Round each number of list to most near number in another list

Question

Suppose I have a certain list x with numbers, and another list y with other numbers. Elements of y should be elements of x, but due to noise in measurements, they are kind of different. I want to find, for each value of y, the value of x that is the nearest to it.

I can do this with some loops and check, for each element y[i], which element x[j] minimizes abs(x[j]-y[i]), but I'm pretty sure there is a much easier, cleaner way to do this. The lists could be huge so I'm looking for efficient code here.

The code I've written so far is:

x_in = [1.1, 2.2, 3, 4, 6.2]
y_in = [0.9, 2, 1.9, 6, 5, 6, 6.2, 0.5, 0, 3.1]
desired_output = [1.1, 2.2, 2.2, 6.2, 4, 6.2, 6.2, 1.1, 1.1, 3]

y_out = []

for y in y_in:
    aux = [abs(l - y) for l in x_in]
    mn,idx = min( (aux[i],i) for i in range(len(aux)) )
    y_out.append(x_in[idx])

>>> y_out == desired_output
True

But I don't know if there is a more efficient way to do this...

EDIT:

Due to my ignorance, I forgot to clarify something that may be of relevance according to the comments I've recieved.

• The x list is sorted.
• x is the only list that can have a pretty big size: between 500,000 and 1,000,000 elements, in general. y will in general be really small, less than 10 elements.

Answer 1

Given that x is sorted, the most efficient way to do this is using bisect to search for the closest value. Just create a list of mid points between the x values and run bisect on those:

In [69]: mid_points = [(x1+x2)/2 for x1, x2 in zip(x[1:], x[:-1])]

In [70]: mid_points
Out[70]: [1.5, 2.5, 3.5, 4.5]

In [72]: [x[bisect.bisect(mid_points, v)] for v in y]
Out[72]: [1, 1, 4, 5, 2]

This will run in O(Mlog(N)+N) time where `M=len(y), N=len(x)

(For python2 do from __future__ import division or use float(x1+x2)/2 in the mid_points calculation)

Answer 2

You can do this quickly with a lambda function and list comprehension:

[min(x, key=lambda x:abs(x-a)) for a in y]

This will work with floats, integers, etc.

Answer 3

So this is something quick I made up that just gets all the differences and than sorts them from least to greatest. Takes the lowest difference, and goes from there.

x = [1, 2, 3, 4, 5]
y = [1.1, 1.2, 3.6, 6.2, 2.1]

for y_index in range(len(y)):
    value_and_index= {}
    for x_index in range(len(x)):
        difference= y[y_index]-x[x_index]
        difference= difference*-1 if difference<0 else difference
        value_and_index[difference]= x_index
    y[y_index]= x[value_and_index[sorted(value_and_index.keys())[0]]]

print y # [1, 1, 4, 5, 2]

Hope this helps, happy coding!

Answer 4

My attempt:

First I sort the X array (if it isn't already sorted). The loop goes through each y and compute absolute value for each x, until this absolute value is higher than previous, then stop the for loop (because array X is sorted):

x = sorted([1, 2, 3, 4, 5])
y = [1.1, 1.2, 3.6, 6.2, 2.1]

out = []
while y:
    current_value = y.pop()
    current_min = float('inf')
    current_x_value = None
    for v in x:
        temp_min = abs(current_value - v)
        if temp_min < current_min:
            current_min = temp_min
            current_x_value = v
        if temp_min > current_min:  # no need to iterate further, X is sorted
            break
    out.insert(0, current_x_value)
print(out)

Outputs:

[1, 1, 4, 5, 2]

Answer 5

If x is sorted, use bisect:

import bisect 
test_out=[]
max_x=max(x)
min_x=min(x)
for f in y:
    if f>=max_x:
        idx=-1
    elif f<=min_x:
        idx=0
    else:
        idx=bisect.bisect_left(x,f)
        if abs(x[idx-1]-f)<abs(x[idx]-f):
            idx-=1
    test_out.append(x[idx])

>>> test_out==desired_output
True

Answer 6

With next assumptions:

• order of results doesn't matter,

• we are using Python 3.3+.

pretty straightforward solution may look like

from itertools import repeat


def evaluate(expected_values, measurements):
    if not expected_values:
        raise ValueError('Expected values should be a non-empty sequence.')
    expected_values = sorted(expected_values)
    measurements = sorted(measurements)
    expected_iter = iter(expected_values)
    left_value = next(expected_iter)
    try:
        right_value = next(expected_iter)
    except StopIteration:
        # there is only one expected value
        yield from repeat(left_value,
                          len(measurements))
        return
    for evaluated_count, measurement in enumerate(measurements):
        while measurement > right_value:
            try:
                left_value, right_value = right_value, next(expected_iter)
            except StopIteration:
                # rest of the measurements are closer to max expected value
                yield from repeat(right_value,
                                  len(measurements) - evaluated_count)
                return

        def key(expected_value):
            return abs(expected_value - measurement)

        yield min([left_value, right_value],
                  key=key)

For Python3.3- we can replace

yield from repeat(object_, times)

with for-loop like

for _ in range(times):
    yield object_

Test

>>> x_in = [1.1, 2.2, 3, 4, 6.2]
>>> y_in = [0.9, 2, 1.9, 6, 5, 6, 6.2, 0.5, 0, 3.1, 7.6, 10.4]
>>> y_out = list(evaluate(x_in, y_in))
>>> y_out
[1.1, 1.1, 1.1, 2.2, 2.2, 3, 4, 6.2, 6.2, 6.2, 6.2, 6.2]