digitize(x, bins, right=False)¶
Return the indices of the bins to which each value in input tensor belongs.
ireturned is such that
bins[i-1] <= x < bins[i]if bins is monotonically increasing, or
bins[i-1] > x >= bins[i]if bins is monotonically decreasing. If values in x are beyond the bounds of bins, 0 or
len(bins)is returned as appropriate. If right is True, then the right bin is closed so that the index
iis such that
bins[i-1] < x <= bins[i]or
bins[i-1] >= x > bins[i]if bins is monotonically increasing or decreasing, respectively.
- x : array_like
- Input tensor to be binned.
- bins : array_like
- Array of bins. It has to be 1-dimensional and monotonic.
- right : bool, optional
- Indicating whether the intervals include the right or the left bin edge. Default behavior is (right==False) indicating that the interval does not include the right edge. The left bin end is open in this case, i.e., bins[i-1] <= x < bins[i] is the default behavior for monotonically increasing bins.
- out : Tensor of ints
- Output tensor of indices, of same shape as x.
- If bins is not monotonic.
- If the type of the input is complex.
bincount, histogram, unique, searchsorted
If values in x are such that they fall outside the bin range, attempting to index bins with the indices that digitize returns will result in an IndexError.
mt.digitize is implemented in terms of mt.searchsorted. This means that a binary search is used to bin the values, which scales much better for larger number of bins than the previous linear search. It also removes the requirement for the input array to be 1-dimensional.
>>> import mars.tensor as mt
>>> x = mt.array([0.2, 6.4, 3.0, 1.6]) >>> bins = mt.array([0.0, 1.0, 2.5, 4.0, 10.0]) >>> inds = mt.digitize(x, bins) >>> inds.execute() array([1, 4, 3, 2])
>>> x = mt.array([1.2, 10.0, 12.4, 15.5, 20.]) >>> bins = mt.array([0, 5, 10, 15, 20]) >>> mt.digitize(x,bins,right=True).execute() array([1, 2, 3, 4, 4]) >>> mt.digitize(x,bins,right=False).execute() array([1, 3, 3, 4, 5])