4

此处已针对排序列表提出了类似的问题，但使用的解决方案bisect不适用于保留排序列表。

假设我有一个列表，以相反的顺序排序，以中间元素为键，

my_list = [[3,0.99,1], [2,0.98,54], [10,.85,4], [1,0.7,10], [12,0.69,31], [12,0.65,43], [1.56,0] ....]

我想在中间元素上应用一系列阈值，它在一个单独的排序列表中，比如说

threshold = [0.97, 0.90, 0.83, 0.6]

我试图找出小于阈值的第一个元素的索引。在上面的例子中它应该返回，

index_list = [2, 2, 3, 6]

建议如何以最快的方式完成？

4

5 回答 5

4

根据@ gnibbler的这个好答案，您可以自己重写代码以满足您的需要bisect

我稍微修改了@ gnibbler的代码，以便可以在您的情况下使用

一个优化是，由于你的阈值也是排序的，我们不需要每次都搜索整个列表，而是从最后一个结果索引开始

def reverse_binary_search(a, x, lo=0, hi=None):
    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi: 
        mid = (lo+hi)/2
        if x > a[mid][4]:
            hi = mid 
        else:
            lo = mid+1
    return lo

my_list = [[3,0.99,1], [2,0.98,54], [10,.85,4], [1,0.7,10], [12,0.69,31], [12,0.65,43], [1.56,0]]
threshold = [0.97, 0.90, 0.83, 0.6]

index_list = []
last_index = 0
for t in threshold:
    last_index = reverse_binary_search(my_list, t, last_index) # next time start search from last_index
    index_list.append(last_index)

感谢@PhilCooper的宝贵建议。这是他提出的使用生成器的代码：

def reverse_binary_search(a, threshold):
    lo = 0
    for t in threshold:
        if lo < 0:
            raise ValueError('lo must be non-negative')
        hi = len(a)
        while lo < hi: 
            mid = (lo+hi)/2
            if t > a[mid][6]:
                hi = mid 
            else:
                lo = mid+1
        yield lo

my_list = [[3,0.99,1], [2,0.98,54], [10,.85,4], [1,0.7,10], [12,0.69,31], [12,0.65,43], [1.56,0]]
threshold = [0.97, 0.90, 0.83, 0.6]

index_list = list(reverse_binary_search(my_list, threshold))
于 2012-07-06T21:42:21.767 回答
1

使用 numpy，我认为它看起来比纯 python 实现更干净，并且几乎可以肯定会更快：


import numpy as np
arr = np.array([[3,0.99,1], [2,0.98,54], [10,.85,4], [1,0.7,10], [12,0.69,31], [12,0.65,43], [10,0.50, 24]])
thresholds = [0.97, 0.90, 0.83, 0.60]
idx = [np.min(np.where(arr[:,1] < i)) for i in thresholds if np.where(arr[:,1] < i)[0].size > 0]
print idx
[2, 2, 3, 6]

于 2012-07-06T22:01:43.170 回答
0

尝试以下操作：

threshold = [0.97, 0.90, 0.83, 0.6]
my_list = [[3,0.99,1], [2,0.98,54], [10,.85,4], [1,0.7,10], [12,0.69,31], [12,0.65,43], [1,.56,0]]
threshold = [0.97, 0.90, 0.83, 0.6]

index_list = []
ti = 0
for i, item in enumerate(my_list):
    if item[1] >= threshold[ti]:
        continue
    while ti < len(threshold) and item[1] < threshold[ti]:
        index_list.append(i)
        ti += 1
于 2012-07-06T21:23:38.057 回答
0

我想你应该拿到钥匙然后倒车。然后二等分就可以了

from bisect import bisect_left

keys = [vals[1] for vals in my_list]
keys.reverse()
mylen = len(my_list)
[mylen-bisect_left(keys,t) for t in threshold]

如果你已经有 numpy：

my_array = np.array([[3,0.99,1], [2,0.98,54], [10,.85,4], [1,0.7,10], [12,0.69,31], [12,0.65,43], [10,0.50, 24]])
thresholds = [0.97, 0.90, 0.83, 0.60]

my_array.shape[0]-arr[::-1,1].searchsorted(threshold)
于 2012-07-06T21:25:36.747 回答
0 
import bisect
my_list_2  = sorted(my_list, key=lambda x:x[1])
for x in threshold:
    len(my_list) - bisect.bisect([z[1] for z in my_list_2], x)
于 2012-07-06T21:44:58.057 回答