numpy:在排序列表中，找到每个唯一值的第一个和最后一个索引 [英] numpy: in a sorted list, find the first and the last index for each unique value

问题描述

具有排序列表，任何人如何找到(使用numpy)每个唯一值的第一个和最后一个索引?

Having a sorted list, how can anyone find (using numpy) the first and the last index for each unique value?

示例:

初始排序列表:

>>> import numpy as np
>>> initial_list = np.array([1, 3, 2, 3, 0, 3, 0, 1, 0])
>>> initial_list.sort()

>>> initial_list
array([0, 0, 0, 1, 1, 2, 3, 3, 3])

其结果将是:

第一:[0，0，0，3，3，5，6，6，6]

first: [ 0, 0, 0, 3, 3, 5, 6, 6, 6 ]

最后:[2，2，2，4，4，5，8，8，8]

last: [ 2, 2, 2, 4, 4, 5, 8, 8, 8 ]

提前谢谢

推荐答案

这是一种利用输入数据的排序特性的方法，该方法利用了非常有效的NumPy array-slicing和其他NumPy函数-

Here's one approach leveraging the sorted nature of input data, making use of the very efficient NumPy array-slicing and other NumPy functions -

def start_stop_arr(initial_list):
    a = np.asarray(initial_list)
    mask = np.concatenate(([True], a[1:] != a[:-1], [True]))
    idx = np.flatnonzero(mask)
    l = np.diff(idx)
    start = np.repeat(idx[:-1], l)
    stop = np.repeat(idx[1:]-1, l)
    return start, stop

级联重复可以进一步提高性能-

Further performance boost is possible with concatenated repetitions -

def start_stop_arr_concat_repeat(initial_list):
    a = np.asarray(initial_list)
    mask = np.concatenate(([True], a[1:] != a[:-1], [True]))
    idx = np.flatnonzero(mask)
    l = np.diff(idx)
    idx2 = np.concatenate((idx[:-1,None], (idx[1:,None]-1)),axis=1)
    ss = np.repeat(idx2, l, axis=0)
    return ss[:,0], ss[:,1]

样品运行-

In [38]: initial_list
Out[38]: array([0, 0, 0, 1, 1, 2, 3, 3, 3])

In [39]: start_stop_arr(initial_list)
Out[39]: (array([0, 0, 0, 3, 3, 5, 6, 6, 6]), array([2, 2, 2, 4, 4, 5, 8, 8, 8]))

运行时测试-

其他方法-

# @Mohammed Elmahgiubi's soln
def reversed_app(initial_list): # input expected is a list
    reversed_initial_list = list(reversed(initial_list))
    first = [initial_list.index(i) for i in initial_list]
    last = list(reversed([(len(initial_list) - 
                           (reversed_initial_list.index(i) + 1)) 
                            for i in reversed_initial_list]))
    return first, last

def unique_app(a): # @B. M.'s soln
    _,ind1,inv1,cou1 = np.unique(a, return_index=True, return_inverse=True, 
                                 return_counts=True)
    return ind1[inv1],(ind1+cou1-1)[inv1]

时间-

案例1:较小的数据集

In [295]: initial_list = np.random.randint(0,1000,(10000))
     ...: initial_list.sort()

In [296]: input_list = initial_list.tolist()

In [297]: %timeit reversed_app(input_list)
1 loop, best of 3: 789 ms per loop

In [298]: %timeit unique_app(initial_list)
1000 loops, best of 3: 353 µs per loop

In [299]: %timeit start_stop_arr(initial_list)
10000 loops, best of 3: 96.3 µs per loop

案例2:更大的数据集

In [438]: initial_list = np.random.randint(0,100000,(1000000))
     ...: initial_list.sort()

In [439]: %timeit unique_app(initial_list) # @B. M.'s soln
10 loops, best of 3: 53 ms per loop

In [440]: %timeit start_stop_arr(initial_list)
100 loops, best of 3: 9.64 ms per loop

In [441]: %timeit start_stop_arr_concat_repeat(initial_list)
100 loops, best of 3: 6.76 ms per loop

这篇关于numpy:在排序列表中，找到每个唯一值的第一个和最后一个索引的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！