Python / Numpy - 快速查找最接近某些值的数组中的索引 [英] Python/Numpy - Quickly Find the Index in an Array Closest to Some Value

问题描述

我有一个值数组t，它总是按递增顺序排列（但不总是均匀间隔）。我有另一个单值x。我需要在t中找到索引，使得t [index]最接近x。对于x<该函数必须返回零。 t.min（）和x> t.max（）的最大索引（或-1）。

I have an array of values, t, that is always in increasing order (but not always uniformly spaced). I have another single value, x. I need to find the index in t such that t[index] is closest to x. The function must return zero for x < t.min() and the max index (or -1) for x > t.max().

我写了两个函数来执行此操作。在这个简单的定时测试中，第一个f1更快。但我喜欢第二个只是一行。这个计算将在一个大型数组上完成，可能每秒很多次。

I've written two functions to do this. The first one, f1, is MUCH quicker in this simple timing test. But I like how the second one is just one line. This calculation will be done on a large array, potentially many times per second.

任何人都可以提出一些其他功能与第一个功能相当，但具有更清晰的代码？比第一个更快的东西怎么样（速度最重要）？

Can anyone come up with some other function with comparable timing to the first but with cleaner looking code? How about something quicker then the first (speed is most important)?

谢谢！

代码：

import numpy as np
import timeit

t = np.arange(10,100000)         # Not always uniform, but in increasing order
x = np.random.uniform(10,100000) # Some value to find within t

def f1(t, x):
   ind = np.searchsorted(t, x)   # Get index to preserve order
   ind = min(len(t)-1, ind)      # In case x > max(t)
   ind = max(1, ind)             # In case x < min(t)
   if x < (t[ind-1] + t[ind]) / 2.0:   # Closer to the smaller number
      ind = ind-1
   return ind

def f2(t, x):
   return np.abs(t-x).argmin()

print t,           '\n', x,           '\n'
print f1(t, x),    '\n', f2(t, x),    '\n'
print t[f1(t, x)], '\n', t[f2(t, x)], '\n'

runs = 1000
time = timeit.Timer('f1(t, x)', 'from __main__ import f1, t, x')
print round(time.timeit(runs), 6)

time = timeit.Timer('f2(t, x)', 'from __main__ import f2, t, x')
print round(time.timeit(runs), 6)

推荐答案

这似乎更快（对我来说，Python 3.2-win32，numpy 1.6.0）：

This seems much quicker (for me, Python 3.2-win32, numpy 1.6.0):

from bisect import bisect_left
def f3(t, x):
    i = bisect_left(t, x)
    if t[i] - x > 0.5:
        i-=1
    return i

---

输出：

---

Output:

[   10    11    12 ..., 99997 99998 99999]
37854.22200356027
37844
37844
37844
37854
37854
37854
f1 0.332725
f2 1.387974
f3 0.085864

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