1

I want my inserted array to replace as many elements in the target array as there are in this replacing array, just like this:

x = np.array([1, 2, 3, 4, 5])
y = np.array([6, 7, 8])
# Doing unknown stuff
x = array([6, 7, 8, 4, 5])

Is there some numpy method or something tricky to implement this?

I would also like to know if this can be done despite of how the lengths of the arrays relate to each other, for instance like this:

x = np.array([1, 2])
y = np.array([6, 7, 8])
# Doing unknown stuff
x = array([6, 7])
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3 Answers 3

Reset to default
4

Another solution

x[:len(y)] = y[:len(x)]

First case

>>> x
array([6, 7, 8, 4, 5])

Second case

>>> x
array([6, 7])
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2 Comments

Perfect! Thanks!
What do you mean by "not the same"? The memory address of x will not be modified only the replaced cells.
2

Here is one solution:

x = np.append(y, x[len(y):])
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1 Comment

Thanks, still please check out the second question I have appended
2

Numpy is relatively fast:

   x = np.array([1, 2, 3, 4, 5])
   y = np.array([6, 7, 8])
   xnew = np.concatenate((y, x[len(y):]))

>>> xnew
array([6, 7, 8, 4, 5])
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1 Comment

Thanks, yet the question has been updated

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