python: round float values to interval limits / grid

I've got an array of (random) floating point numbers. I want to round each value up to a limit of an arbitrary grid. See the following example:

import numpy as np
np.random.seed(1)

# setup
sample = np.random.normal(loc=20, scale=6, size=10)
intervals = [-np.inf, 10, 12, 15, 18, 21, 25, 30, np.inf]

# round each interval up
for i in range(len(intervals) - 1):
sample[np.logical_and(sample > intervals[i], sample <= intervals[i+1])] = intervals[i+1]


This results in:

[ 30. 18. 18. 15. 30. 10. inf 18. 25. 21.]


My question is: How can I avoid the for-loop? I'm sure there's some way using numpy's array magic that I don't see right now.

4 Answers
4

If intervals is sorted, you can use np.searchsorted:

intervals

np.searchsorted

np.array(intervals)[np.searchsorted(intervals, sample)]
# array([ 30., 18., 18., 15., 30., 10., inf, 18., 25., 21.])


searchsorted returns the index of the interval where the element belongs to:

searchsorted

np.searchsorted(intervals, sample)
# array([7, 4, 4, 3, 7, 1, 8, 4, 6, 5])


The default side='left' returns the smallest index of such interval and the result falls into the left open, right close scenario.

side='left'

Did not run a check but:

from bisect import bisect

for index, value in enumerate(sample):
sample[index] = intervals[ bisect( intervals, value)]


You can use Pandas cut():

cut()

import pandas as pd

pd.cut(sample, intervals, labels=intervals[1:]).tolist()


If values is a 1D arrays with your values, you could do something like

values

diff = values < intervals[:, None]
t = np.argmax(diff, axis=0)
new_values = intervals[t]


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