[英]Numpy: Fastest way to insert value into array such that array's in order
Suppose I have an array my_array
and a singular value my_val
. 假设我有一个数组my_array
和一个奇异值my_val
。 (Note that my_array
is always sorted). (请注意, my_array
始终排序)。
my_array = np.array([1, 2, 3, 4, 5])
my_val = 1.5
Because my_val
is 1.5, I want to put it in between 1 and 2, giving me the array [1, 1.5, 2, 3, 4, 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]
my_val
[1, 1.5, 2, 3, 4, 5]
。
My question is: What's the fastest way (ie in microseconds) of producing the ordered output array as my_array
grows arbitrarily large? 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?
The original way I though of was concatenating the value to the original array and then sorting: 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. ]
I know that np.concatenate
is fast but I'm unsure how np.sort
would scale as my_array
grows, even given that my_array
will always be sorted. 我知道np.concatenate
很快但我不确定np.sort
如何随着my_array
增长而扩展,即使my_array
总是会被排序。
Edit: 编辑:
I've compiled the times for the various methods listed at the time an answer was accepted: 我已经为接受答案时列出的各种方法编制了时间:
Input: 输入:
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))
Output: 输出:
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
And the same for 100 repetitions for an array of length 1,000,000: 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 Panzer's method took 0.3931240139911161s
Sanchit Anand's method took 0.40981490723551417s
Use np.searchsorted
to find the insertion point in logarithmic time: 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. ])
Note 1: I recommend looking at @Willem Van Onselm's and @hpaulj's insightful comments. 注1:我建议查看@Willem Van Onselm和@ hpaulj的深刻见解。
Note 2: Using np.insert
as suggested by @Sanchit Anand may be slightly more convenient if all datatypes are matching from the beginning. 注意2:如果所有数据类型从头开始匹配,则使用np.insert
Anand建议的np.insert可能会稍微方便一些。 It is, however, worth mentioning that this convenience comes at the cost of significant overhead: 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 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]
try 尝试
my_array = np.insert(my_array,my_array.searchsorted(my_val),my_val)
[EDIT] make sure that the array is of type float32 or float64, or add a decimal point to any of the list elements while initializing it. [编辑]确保数组的类型为float32或float64,或者在初始化时为任何列表元素添加小数点。
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