Normalizing a list of numbers in Python

Question

I need to normalize a list of values to fit in a probability distribution, i.e. between 0.0 and 1.0.

I understand how to normalize, but was curious if Python had a function to automate this.

I'd like to go from:

raw = [0.07, 0.14, 0.07]  

to

normed = [0.25, 0.50, 0.25]

Answer 1

How long is the list you're going to normalize?

def psum(it):
    "This function makes explicit how many calls to sum() are done."
    print "Another call!"
    return sum(it)

raw = [0.07,0.14,0.07]
print "How many calls to sum()?"
print [ r/psum(raw) for r in raw]

print "\nAnd now?"
s = psum(raw)
print [ r/s for r in raw]

# if one doesn't want auxiliary variables, it can be done inside
# a list comprehension, but in my opinion it's quite Baroque    
print "\nAnd now?"
print [ r/s  for s in [psum(raw)] for r in raw]

Output

# How many calls to sum()?
# Another call!
# Another call!
# Another call!
# [0.25, 0.5, 0.25]
# 
# And now?
# Another call!
# [0.25, 0.5, 0.25]
# 
# And now?
# Another call!
# [0.25, 0.5, 0.25]

Answer 2

There isn't any function in the standard library (to my knowledge) that will do it, but there are absolutely modules out there which have such functions. However, its easy enough that you can just write your own function:

def normalize(lst):
    s = sum(lst)
    return map(lambda x: float(x)/s, lst)

Sample output:

>>> normed = normalize(raw)
>>> normed
[0.25, 0.5, 0.25]

Answer 3

if your list has negative numbers, this is how you would normalize it

a = range(-30,31,5)
norm = [(float(i)-min(a))/(max(a)-min(a)) for i in a]

Answer 4

Use :

norm = [float(i)/sum(raw) for i in raw]

to normalize against the sum to ensure that the sum is always 1.0 (or as close to as possible).

use

norm = [float(i)/max(raw) for i in raw]

to normalize against the maximum

Answer 5

If you consider using numpy, you can get a faster solution.

import random, time
import numpy as np

a = random.sample(range(1, 20000), 10000)
since = time.time(); b = [i/sum(a) for i in a]; print(time.time()-since)
# 0.7956490516662598

since = time.time(); c=np.array(a);d=c/sum(a); print(time.time()-since)
# 0.001413106918334961

Answer 6

try:

normed = [i/sum(raw) for i in raw]

normed
[0.25, 0.5, 0.25]