How to normalize and inverse a vector so it sums to 1 ?

Question

I understand how normalization works. You sum up the individual values of the vector, you divide each value by the sum, and voila... they sum to 1.

Why doesn't it work when you subtract them from 1? If all values sum to 1 after normalization, then shouldn't the ratios work out so (1 - X_1).sum() == 1 ?

What am I doing wrong? I want basically want the smaller values to have a higher weight in the end. Where a value of 0.0 would be 100% (i.e. 1.0).

# Python 3 with NumPy
X = np.array([[ 58.50853002,  74.73077551,  54.46120887,  55.55526553],
              [ 68.14133201,  22.2475803 ,  88.79126866,  86.24927424],
              [ 43.86150599,  75.99344646,  81.90051932,  50.66885662],
              [ 74.81149378,  82.86920509,  36.75953127,  58.42956957]])

# Get 2nd row
X_1 = X[1,:]
# array([ 68.14133201,  22.2475803 ,  88.79126866,  86.24927424])

# Get ratios for row
ratios = X_1/X_1.sum()
#array([ 0.25672106,  0.0838173 ,  0.33451927,  0.32494236])

# Sum to 1
ratios.sum()
# 1.0

# Shouldn't this work? 
(1 - ratios)
# array([ 0.74327894,  0.9161827 ,  0.66548073,  0.67505764])

# But it doesn't...
(1 - ratios).sum()
#3.0

Answer 1

Let me try to understand what you are doing:

You sum up the individual values of the vector, you divide each value by the sum, and voila... they sum to 1.

First the sum: $$ S(x) = \sum_i x_i $$ Then the described normalization: $$ x' = x / S(x) $$ So $$ S(x') = \sum_i x_i' = \sum_i \frac{x_i}{S(x)} = \frac{1}{S(x)} \sum_i x_i = \frac{1}{S(x)} S(x) = 1 $$ OK, the normalized vector $x'$ has unit sum. (Beware: if $S(x)$ was zero, you are in for a surprise)

Why doesn't it work when you subtract them from 1? If all values sum to 1 after normalization, then shouldn't the ratios work out so (1 - X_1).sum() == 1 ?

Fine, if we have the new vector $y$ with $$ y_i = 1 - x_i' $$ then we get $$ S(y_i) = \sum_i y_i = \sum_i (1-x_i') = \left( \sum_i 1 \right) - \left( \sum_i x_i' \right) = n - S(x') = n - 1 $$ So unless $n=2$ (two components in the vector), this would not return a vector $y$ with sum $1$.

Answer 2

There is no reason why you should expect 1 - ratios to lead to a vector where the sum of the elements is $1$.

In particular, if you have the initial array $[a,b,c,d]$ with $a + b + c + d = 1$, then $1 - [a,b,c,d] = [1-a, 1-b, 1-c, 1-d]$ (where I use equality in the numpy sense). When you sum these elements together, you get $1-a + 1-b + 1-c + 1-d = 4 - (a + b + c + d) = 4 - 1 = 3$.

So you will always get $(\text{dimension of vector} - 1)$ after this operation.