Python中的numpy.hypot()

这个数学函数帮助用户计算直角三角形的斜边，给定其边和垂线。计算结果等同于sqrt(x12 + x22)，按元素计算。
语法 :

numpy.exp2(arr1, arr2[, out]) = ufunc 'hypot') : 

参数 :

arr1, arr2 : [array_like] 三角形的腿（边和垂直线）。
out : [ndarray, optional] 输出结果的数组。

返回 :

具有直角三角形斜边的阵列。

代码#1 :

# Python3 program explaining
# hypot() function
  
import numpy as np
  
leg1 = [12, 3, 4, 6]
print ("leg1 array : ", leg1)
  
  
leg2 = [5, 4, 3, 8]
print ("leg2 array : ", leg2)
  
result = np.hypot(leg1, leg2)
print("\nHypotenuse is as follows :")
print(result)

输出 :

leg1 array :  [12, 3, 4, 6]
leg2 array :  [5, 4, 3, 8]

Hypotenuse is as follows :
[ 13.   5.   5.  10.]

代码#2：使用二维数组工作

# Python3 program explaining
# hypot() function
  
import numpy as np
  
leg1 = np.random.rand(3, 4)
print ("leg1 array : \n", leg1)
  
leg2 = np.ones((3, 4))
print ("leg2 array : \n", leg2)
  
result = np.hypot(leg1, leg2)
print("\nHypotenuse is as follows :")
print(result)

输出 :

leg1 array : 
 [[ 0.57520509  0.12043366  0.50011671  0.13800957]
 [ 0.0528084   0.17827692  0.44236813  0.87758732]
 [ 0.94926413  0.47816742  0.46111934  0.63728903]]
leg2 array : 
 [[ 1.  1.  1.  1.]
 [ 1.  1.  1.  1.]
 [ 1.  1.  1.  1.]]

Hypotenuse is as follows :
[[ 1.15362944  1.00722603  1.11808619  1.0094784 ]
 [ 1.00139339  1.01576703  1.09347591  1.33047342]
 [ 1.37880469  1.10844219  1.10119528  1.18580661]]

代码3：等同于sqrt(x1)2 + x22），从元素上看。

# Python3 program explaining
# hypot() function
  
import numpy as np
  
leg1 = np.random.rand(3, 4)
print ("leg1 array : \n", leg1)
  
leg2 = np.ones((3, 4))
print ("leg2 array : \n", leg2)
  
result = np.sqrt((leg1 * leg1) + (leg2 * leg2))
print("\nHypotenuse is as follows :")
print(result)

输出 :

leg1 array : 
 [[ 0.7015073   0.89047987  0.1595603   0.27557254]
 [ 0.67249153  0.16430312  0.70137114  0.48763522]
 [ 0.68067777  0.52154819  0.04339669  0.2239366 ]]
leg2 array : 
 [[ 1.  1.  1.  1.]
 [ 1.  1.  1.  1.]
 [ 1.  1.  1.  1.]]

Hypotenuse is as follows :
[[ 1.15362944  1.00722603  1.11808619  1.0094784 ]
 [ 1.00139339  1.01576703  1.09347591  1.33047342]
 [ 1.37880469  1.10844219  1.10119528  1.18580661]]