Python：反向排序列表中第一個元素的索引小於閾值

Question

類似的問題在這里被要求排序列表，但解決方案使用了bisect ，它不適用於保留排序列表。

假設我有一個列表，按相反順序排序，鍵入中間元素，

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] ....]

我想在中間元素上應用一系列閾值，例如，在單獨的排序列表中

threshold = [0.97, 0.90, 0.83, 0.6]

我試圖找出小於閾值的第一個元素的索引。 在上面的例子中它應該返回，

index_list = [2, 2, 3, 6]

建議如何以最快的方式完成？

Answer 1

根據@ gnibbler的這個好答案 ，你可以自己重寫bisect代碼以滿足你的需要

我稍微修改了@ gnibbler中的代碼，以便它可以在你的情況下使用

優化是因為您的閾值也是排序的，我們不需要每次搜索整個列表，而是從最后的結果索引開始

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)

感謝@ PhilCooper提出的寶貴建議。 以下是他提出的使用生成器的代碼：

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))

Answer 2

使用numpy，我認為它看起來比純python實現更清晰，並且幾乎肯定會更快：

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]

Answer 3

請嘗試以下方法：

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

Answer 4

我在想你應該得到鑰匙並反轉。 然后bisecet沒問題

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]

如果你已經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)

Answer 5

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)