如何在numpy数组中找到最近的值?例子:
np.find_nearest(array, value)
如何在numpy数组中找到最近的值?例子:
np.find_nearest(array, value)
当前回答
这是在向量数组中找到最近向量的扩展。
import numpy as np
def find_nearest_vector(array, value):
idx = np.array([np.linalg.norm(x+y) for (x,y) in array-value]).argmin()
return array[idx]
A = np.random.random((10,2))*100
""" A = array([[ 34.19762933, 43.14534123],
[ 48.79558706, 47.79243283],
[ 38.42774411, 84.87155478],
[ 63.64371943, 50.7722317 ],
[ 73.56362857, 27.87895698],
[ 96.67790593, 77.76150486],
[ 68.86202147, 21.38735169],
[ 5.21796467, 59.17051276],
[ 82.92389467, 99.90387851],
[ 6.76626539, 30.50661753]])"""
pt = [6, 30]
print find_nearest_vector(A,pt)
# array([ 6.76626539, 30.50661753])
其他回答
如果你有很多值需要搜索(值可以是多维数组),下面是@Dimitri的快速向量化解决方案:
# `values` should be sorted
def get_closest(array, values):
# make sure array is a numpy array
array = np.array(array)
# get insert positions
idxs = np.searchsorted(array, values, side="left")
# find indexes where previous index is closer
prev_idx_is_less = ((idxs == len(array))|(np.fabs(values - array[np.maximum(idxs-1, 0)]) < np.fabs(values - array[np.minimum(idxs, len(array)-1)])))
idxs[prev_idx_is_less] -= 1
return array[idxs]
基准
>使用@Demitri的解决方案比使用for循环快100倍”
>>> %timeit ar=get_closest(np.linspace(1, 1000, 100), np.random.randint(0, 1050, (1000, 1000)))
139 ms ± 4.04 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
>>> %timeit ar=[find_nearest(np.linspace(1, 1000, 100), value) for value in np.random.randint(0, 1050, 1000*1000)]
took 21.4 seconds
也许对ndarray有帮助:
def find_nearest(X, value):
return X[np.unravel_index(np.argmin(np.abs(X - value)), X.shape)]
如果你不想使用numpy,可以这样做:
def find_nearest(array, value):
n = [abs(i-value) for i in array]
idx = n.index(min(n))
return array[idx]
所有的答案都有助于收集信息来编写高效的代码。但是,我已经编写了一个小的Python脚本来针对各种情况进行优化。如果提供的数组已排序,则将是最佳情况。如果搜索一个指定值的最近点的索引,那么对半模块是最省时的。当一个索引对应一个数组时,numpy searchsorted是最有效的。
import numpy as np
import bisect
xarr = np.random.rand(int(1e7))
srt_ind = xarr.argsort()
xar = xarr.copy()[srt_ind]
xlist = xar.tolist()
bisect.bisect_left(xlist, 0.3)
In[63]: %时间平分。bisect_left (xlist, 0.3) CPU次数:user 0ns, sys: 0ns, total: 0ns 壁时间:22.2µs
np.searchsorted(xar, 0.3, side="left")
In [64]: %time np。Searchsorted (xar, 0.3, side="left") CPU次数:user 0ns, sys: 0ns, total: 0ns 壁时间:98.9µs
randpts = np.random.rand(1000)
np.searchsorted(xar, randpts, side="left")
%的时间np。Searchsorted (xar, randpts, side="left") CPU次数:用户4ms, sys: 0ns, total: 4ms 壁时间:1.2 ms
如果我们遵循乘法规则,那么numpy应该花费~100 ms,这意味着快了~83倍。
import numpy as np
def find_nearest(array, value):
array = np.array(array)
z=np.abs(array-value)
y= np.where(z == z.min())
m=np.array(y)
x=m[0,0]
y=m[1,0]
near_value=array[x,y]
return near_value
array =np.array([[60,200,30],[3,30,50],[20,1,-50],[20,-500,11]])
print(array)
value = 0
print(find_nearest(array, value))