Your iterator is not creating indices, but the actual elements in the array

Use with AIfor label in labels_class

In the above label is not an index, but the actual element you are trying to change

You can do something like this:

Use with AIfor i, label in enumerate(labels_class):
     labels_class[i] = 1 if (label == classifier) else 0

As others have said, using vectorization is typically better/faster/nicer/...

But if you have good reasons to use iteration, you can of course do it.

I just copied this from the official documentation:

>>> a = np.arange(6).reshape(2,3)
>>> a
array([[0, 1, 2],
       [3, 4, 5]])
>>> for x in np.nditer(a, op_flags=['readwrite']):
...     x[...] = 2 * x
...
>>> a
array([[ 0,  2,  4],
       [ 6,  8, 10]])

There is a numpy function for this, numpy.full or numpy.full_like:

>>> x = np.array(([1,2,2], [1,4,3]))
>>> np.full(x.shape, 10)

array([[10, 10, 10],
       [10, 10, 10]])
# OR,
>>> np.full_like(x, 10)

array([[10, 10, 10],
       [10, 10, 10]])

If you want to iterate you can either use itertools.product:

>>> from itertools import product

>>> def Iter_Replace(x):
        indices = product(*map(range, x.shape))
        for index in indices:
            x[tuple(index)] = 10
        return x
>>> x = np.array([[1,2,2], [1,4,3]])
>>> Iter_Replace(x)

array([[10, 10, 10],
       [10, 10, 10]])

Or, use np.nditer

>>> x = np.array([[1,2,2], [1,4,3]])

>>> for index in np.ndindex(x.shape):
        x[index] = 10
    
>>> x

array([[10, 10, 10],
       [10, 10, 10]])

You cannot mutate the length of a numpy array, because numpy assigns the required memory for an array upon its creation.

With

array = np.delete(array, idx+1)

You are creating a new array on the right hand side of the = and reassign the name array.

The return value for enumerate(np.nditer(array)) has already been created at that point and won't recognize that the name array has been rebound.

In principle, you can iterate over a sequence and mutate its length at the same time (generally not a good idea). The object just needs to have methods that let you mutate its length (like lists, for example).

Consider:

>>> l = [5, 4, 3, 2, 1]                                                         
>>> for idx, ele in enumerate(l): 
...:     if ele == 3: 
...:         l.pop(idx) # mutates l 
...:     print(ele) 
...:                                                                            
5
4
3
1
>>> l                                                                           
[5, 4, 2, 1]

Notice that

1. l is mutated.
2. The loop does not print 2 because popping an element reduces the indexes of all the remaining elements by one. Now l[2] == 2, but index 2 has already been visited by the iterator, so the next print-call prints l[3] which is 1.

This proves that mutations to l have effect on subsequent iterations.

There are several issues with your example (some of which may be actual problems, others just typos or over simplifiactions):

import numpy as np     # if you want to use for-loops don't use numpy
rays = np.array(...    # closing parentheses instead of brackets
                       # unequal dimensions row of 5 and row of 6
                            
for i in range(rays):  # rays is not a number, did you mean len(rays[0])?
    for w in range(i): # w is not used anywhere
        estimate = rays[0][i]/(rays[0][i]+rays[1][i]) 
                       # estimate is overwritten at each iteration

The whole point of using numpy is to avoid "manually" iterating through array elements using for-loops. You should think of your result as an operation between matrices (or vectors):

For example (without for-loops):

import numpy as np
rays = np.array([[7.651e-03, 7.284e-03, 5.134e-03, 7.442e-03, 3.035e-03],
                 [2.373e-03, 6.877e-03, 4.809e-03, 2.870e-04, 3.175e-04]])

estimates = rays[0]/(rays[0]+rays[1])
print(estimates)
[0.76326816 0.51437045 0.51634316 0.96286712 0.90529456]

Note that I removed the last value from the second row because numpy requires fixed dimensions (i.e. it cannot have one row with 5 elements and another with 6)

Your nested loop for w in range(i), though you're not doing anything with w, suggests that you may be looking for the ratio between cumulative sums. If that is the case, use the cumsum function from numpy:

estimates = np.cumsum(rays[0])/np.cumsum(rays[0]+rays[1])
print(estimates)
[0.76326816 0.61753153 0.58805087 0.65726163 0.67565445]

I'm assuming your code looks something like this

a1 = np.array([0, 0, 0])
update = [0.6402942957283034, 0.6017736334918196, 0.6096552413265736]
for i in range(0,3):
    a1[i] = update[i]
print(a1)

which indeed prints array([0 0 0])

The reason you are seeing this behavior is because unlike python lists, numpy arrays must contain values of the same type. When you created the array a1, you did not specify the type explicitly thus creating an int64 array. Try doing

print(a1.dtype)

update is a float64 array where all the numbers are less than 1. When you convert a float < 1 to int, it becomes 0.

The solution to your problem is to explicitly specify the type of array a1.

a1 = np.array([0, 0, 0], dtype='float64')

On a different note, why do you copy element by element? You could have done

a1 = np.asarray(update)

This is not related with numpy matrix, but how python deal with your

i = i - y

i - y produces a new reference of an array. When you assigns it to name i, so i is not referred to the one it was before, but the newly created array.

The following code will meet your purpose

for idx, i in enumerate(m):
    m[idx] = i - y