同一方程式中变量的多个值

Question

A = [18.0,10.0]; B = [13.0,15.0]; C = [10.5,12.0];

these are the variables and think about function like

def hlf(A,B,C):
       return A**(-1.0/2.0)-0.2*B-43+C
print "T:"
hlf(A,B,C)

Firstly, I want to use first values of the AB and C in the equation. After I want to use second values. How can I do this ?

Answer 1

map + list

Note map can take multiple iterable arguments:

res = map(hlf, A, B, C)

[-34.86429773960448, -33.68377223398316]

In Python 2.7, map returns a list . In Python 3.x map returns an iterator, so you can either iterate lazily or exhaust via list , ie list(map(hfl, A, B, C)) .

Reference :

map ( function , iterable , ...)

...If additional iterable arguments are passed, function must take that many arguments and is applied to the items from all iterables in parallel.

zip + list comprehension

You can use zip within a list comprehension. For clarity, you should avoid naming your arguments the same as your variables.

A = [18.0,10.0]; B = [13.0,15.0]; C = [10.5,12.0]; 

def hlf(x, y, z):
    return x**(-1.0/2.0) - 0.2*y - 43 + z

res = [hlf(*vars) for vars in zip(A, B, C)]

[-34.86429773960448, -33.68377223398316]

Answer 2

Vectorize with Numpy. Best Performace

Normally its much better try to vectorize this kind of operations with numpy, because the best performance results. When you vectorize instead to use a loop, you are using all your cores, and its the fastest solution. You should vectorize the operation with numpy. Something like this:

import numpy as np

A = [18.0,10.0]; B = [13.0,15.0]; C = [10.5,12.0];
a = np.array(A)
b = np.array(B)
c = np.array(C)

And now your function with the new vectors like arguments:

def hlf(a_vector,b_vector,c_vector):
    return a_vector**(-1.0/2.0)-0.2*b_vector-43+c_vector

And finally call your new function vectorized:

print (hlf(a_vector = a,b_vector = b,c_vector = c))

Output:

>>> array([-34.86429774, -33.68377223])

Answer 3

If you want to keep your function as is, you should call it N times with:

for i in range(N):
    result = hlf(A[i], B[i], C[i])
    print(result)

Another interesting method is to make a generator with your function:

A = [18.0,10.0]
B = [13.0,15.0]
C = [10.5,12.0];

def hlf(*args):
    i=0
    while i < len(args[0]):
        yield args[0][i]**(-1.0/2.0) - 0.2*args[1][i] - 43 + args[2][i]
        i += 1

results = hlf(A, B, C)
for r in results:
    print(r)

Output:

-34.86429773960448
-33.68377223398316

Last one is rather edicational if you want to practice python generators.