Python，在随机索引处向向量添加值的最有效方法是什么？

Question

I am trying to find the best way to execute the following code. in practice i would be adding specific values at specific indices, not adding 1.

#this is a simplified code for illustration purpose and must not be optimized

from random import randint, random
from numpy import zeros

n = 10
N = 1000

indicies = [randint(0, n-1) for p in range(0, N)]
values = [random.random() for p in range(0, N)]
x = zeros(n)
y = zeros(n)



#the issue is here. I want to do this as efficiently as possible :
for i in range(N):
    x[indicies[i]]+= values[i]

#I tried this but it does not work.
y[indicies]+= values


print(x)
print(y)

here is what i get by substituting the values vector by a vector of ones:

#correct result
[103.  95. 109. 113.  79. 105.  90.  99. 105. 102.]
#failed attempt to go faster
[1. 1. 1. 1. 1. 1. 1. 1. 1. 1.]

I want to get x. Obviously my attempt with y has failed.

For those interested, i am computing conditional expectations in Monte Carlo, projecting values on a (time x space) grid. So for 252 time steps of my stochastic process and 2^20 MC paths, i would be doing 250 * 2 million operations. In practice the diffusion of the process is done by batches of 2^10 MC paths and efficiently vectorized but when i am projecting, i have not found a way to do it other than time step by time step and path by path.

This step of the algorithm take up 98% of the computation time so i must be doing something wrong.

Thanks for your help.

Edit:

Given the first answers i must clarify my issue. I already have a vector of indexes, and a vector of values. The issue is not with generating random numbers. I am looking for an efficient way to increment my target vector:

for i in range(N):
    target[indexes[i]]+=values[i]

Edit 2:

The algorithm works, here is an illustration of what i get. it is just too slow:

look at it, it is quite cool

Answer 1

First, since you're using numpy , use np.random.randint for x :

x = np.random.randint(0, n-1, N)

Then to add random values to x , just use another random vector and add it to x :

x + np.random.randint(low, high, N)

RE: edit 2.

Since you seem dead set on this approach, here's a possible solution, assuming I'm understanding you correctly:

>>> indices = [0, 0, 4, 0, 1, 1, 2, 3, 4]
>>> values = [0, 0, 0, 0, 0]
>>> for idx in set(indices):
...     values[idx] += indices.count(idx)
...
>>> values
[3, 2, 1, 1, 2]

Answer 2

Use np.random.randint ( docs ).

How you might use it:

n = 10
N = 1000
z = np.random.randint(0, N, size=n)

output: z = [292 237 838 860 590 933 29 154 168 606]

Answer 3

accumulate 1000 random choices

n = 10
N = 1000

indice = [random.randint(0, n-1) for p in range(0, N)]
values = [random.random() for p in range(0, N)]

x = zeros(n)
y = zeros(n)

def GetRandomChoice():
    index=random.choice(indice)
    value=values[index]
    return index,value

for i in range(N):
    index,value=GetRandomChoice()
    x[index]+= value
    index,value=GetRandomChoice()
    y[index]+= value

print(x,y)
output:
[  4.98649477  13.4498052  100.04144429  77.06294844  
   80.52940448 9.50733693  41.0324075   22.74005395  
   48.31112763  83.95712759]
[  5.24090776  14.72399728  81.06806692  92.87073274  
   75.9711363  8.36970687  37.00243891  18.15536565  
   58.40599011 116.76795906]