np.random.choice: probabilities do not sum to 1

Question

how can I use np.random.choice here? there is p that calculate by some opertation, like :

 p=[  1.42836755e-01,   1.42836735e-01  , 1.42836735e-01,   1.42836735e-01
,   4.76122449e-05,   1.42836735e-01  , 4.76122449e-05  , 1.42836735e-01,
   1.42836735e-01,   4.76122449e-05]

usually sum p is not exact equal to 1:

>>> sum(p)
1.0000000017347

I want to make random choice by probabilities=p:

>>> np.random.choice([1,2,3,4,5,6,7,8,9, 10], 4, p=p, replace=False)
array([4, 3, 2, 9])

this work here! but in the program it has an error :

Traceback (most recent call last):
    indexs=np.random.choice(range(len(population)), population_number, p=p, replace=False)
  File "mtrand.pyx", line 1141, in mtrand.RandomState.choice (numpy/random/mtrand/mtrand.c:17808)
ValueError: probabilities do not sum to 1

if I print the p:

[  4.17187500e-05   2.49937500e-01   4.16562500e-05   4.16562500e-05
   2.49937500e-01   4.16562500e-05   4.16562500e-05   4.16562500e-05
   2.49937500e-01   2.49937500e-01]

but it works, in python shell by this p:

>>> p=[  4.17187500e-05 ,  2.49937500e-01   ,4.16562500e-05  , 4.16562500e-05,
   2.49937500e-01  , 4.16562500e-05  , 4.16562500e-05  , 4.16562500e-05,
   2.49937500e-01   ,2.49937500e-01]
>>> np.random.choice([1,2,3,4,5,6,7,8,9, 10], 4, p=p, replace=False)
array([ 9, 10,  2,  5])

UPDATE I have tested it by precision=15:

 np.set_printoptions(precision=15)
 print(p)
[  2.499375625000002e-01   2.499375000000000e-01   2.499375000000000e-01
   4.165625000000000e-05   4.165625000000000e-05   4.165625000000000e-05
   4.165625000000000e-05   4.165625000000000e-05   2.499375000000000e-01
   4.165625000000000e-05]

testing:

>>> p=np.array([  2.499375625000002e-01   ,2.499375000000000e-01   ,2.499375000000000e-01,
   4.165625000000000e-05   ,4.165625000000000e-05,   4.165625000000000e-05,
   4.165625000000000e-05  , 4.165625000000000e-05 ,  2.499375000000000e-01,
   4.165625000000000e-05])
>>> np.sum(p)
1.0000000000000002

how fix this to use np.random.choice ?

Answer 1

Convert it to float64:

p = np.asarray(p).astype('float64')
p = p / np.sum(p)
np.random.choice([1,2,3,4,5,6,7,8,9, 10], 4, p=p, replace=False)

This was inspired by another post: How can I avoid value errors when using numpy.random.multinomial?

Answer 2

This is a known issue with numpy. The random choice function checks for the sum of the probabilities using a given tolerance (here the source)

The solution is to normalize the probabilities by dividing them by their sum if the sum is close enough to 1

Example:

>>> p=[  1.42836755e-01,   1.42836735e-01  , 1.42836735e-01,   1.42836735e-01
,   4.76122449e-05,   1.42836735e-01  , 4.76122449e-05  , 1.42836735e-01,
   1.42836735e-01,   4.79122449e-05]
>>> sum(p) 
1.0000003017347 # over tolerance limit
>>> np.random.choice([1,2,3,4,5,6,7,8,9, 10], 4, p=p, replace=False)

Traceback (most recent call last):
  File "<pyshell#23>", line 1, in <module>
    np.random.choice([1,2,3,4,5,6,7,8,9, 10], 4, p=p, replace=False)
  File "mtrand.pyx", line 1417, in mtrand.RandomState.choice (numpy\random\mtrand\mtrand.c:15985)
ValueError: probabilities do not sum to 1

With normalization:

>>> p = np.array(p)
>>> p /= p.sum()  # normalize
>>> np.random.choice([1,2,3,4,5,6,7,8,9, 10], 4, p=p, replace=False)
array([8, 4, 1, 6])