做外贸网站市场分析,网站建设列表横向,外贸营销型网站建设平台,资源分享论坛wordpress假设我有一个数组my_array和一个奇异值my_val. (请注意,my_array始终排序).
my_array np.array([1, 2, 3, 4, 5])
my_val 1.5
因为my_val是1.5,我想把它放在1和2之间,给我数组[1,1.5,2,3,4,5].
我的问题是#xff1a;当my_array任意增大时,生成有序输出数组的最快方式(即以微…假设我有一个数组my_array和一个奇异值my_val. (请注意,my_array始终排序).
my_array np.array([1, 2, 3, 4, 5])
my_val 1.5
因为my_val是1.5,我想把它放在1和2之间,给我数组[1,1.5,2,3,4,5].
我的问题是当my_array任意增大时,生成有序输出数组的最快方式(即以微秒为单位)是什么
我原来的方式是将值连接到原始数组然后排序
arr_out np.sort(np.concatenate((my_array, np.array([my_val]))))
[ 1. 1.5 2. 3. 4. 5. ]
我知道np.concatenate很快但我不确定np.sort如何随着my_array的增长而扩展,即使my_array总是会被排序.
编辑
我已经为接受答案时列出的各种方法编制了时间
输入
import timeit
timeit_setup import numpy as np\n \
my_array np.array([i for i in range(1000)], dtypenp.float64)\n \
my_val 1.5
num_trials 1000
my_time timeit.timeit(
np.sort(np.concatenate((my_array, np.array([my_val])))),
setuptimeit_setup, numbernum_trials
)
pauls_time timeit.timeit(
idx my_array.searchsorted(my_val)\n
np.concatenate((my_array[:idx], [my_val], my_array[idx:])),
setuptimeit_setup, numbernum_trials
)
sanchit_time timeit.timeit(
np.insert(my_array, my_array.searchsorted(my_val), my_val),
setuptimeit_setup, numbernum_trials
)
print(Times for 1000 repetitions for array of length 1000:)
print(My method took {}s.format(my_time))
print(Paul Panzers method took {}s.format(pauls_time))
print(Sanchit Anands method took {}s.format(sanchit_time))
输出
Times for 1000 repetitions for array of length 1000:
My method took 0.017865657746239747s
Paul Panzers method took 0.005813951002013821s
Sanchit Anands method took 0.014003945532323987s
对于长度为1,000,000的数组,重复100次
Times for 100 repetitions for array of length 1000000:
My method took 3.1770704101754195s
Paul Panzers method took 0.3931240139911161s
Sanchit Anands method took 0.40981490723551417s
解决方法:
使用np.searchsorted以对数时间查找插入点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我建议查看Willem Van Onselm和 hpaulj的深刻见解.
注意2如果所有数据类型从头开始匹配,则使用Sanchit Anand建议的np.insert可能会稍微方便一些.然而,值得一提的是,这种便利是以巨大的开销为代价的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 repeatkwds dict(globalsglobals(), number100000)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]
标签python,sorting,concatenation,numpy
来源 https://codeday.me/bug/20190527/1162537.html
