numpy.average

numpy.average(a, axis=None, weights=None, returned=False)

Compute the weighted average along the specified axis.

Parameters :

a : array_like

Array containing data to be averaged. If a is not an array, a conversion is attempted.

axis : int, optional

Axis along which to average a. If None, averaging is done over the flattened array.

weights : array_like, optional

An array of weights associated with the values in a. Each value in a contributes to the average according to its associated weight. The weights array can either be 1-D (in which case its length must be the size of a along the given axis) or of the same shape as a. If weights=None, then all data in a are assumed to have a weight equal to one.

returned : bool, optional

Default is False. If True, the tuple (average, sum_of_weights) is returned, otherwise only the average is returned. If weights=None, sum_of_weights is equivalent to the number of elements over which the average is taken.

Returns :

average, [sum_of_weights] : {array_type, double}

Return the average along the specified axis. When returned is True, return a tuple with the average as the first element and the sum of the weights as the second element. The return type is Float if a is of integer type, otherwise it is of the same type as a. sum_of_weights is of the same type as average.

Raises :

ZeroDivisionError :

When all weights along axis are zero. See numpy.ma.average for a version robust to this type of error.

TypeError :

When the length of 1D weights is not the same as the shape of a along axis.

See also

mean

ma.average

average for masked arrays

Examples

>>> data = range(1,5)
>>> data
[1, 2, 3, 4]
>>> np.average(data)
2.5
>>> np.average(range(1,11), weights=range(10,0,-1))
4.0

>>> data = np.arange(6).reshape((3,2))
>>> data
array([[0, 1],
       [2, 3],
       [4, 5]])
>>> np.average(data, axis=1, weights=[1./4, 3./4])
array([ 0.75,  2.75,  4.75])
>>> np.average(data, weights=[1./4, 3./4])
...
TypeError: Axis must be specified when shapes of a and weights differ.