxis) combined with broadcasting:

In [6]: data - vector[:,None]
Out[6]:
array([[0, 0, 0],
       [0, 0, 0],
       [0, 0, 0]])

In [7]: data / vector[:,None]
Out[7]:
array([[1, 1, 1],
       [1, 1, 1],
       [1, 1, 1]])

As has been mentioned, slicing with None or with np.newaxes is a great way to do this. Another alternative is to use transposes and broadcasting, as in

(data.T - vector).T

and

(data.T / vector).T

For higher dimensional arrays you may want to use the swapaxes method of NumPy arrays or the NumPy rollaxis function. There really are a lot of ways to do this.

For a fuller explanation of broadcasting, see http://docs.scipy.org/doc/numpy/user/basics.broadcasting.html

Pythonic way to do this is ...

np.divide(data.T,vector).T

This takes care of reshaping and also the results are in floating point format. In other answers results are in rounded integer format.

#NOTE: No of columns in both data and vector should match

JoshAdel's solution uses np.newaxis to add a dimension. An alternative is to use reshape() to align the dimensions in preparation for broadcasting.

data = np.array([[1,1,1],[2,2,2],[3,3,3]])
vector = np.array([1,2,3])

data
# array([[1, 1, 1],
#        [2, 2, 2],
#        [3, 3, 3]])
vector
# array([1, 2, 3])

data.shape
# (3, 3)
vector.shape
# (3,)

data / vector.reshape((3,1))
# array([[1, 1, 1],
#        [1, 1, 1],
#        [1, 1, 1]])

Performing the reshape() allows the dimensions to line up for broadcasting:

data:            3 x 3
vector:              3
vector reshaped: 3 x 1

Note that data/vector is ok, but it doesn't get you the answer that you want. It divides each column of array (instead of each row) by each corresponding element of vector. It's what you would get if you explicitly reshaped vector to be 1x3 instead of 3x1.

data / vector
# array([[1, 0, 0],
#        [2, 1, 0],
#        [3, 1, 1]])
data / vector.reshape((1,3))
# array([[1, 0, 0],
#        [2, 1, 0],
#        [3, 1, 1]])