Adding a number to each index of an array and creating a list of new arrays

Question

I have a numpy array A = np.array([1,2,3]). I want to add 1 to each element of this array, and return an array with each addition, separately:

My desired output would be:

list1 = [[2,2,3][1,3,3][1,2,4]]

I have tried the np.ufunc method to add my arrays, and using a normal list but both methods add the arrays/lists cumulatively:

In[1]:    list1 = []
          A = np.array([1,2,3])
          for i in range(len(A)):
              np.add.at(A, [i,], 1)
              list1.append(A)
              print(list1)

Out[1]:   [array([2, 2, 3])]
          [array([2, 3, 3]), array([2, 3, 3])]
          [array([2, 3, 4]), array([2, 3, 4]), array([2, 3, 4])]

This seems like something that needs to be done outside the for loop, but I'm not sure what. Where am I going wrong?

Can you clarify what you need help with? I'm not quite sure I understand what's going wrong... — 12944qwerty
– 12944qwerty, Commented Apr 19, 2021 at 12:30
I want to take the array [1,2,3] and add 1 to every index of this array and store them, separately in a new list. So I want list 1 = [[2,2,3],[1,3,3],[1,2,4]]. But I am getting list1 = [[2,3,4],[2,3,4],[2,3,4]] instead, where the arrays are added each time — pkfly103
– pkfly103, Commented Apr 19, 2021 at 12:32
Oh, ok. You're appending it into a new list. Have you tried printing A directly instead of appending it to a list and printing the list? — 12944qwerty
– 12944qwerty, Commented Apr 19, 2021 at 12:33
You said "return an array", then called it list1, showed something invalid, now say "in a new list". Can you make up your mind? — Manuel
– Manuel, Commented Apr 19, 2021 at 12:36

Manuel · Accepted Answer · 2021-04-19 12:32:08Z

3

>>> A + np.eye(A.size)
array([[2., 2., 3.],
       [1., 3., 3.],
       [1., 2., 4.]])

answered Apr 19, 2021 at 12:32

Manuel

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Sid · Accepted Answer · 2021-04-19 12:34:20Z

1

Loop through the list.

a = [1,2,3]
out = []
for c, n in enumerate(a):
    newlst = []
    for c2, v in enumerate(a):
        if not c2 == c:
            newlst.append(v)
        else:
            newlst.append(v+1)
    out.append(newlst)
print(out)

Output:

[[2, 2, 3], [1, 3, 3], [1, 2, 4]]

answered Apr 19, 2021 at 12:34

Sid

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