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correlate(a, v, mode='valid')

This function computes the correlation as generally defined in signal processing texts:

c_{av}[k] = sum_n a[n+k] * conj(v[n])

with a and v sequences being zero-padded where necessary and conj being the conjugate.

Notes

The definition of correlation above is not unique and sometimes correlation may be defined differently. Another common definition is:

c'_{av}[k] = sum_n a[n] conj(v[n+k])

which is related to c_{av}[k] by c'_{av}[k] = c_{av}[-k] .

numpy.correlate may perform slowly in large arrays (i.e. n = 1e5) because it does not use the FFT to compute the convolution; in that case, scipy.signal.correlate might be preferable.

Parameters

a, v : array_like: Input sequences.
mode : {'valid', 'same', 'full'}, optional: Refer to the convolve docstring. Note that the default is 'valid', unlike convolve , which uses 'full'.
old_behavior : bool: :None:None:`old_behavior` was removed in NumPy 1.10. If you need the old behavior, use :None:None:`multiarray.correlate`.

Returns

out : ndarray: Discrete cross-correlation of a and v.

Cross-correlation of two 1-dimensional sequences.

Examples

>>> np.correlate([1, 2, 3], [0, 1, 0.5])
array([3.5])

>>> np.correlate([1, 2, 3], [0, 1, 0.5], "same")
array([2. ,  3.5,  3. ])

>>> np.correlate([1, 2, 3], [0, 1, 0.5], "full")
array([0.5,  2. ,  3.5,  3. ,  0. ])

Using complex sequences:

>>> np.correlate([1+1j, 2, 3-1j], [0, 1, 0.5j], 'full')
array([ 0.5-0.5j,  1.0+0.j ,  1.5-1.5j,  3.0-1.j ,  0.0+0.j ])

Note that you get the time reversed, complex conjugated result when the two input sequences change places, i.e., c_{va}[k] = c^{*}_{av}[-k] :

>>> np.correlate([0, 1, 0.5j], [1+1j, 2, 3-1j], 'full')
array([ 0.0+0.j ,  3.0+1.j ,  1.5+1.5j,  1.0+0.j ,  0.5+0.5j])

See :

Back References

The following pages refer to to this document either explicitly or contain code examples using this.

numpy.ma.core.correlate numpy.correlate matplotlib.pyplot.acorr matplotlib.axes._axes.Axes.xcorr matplotlib.pyplot.xcorr scipy.signal._max_len_seq.max_len_seq matplotlib.axes._axes.Axes.acorr

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