You can multiply numpy arrays by scalars and it just works.

>>> import numpy as np
>>> np.array([1, 2, 3]) * 2
array([2, 4, 6])
>>> np.array([[1, 2, 3], [4, 5, 6]]) * 2
array([[ 2,  4,  6],
       [ 8, 10, 12]])

This is also a very fast and efficient operation. With your example:

>>> a_1 = np.array([1.0, 2.0, 3.0])
>>> a_2 = np.array([[1., 2.], [3., 4.]])
>>> b = 2.0
>>> a_1 * b
array([2., 4., 6.])
>>> a_2 * b
array([[2., 4.],
       [6., 8.]])

 you can do this in two simple steps using NumPy:

>>> # multiply column 2 of the 2D array, A, by 5.2
>>> A[:,1] *= 5.2

>>> # assuming by 'cumulative sum' you meant the 'reduced' sum:
>>> A[:,1].sum()

>>> # if in fact you want the cumulative sum (ie, returns a new column)
>>> # then do this for the second step instead:
>>> NP.cumsum(A[:,1])

with some mocked data:

>>> A = NP.random.rand(8, 5)
>>> A
  array([[ 0.893,  0.824,  0.438,  0.284,  0.892],
         [ 0.534,  0.11 ,  0.409,  0.555,  0.96 ],
         [ 0.671,  0.817,  0.636,  0.522,  0.867],
         [ 0.752,  0.688,  0.142,  0.793,  0.716],
         [ 0.276,  0.818,  0.904,  0.767,  0.443],
         [ 0.57 ,  0.159,  0.144,  0.439,  0.747],
         [ 0.705,  0.793,  0.575,  0.507,  0.956],
         [ 0.322,  0.713,  0.963,  0.037,  0.509]])

>>> A[:,1] *= 5.2

>>> A
  array([[ 0.893,  4.287,  0.438,  0.284,  0.892],
         [ 0.534,  0.571,  0.409,  0.555,  0.96 ],
         [ 0.671,  4.25 ,  0.636,  0.522,  0.867],
         [ 0.752,  3.576,  0.142,  0.793,  0.716],
         [ 0.276,  4.255,  0.904,  0.767,  0.443],
         [ 0.57 ,  0.827,  0.144,  0.439,  0.747],
         [ 0.705,  4.122,  0.575,  0.507,  0.956],
         [ 0.322,  3.71 ,  0.963,  0.037,  0.509]])

>>> A[:,1].sum()
  25.596156138451427

just a few simple rules are required to grok element selection (indexing) in NumPy:

• NumPy, like Python, is 0-based, so eg, the "1" below refers to the second column

• commas separate the dimensions inside the brackets, so [rows, columns], eg, A[2,3] means the item ("cell") at row three, column four

• a colon means all of the elements along that dimension, eg, A[:,1] creates a view of A's column 2; A[3,:] refers to the fourth row

You could use None (or np.newaxis) to expand A to match B:

>>> A = np.arange(10)
>>> B = np.random.random((10,3,5))
>>> C0 = np.array([A[i]*B[i,:,:] for i in range(len(A))])
>>> C1 = A[:,None,None] * B
>>> np.allclose(C0, C1)
True

But this will only work for the 2 case. Borrowing from @ajcr, with enough transposes we can get implicit broadcasting to work for the general case:

>>> C3 = (A * B.T).T
>>> np.allclose(C0, C3)
True

Alternatively, you could use einsum to provide the generality. In retrospect it's probably overkill here compared with the transpose route, but it's handy when the multiplications are more complicated.

>>> C2 = np.einsum('i,i...->i...', A, B)
>>> np.allclose(C0, C2)
True

and

>>> B = np.random.random((10,4))
>>> D0 = np.array([A[i]*B[i,:] for i in range(len(A))])
>>> D2 = np.einsum('i,i...->i...', A, B)
>>> np.allclose(D0, D2)
True

Simplest solution

Use numpy.dot or a.dot(b). See the documentation here.

>>> a = np.array([[ 5, 1 ,3], 
                  [ 1, 1 ,1], 
                  [ 1, 2 ,1]])
>>> b = np.array([1, 2, 3])
>>> print a.dot(b)
array([16, 6, 8])

This occurs because numpy arrays are not matrices, and the standard operations *, +, -, / work element-wise on arrays.

Note that while you can use numpy.matrix (as of early 2021) where * will be treated like standard matrix multiplication, numpy.matrix is deprecated and may be removed in future releases.. See the note in its documentation (reproduced below):

It is no longer recommended to use this class, even for linear algebra. Instead use regular arrays. The class may be removed in the future.

Thanks @HopeKing.

Other Solutions

Also know there are other options:

• As noted below, if using python3.5+ and numpy v1.10+, the @ operator works as you'd expect:

  >>> print(a @ b)
  array([16, 6, 8])
  

• If you want overkill, you can use numpy.einsum. The documentation will give you a flavor for how it works, but honestly, I didn't fully understand how to use it until reading this answer and just playing around with it on my own.

  >>> np.einsum('ji,i->j', a, b)
  array([16, 6, 8])
  

• As of mid 2016 (numpy 1.10.1), you can try the experimental numpy.matmul, which works like numpy.dot with two major exceptions: no scalar multiplication but it works with stacks of matrices.

  >>> np.matmul(a, b)
  array([16, 6, 8])
  

• numpy.inner functions the same way as numpy.dot for matrix-vector multiplication but behaves differently for matrix-matrix and tensor multiplication (see Wikipedia regarding the differences between the inner product and dot product in general or see this SO answer regarding numpy's implementations).

  >>> np.inner(a, b)
  array([16, 6, 8])
  
  # Beware using for matrix-matrix multiplication though!
  >>> b = a.T
  >>> np.dot(a, b)
  array([[35,  9, 10],
         [ 9,  3,  4],
         [10,  4,  6]])
  >>> np.inner(a, b) 
  array([[29, 12, 19],
         [ 7,  4,  5],
         [ 8,  5,  6]])
  

• If you have multiple 2D arrays to dot together, you may consider the np.linalg.multi_dot function, which simplifies the syntax of many nested np.dots. Note that this only works with 2D arrays (i.e. not for matrix-vector multiplication).

    >>> np.dot(np.dot(a, a.T), a).dot(a.T)
    array([[1406,  382,  446],
           [ 382,  106,  126],
           [ 446,  126,  152]])
    >>> np.linalg.multi_dot((a, a.T, a, a.T))
    array([[1406,  382,  446],
           [ 382,  106,  126],
           [ 446,  126,  152]])
  

Rarer options for edge cases

• If you have tensors (arrays of dimension greater than or equal to one), you can use numpy.tensordot with the optional argument axes=1:

  >>> np.tensordot(a, b, axes=1)
  array([16,  6,  8])
  

• Don't use numpy.vdot if you have a matrix of complex numbers, as the matrix will be flattened to a 1D array, then it will try to find the complex conjugate dot product between your flattened matrix and vector (which will fail due to a size mismatch n*m vs n).