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
我认为最python的方式是:
num = 65 # Input number
array = np.random.random((10))*100 # Given array
nearest_idx = np.where(abs(array-num)==abs(array-num).min())[0] # If you want the index of the element of array (array) nearest to the the given number (num)
nearest_val = array[abs(array-num)==abs(array-num).min()] # If you directly want the element of array (array) nearest to the given number (num)
这是基本代码。你可以把它作为一个函数来使用
其他回答
下面是@Ari Onasafari的scipy版本,回答“在向量数组中找到最近的向量”
In [1]: from scipy import spatial
In [2]: import numpy as np
In [3]: A = np.random.random((10,2))*100
In [4]: A
Out[4]:
array([[ 68.83402637, 38.07632221],
[ 76.84704074, 24.9395109 ],
[ 16.26715795, 98.52763827],
[ 70.99411985, 67.31740151],
[ 71.72452181, 24.13516764],
[ 17.22707611, 20.65425362],
[ 43.85122458, 21.50624882],
[ 76.71987125, 44.95031274],
[ 63.77341073, 78.87417774],
[ 8.45828909, 30.18426696]])
In [5]: pt = [6, 30] # <-- the point to find
In [6]: A[spatial.KDTree(A).query(pt)[1]] # <-- the nearest point
Out[6]: array([ 8.45828909, 30.18426696])
#how it works!
In [7]: distance,index = spatial.KDTree(A).query(pt)
In [8]: distance # <-- The distances to the nearest neighbors
Out[8]: 2.4651855048258393
In [9]: index # <-- The locations of the neighbors
Out[9]: 9
#then
In [10]: A[index]
Out[10]: array([ 8.45828909, 30.18426696])
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])
我认为最python的方式是:
num = 65 # Input number
array = np.random.random((10))*100 # Given array
nearest_idx = np.where(abs(array-num)==abs(array-num).min())[0] # If you want the index of the element of array (array) nearest to the the given number (num)
nearest_val = array[abs(array-num)==abs(array-num).min()] # If you directly want the element of array (array) nearest to the given number (num)
这是基本代码。你可以把它作为一个函数来使用
这是unutbu答案的矢量化版本:
def find_nearest(array, values):
array = np.asarray(array)
# the last dim must be 1 to broadcast in (array - values) below.
values = np.expand_dims(values, axis=-1)
indices = np.abs(array - values).argmin(axis=-1)
return array[indices]
image = plt.imread('example_3_band_image.jpg')
print(image.shape) # should be (nrows, ncols, 3)
quantiles = np.linspace(0, 255, num=2 ** 2, dtype=np.uint8)
quantiled_image = find_nearest(quantiles, image)
print(quantiled_image.shape) # should be (nrows, ncols, 3)
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))