Python，numpy.array切片，使用切片更改数组值

Question

I have a Task for numerical integration in which we approximate integral with quadrature formula. My problem is that the task needs me to avoid loops and use vectorized variant, which would be a slice?!

I have np.array object with n values and i have to alter each value of this array using a specific formula. The problem is that the value of this array at point i ist used in the formula to alter the position in. With a for loop it would be easy:

x = np.array([...])
for i in range(0,n):
   x[i]=f(x[i]+a)*b`

(a,b some othe variables) How do i do this with slices? I Have to do this for all elements of the array so it would be something like:

x[:]=f(x[???]+a)*b

And how do i get the right position from my array in to the formula? A slicing instruction like x[:] just runs through my whole object. Is there a way to somehow save the index i am currently at? I tried to search but found nothing. The other problem is that i do not even know how to properly put the search request...

Answer 1

You may be confusing two issues

• modifying all elements of an array
• calculating values for all elements of an array

In

x = np.array([...])
for i in range(0,n):
   x[i]=f(x[i]+a)*b`

you change elements of x one by one, and also pass them one by one to f .

x[:] = ... lets you change all elements of x at once, but the source (the right hand side of the equation) has to generate all those values. But usually you don't need to assign values. Instead just use x = ... . It's just as fast and memory efficient.

Using x[:] on the RHS does nothing for you. If x is a list this makes a copy; if x is an array is just returns a view , an array with the same values.

The key question is, what does your f(...) function accept? If it uses operations like + , * and functions like np.sin , you can give it an array, and it will return an array.

But if it only works with scalars (that includes using functions like math.sin ), the you have to feed it scalars, ie x[i] .

Let's try to unpack that comment (which might be better as an edit to the original question)

I have an interval which has to be cut in picies.

x = np.linspace(start,end,pieceAmount)
function f
quadrature formula
b (weights or factors)
c (function values)
b1*f(x[i]+c1)+...+bn*f(x[i]+cn)

For example

In [1]: x = np.arange(5)
In [2]: b = np.arange(3)
In [6]: c = np.arange(4,7)*.1

We can do the x[i]+c for all x and c with broadcasting

In [7]: xc = x + c[:,None]
In [8]: xc
Out[8]: 
array([[ 0.4,  1.4,  2.4,  3.4,  4.4],
       [ 0.5,  1.5,  2.5,  3.5,  4.5],
       [ 0.6,  1.6,  2.6,  3.6,  4.6]])

If f is a function like np.sin that takes any array, we can pass xc to that, getting back a like sized array.

Again with broadcasting we can do the b[n]*f(x[i]+c[n]) calculation

In [9]: b[:,None]* np.sin(xc)
Out[9]: 
array([[ 0.        ,  0.        ,  0.        , -0.        , -0.        ],
       [ 0.47942554,  0.99749499,  0.59847214, -0.35078323, -0.97753012],
       [ 1.12928495,  1.99914721,  1.03100274, -0.88504089, -1.98738201]])

and then we can sum, getting back an array shaped just like x :

In [10]: np.sum(_, axis=0)
Out[10]: array([ 1.60871049,  2.99664219,  1.62947489, -1.23582411, -2.96491212])

That's the dot or matrix product:

In [11]: b.dot(np.sin(xc))
Out[11]: array([ 1.60871049,  2.99664219,  1.62947489, -1.23582411, -2.96491212])

And as I noted earlier we can complete the action with

x = b.dot(f(x+c[:,None])

The key to a simple expression like this is f taking an array.