假設(shè)我有一個數(shù)組my_array和一個奇異值my_val. (請注意,my_array始終排序).

my_array = np.array([1, 2, 3, 4, 5])

my_val = 1.5

因為my_val是1.5,我想把它放在1和2之間,給我數(shù)組[1,1.5,2,3,4,5].

我的問題是：當(dāng)my_array任意增大時,生成有序輸出數(shù)組的最快方式(即以微秒為單位)是什么？

我原來的方式是將值連接到原始數(shù)組然后排序：

arr_out = np.sort(np.concatenate((my_array, np.array([my_val]))))

[ 1. 1.5 2. 3. 4. 5. ]

我知道np.concatenate很快但我不確定np.sort如何隨著my_array的增長而擴展,即使my_array總是會被排序.

編輯：

我已經(jīng)為接受答案時列出的各種方法編制了時間：

輸入：

import timeit

timeit_setup = 'import numpy as np\n' \

'my_array = np.array([i for i in range(1000)], dtype=np.float64)\n' \

'my_val = 1.5'

num_trials = 1000

my_time = timeit.timeit(

'np.sort(np.concatenate((my_array, np.array([my_val]))))',

setup=timeit_setup, number=num_trials

)

pauls_time = timeit.timeit(

'idx = my_array.searchsorted(my_val)\n'

'np.concatenate((my_array[:idx], [my_val], my_array[idx:]))',

setup=timeit_setup, number=num_trials

)

sanchit_time = timeit.timeit(

'np.insert(my_array, my_array.searchsorted(my_val), my_val)',

setup=timeit_setup, number=num_trials

)

print('Times for 1000 repetitions for array of length 1000:')

print("My method took {}s".format(my_time))

print("Paul Panzer's method took {}s".format(pauls_time))

print("Sanchit Anand's method took {}s".format(sanchit_time))

輸出：

Times for 1000 repetitions for array of length 1000:

My method took 0.017865657746239747s

Paul Panzer's method took 0.005813951002013821s

Sanchit Anand's method took 0.014003945532323987s

對于長度為1,000,000的數(shù)組,重復(fù)100次：

Times for 100 repetitions for array of length 1000000:

My method took 3.1770704101754195s

Paul Panzer's method took 0.3931240139911161s

Sanchit Anand's method took 0.40981490723551417s

解決方法:

使用np.searchsorted以對數(shù)時間查找插入點：

>>> idx = my_array.searchsorted(my_val)

>>> np.concatenate((my_array[:idx], [my_val], my_array[idx:]))

array([1. , 1.5, 2. , 3. , 4. , 5. ])

注1：我建議查看@Willem Van Onselm和@ hpaulj的深刻見解.

注意2：如果所有數(shù)據(jù)類型從頭開始匹配,則使用@Sanchit Anand建議的np.insert可能會稍微方便一些.然而,值得一提的是,這種便利是以巨大的開銷為代價的：

>>> def f_pp(my_array, my_val):

... idx = my_array.searchsorted(my_val)

... return np.concatenate((my_array[:idx], [my_val], my_array[idx:]))

...

>>> def f_sa(my_array, my_val):

... return np.insert(my_array, my_array.searchsorted(my_val), my_val)

...

>>> my_farray = my_array.astype(float)

>>> from timeit import repeat

>>> kwds = dict(globals=globals(), number=100000)

>>> repeat('f_sa(my_farray, my_val)', **kwds)

[1.2453778409981169, 1.2268288589984877, 1.2298014000116382]

>>> repeat('f_pp(my_array, my_val)', **kwds)

[0.2728819379990455, 0.2697303680033656, 0.2688361559994519]

標簽：python,sorting,concatenation,numpy

來源： https://codeday.me/bug/20190527/1162537.html

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