Question

I am binning a 2d array (x by y) in Python into the bins of its x value (given in “bins”), using np.digitize:

elements_to_bins = digitize(vals, bins)

where “vals” is a 2d array, i.e.:

 vals = array([[1, v1], [2, v2], ...]). 

elements_to_bins just says what bin each element falls into. What I then want to do is get a list whose length is the number of bins in “bins”, and each element returns the y-dimension of “vals” that falls into that bin. I do it this way right now:

points_by_bins = []
for curr_bin in range(min(elements_to_bins), max(elements_to_bins) + 1):
    curr_indx = where(elements_to_bins == curr_bin)[0]
    curr_bin_vals = vals[:, curr_indx]
    points_by_bins.append(curr_bin_vals)

is there a more elegant/simpler way to do this? All I need is a list of of lists of the y-values that fall into each bin.

thanks.

Answer 1

If I understand your question correctly:

vals = array([[1, 10], [1, 11], [2, 20], [2, 21], [2, 22]])  # Example

(x, y) = vals.T  # Shortcut
bin_limits = range(min(x)+1, max(x)+2)  # Other limits could be chosen
points_by_bin = [ [] for _ in bin_limits ]  # Final result
for (bin_num, y_value) in zip(searchsorted(bin_limits, x, "right"), y):  # digitize() finds the correct bin number
    points_by_bin[bin_num].append(y_value)

print points_by_bin  # [[10, 11], [20, 21, 22]]

Numpy’s fast array operation searchsorted() is used for maximum efficiency. Values are then added one by one (since the final result is not a rectangular array, Numpy cannot help much, for this). This solution should be faster than multiple where() calls in a loop, which force Numpy to re-read the same array many times.