numpy.diff(a, n=1, axis=-1)[source]

Calculate the n-th discrete difference along the given axis.

The first difference is given by out[n] = a[n+1] - a[n] along the given axis, higher differences are calculated by using diff recursively.

a : array_like

Input array

n : int, optional

The number of times values are differenced. If zero, the input is returned as-is.

axis : int, optional

The axis along which the difference is taken, default is the last axis.

diff : ndarray

The n-th differences. The shape of the output is the same as a except along axis where the dimension is smaller by n. The type of the output is the same as the type of the difference between any two elements of a. This is the same as the type of a in most cases. A notable exception is datetime64, which results in a timedelta64 output array.

See also

gradient, ediff1d, cumsum


Type is preserved for boolean arrays, so the result will contain False when consecutive elements are the same and True when they differ.

For unsigned integer arrays, the results will also be unsigned. This should not be surprising, as the result is consistent with calculating the difference directly:

>>> u8_arr = np.array([1, 0], dtype=np.uint8)
>>> np.diff(u8_arr)
array([255], dtype=uint8)
>>> u8_arr[1,...] - u8_arr[0,...]
array(255, np.uint8)

If this is not desirable, then the array should be cast to a larger integer type first:

>>> i16_arr = u8_arr.astype(np.int16)
>>> np.diff(i16_arr)
array([-1], dtype=int16)


>>> x = np.array([1, 2, 4, 7, 0])
>>> np.diff(x)
array([ 1,  2,  3, -7])
>>> np.diff(x, n=2)
array([  1,   1, -10])
>>> x = np.array([[1, 3, 6, 10], [0, 5, 6, 8]])
>>> np.diff(x)
array([[2, 3, 4],
       [5, 1, 2]])
>>> np.diff(x, axis=0)
array([[-1,  2,  0, -2]])
>>> x = np.arange('1066-10-13', '1066-10-16', dtype=np.datetime64)
>>> np.diff(x)
array([1, 1], dtype='timedelta64[D]')

Previous topic


Next topic