假設我有一個數組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,給我數組[1,1.5,2,3,4,5].
我的問題是：當my_array任意增大時,生成有序輸出數組的最快方式(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的數組,重復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：如果所有數據類型從頭開始匹配,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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