Replace Elements In Array With Class Instances

This is similar to this so please read it first to understand what I am trying to do. Now, I want to make the replacement when I have class instances.Something like: import numpy a

Solution 1:

So I add

print(d.shape)
print(d)

and get

2249:~/mypy$ python3 stack42283851.py 
(2, 1, 4)
[[[<__main__.B object at 0xb71d760c> <__main__.B object at 0xb71d7aac>
   <__main__.B object at 0xb71d7acc> <__main__.B object at 0xb71e5cec>]][[<__main__.B object at 0xb391718c> <__main__.B object at 0xb39171ac>
   <__main__.B object at 0xb39171cc> <__main__.B object at 0xb39171ec>]]]

adding a __repr__ to B I get

1231:~/mypy$ python3 stack42283851.py 
(2, 1, 4)
[[[B(100, a) B(11, b) B(300, c) B(33, d)]][[B(45, a) B(65, b) B(77, c) B(88, d)]]]

Adding

import itertools
for a,b in itertools.zip_longest(arr[0,:],b1):
     print(a,b)

produces

1B(100, a)
2B(11, b)
3B(300, c)
4B(33, d)
5None

Changing that to:

newlist = []
for a,b in itertools.zip_longest(arr[0,:],b1):
    if b isnotNone:
        new_b = B(a, b.b)
        last_b = b
    else:
        new_b = B(a, last_b.b)
    newlist.append(new_b)
print(np.array(newlist))

produces

[B(1, a) B(2, b) B(3, c) B(4, d) B(5, d)]

Assign that to b1, and repeat for a[1,:] and b2.

To be a little cleaner I could write that new_b code as a function, and rewrite the loop as a list comprehension.

Yes, I could modify b in place, e.g.

b.a = a

but since I need to create a new B object to replace the None, why bother. I can't add the new B object to the original b1 array. So it is simpler to create a new array via a list.

---

I can do an in-place change of d and b1 with:

def replace(a,b):
    b.a = a
f = np.frompyfunc(replace, 2, 1)
f(arr[:,None,:4], d)      # produces array of None; ignore
print(d)
print(b1)

[[[B(1, a) B(2, b) B(3, c) B(4, d)]]  # chgd d

[[B(6, a) B(7, b) B(8, c) B(9, d)]]]
[B(1, a) B(2, b) B(3, c) B(4, d)]    # chgd b1

I'm just using frompyfunc as a lazy mans way of broadcasting arr against d and iterating over all elements. Note that I have to change arr to match d shape. Also this does not add any new B(). Obviously you can't do that in-place.

---

My B is

classB():def__init__(self, a,b):
        self.a = a
        self.b = b
    def__repr__(self):
        return'B(%s, %s)'%(self.a, self.b)

---

Playing with frompyfunc some more:

getB_b = np.frompyfunc(lambda x: x.b, 1,1)   # fetch b attributes
print(getB_b(d))
#[[['a' 'b' 'c' 'd']]
#
# [['a' 'b' 'c' 'd']]]

mkB = np.frompyfunc(B, 2,1)   # build array of B() with broadcasting
print(mkB(arr, ['a','b','c','d','e']))
# [[B(1, a) B(2, b) B(3, c) B(4, d) B(5, e)]
#  [B(6, a) B(7, b) B(8, c) B(9, d) B(10, e)]]print(mkB(arr[:,:4], getB_b(d[:,0,:])))
# [[B(1, a) B(2, b) B(3, c) B(4, d)]
#  [B(6, a) B(7, b) B(8, c) B(9, d)]]

---

edit for comments

arr1 = np.array([ [1,2],[6,7] ])
newlist = []
for a,b in itertools.zip_longest(arr1[0,:],b1):
    if b is not None:
        new_b = B(a, b.b)
        last_b = b
    else:
        new_b = B(a, last_b.b)
    newlist.append(new_b)
print(np.array(newlist))

produces

[B(1, a) B(2, b) B(None, c) B(None, d)]

When arr is shorter, a will be None (instead of b); so we need to test for that

deffoo(arr,bn):
    newlist = []
    for a,b in itertools.zip_longest(arr,bn):
        print(a,b)
        if a isNone:
            passelse:
            if b isnotNone:
                new_b = B(a, b.b)
                last_b = b
            else:
                new_b = B(a, last_b.b)
            newlist.append(new_b)
    return newlist
print(np.array(foo(arr1[0,:],b1)))  # arr1 shorterprint(np.array(foo(arr[0,:], b2)))  # arr longer 

testing:

1B(1, a)
2B(2, b)
NoneB(3, c)
NoneB(4, d)
[B(1, a) B(2, b)]1B(6, a)
2B(7, b)
3B(8, c)
4B(9, d)
5None[B(1, a) B(2, b) B(3, c) B(4, d) B(5, d)]

Nothing special or magical; just a matter making sure I get the if tests and indentation right.

Solution 2:

Make the replacement block like this. Also note this assumes your arr will be strictly in multiples of 5 and there will be as many blocks like b1,b2 etc as there are blocks in arr as given in the example.

for i,b_arr in enumerate(d):
    temp_arr = []
    for j in range(5):
        if j<len(b_arr)
            temp_arr.append(B(arr[i,j],b_arr[j].b))
        else:
            temp_arr.append(B(arr[i,j],b_arr[-1].b))
     d[i] = np.array(temp_arr) ## not sure if this step is right, not too familiar with numpy.