Nested array duplication [duplicate] - python

This question already has answers here:
List of lists changes reflected across sublists unexpectedly
(17 answers)
Closed 1 year ago.
I'm guessing the problem happens because I append the same array multiple times and any changes made after are made to all the nested arrays. Is there a way to add them to an array and not get that to happen. For what I'm doing I to need about 500 different nested arrays, so doing it manually is a pain.
Output:
Before append
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
after append
[[0, 2, 0, 0], [0, 2, 0, 0], [0, 2, 0, 0], [0, 2, 0, 0]]
What I expected to happen after append:
[[0, 0, 0, 0], [0, 2, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
Code:
testArray = []
nestedTestArray = []
def testarraynestedcreate():
for countforloop1 in range(4):
nestedTestArray.append(0)
testArray.append(nestedTestArray)
testarraynestedcreate()
print(testArray) # before change
testArray[1][1] = 2 # change element in array
print(testArray) # after change )


You can use copy():
testArray.append(nestedTestArray.copy())
This way each list is independent from each other (and not a reference to the first one).

Related

Number changes in all row of array [duplicate]

This question already has answers here:
List of lists changes reflected across sublists unexpectedly
(17 answers)
Closed 3 months ago.
I created a 4x5 2D array using python, and when I wanted to change a number inside it, it automatically changes the number in every row
rows,cols = (4,5)
arr = [[0]*cols]*rows
print (arr)
And this is how the output shows
[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
After I created the array, I decide to change a number in the first row
arr[0][2] = 3
print(arr)
But it appears like this
[[0, 0, 3, 0, 0], [0, 0, 3, 0, 0], [0, 0, 3, 0, 0], [0, 0, 3, 0, 0]]
I checked with it and I still can't find any problem in it. May someone help me with it?

'Multiplying' the list copies the value references repeatedly - which is fine for primitives but not so much for lists, like you've seen. If you want different instances of the list, you could use list comprehension:
rows, cols = (4, 5)
arr = [[0] * cols for _ in range(rows)]
arr[0][2] = 3
print(arr) # [[0, 0, 3, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]

python 2d list problem with assigning value [duplicate]

This question already has answers here:
List of lists changes reflected across sublists unexpectedly
(17 answers)
Closed 2 years ago.
I have a question regarding 2d lists in python after initilizing the list when I assign a value to an index of list I will get all the indexes assigned with the value
I'm really confused
can someone explain why this is happening
also how should I fix this issue
class_list = [[0] * 3] * 10
print(class_list)
class_list[2][1] = 10
print(class_list)
and this is result:
[[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0], [0,
0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]] [[0, 10, 0], [0, 10, 0], [0,
10, 0], [0, 10, 0], [0, 10, 0], [0, 10, 0], [0, 10, 0], [0, 10, 0],
[0, 10, 0], [0, 10, 0]]

Numpy
import numpy as np
class_list = np.array([[0] * 3]*10)
print(class_list)
class_list[2][1] = 10
# class_list.tolist() # if you want to use lists only.
print(class_list)
You will get the expected output.

np.zeros
import numpy as np
class_list = np.zeros((10, 3))
class_list[2][1] = 10

Multiplying a list by a number creates items have relation with the original one? [duplicate]

This question already has answers here:
List of lists changes reflected across sublists unexpectedly
(17 answers)
Closed 9 years ago.
I have a list of List say mysolution:
>>>mySolution
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
>>> mySolution[0][0] = 1
>>> mySolution
[[1, 0, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0]]
Intended output:
[[1, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
why is it that all the 1st elements in my list of list's is being changed to 1?
I would only like to change the first element of the first list to 1.

What matters is how you created your original mysolution list. As it seems, it contains four times the same list which is why changing it once will make it change in all four locations.
To initialize independent zero-filled lists like that, you can do the following:
mysolution = [[0] * 4 for i in range(4)]

It's quite possible that you created the list like this:
mySolution = [0]*4
mySolution = [mySolution]*4
Or equivalently:
mySolution = [[0]*4]*4
Either of the above snippets will create a list with four sublists which are copies of the exact, same sublist, so any modification on one sublist will be reflected on the others - they're one and the same. The solution is to create four different sublists:
mySolution = [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
Or a bit shorter:
mySolution = [[0]*4 for _ in xrange(4)]

Because all the contained lists are actually the same list. When you do:
l = [0, 0, 0, 0]
my_solution = [l, l, l]
Then, my_solution[0], my_solution[1], and my_solution[2] are references to the same object (l).
If you modify the list in one location, it changes everywhere. That is because lists are mutable objects.
Instead, use multiple lists:
l1 = [0, 0, 0, 0]
l2 = [0, 0, 0, 0]
l3 = [0, 0, 0, 0]
my_solution = [l1, l2, l3]
Which will work as intended.

please note that this is doing fine:
mySolution = [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
mySolution[0][0] = 1
print mySolution
>>>
[[1, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
it all depends on how you initialized your solution. this
mySolution = [[0, 0, 0, 0]]*4
mySolution[0][0] = 1
print mySolution
gives
>>>
[[1, 0, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0]]
>>>
because here each array [0, 0, 0, 0] in mySolution is a copy of initialization array [0, 0, 0, 0] in [[0, 0, 0, 0]]*4. if you change first element of first array, copy of it also change.
with this initialization mySolution = [[0, 0, 0, 0] for x in range(4)] you are not copying the array but appending [0,0,0,0] four times, giving the result that you are expecting.

How to create copy of a 3D list in python [duplicate]

This question already has answers here:
How do I clone a list so that it doesn't change unexpectedly after assignment?
(24 answers)
Closed 5 years ago.
I have this code
list1 = [[[0, 0, 0], [0, 0, 0], [0, 0, 0]],
[[0, 0, 0], [0, 0, 0], [0, 0, 0]],
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]]
list2 = list1
list1[0][0][0] = 2
print(list2)
I want to create a copy of list1 in list2 so that when I change the value in list1 it doesn't change the value in list2 also.

You could use deepcopy
import copy
list1 = [[[0, 0, 0], [0, 0, 0], [0, 0, 0]],
[[0, 0, 0], [0, 0, 0], [0, 0, 0]],
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]]
list2 = copy.deepcopy(list1)
list1[0][0][0] = 2
print(list2)

Changing an element in one list changes multiple lists [duplicate]

This question already has answers here:
List of lists changes reflected across sublists unexpectedly
(17 answers)
Closed 9 years ago.
I have a list of List say mysolution:
>>>mySolution
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
>>> mySolution[0][0] = 1
>>> mySolution
[[1, 0, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0]]
Intended output:
[[1, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
why is it that all the 1st elements in my list of list's is being changed to 1?
I would only like to change the first element of the first list to 1.

What matters is how you created your original mysolution list. As it seems, it contains four times the same list which is why changing it once will make it change in all four locations.
To initialize independent zero-filled lists like that, you can do the following:
mysolution = [[0] * 4 for i in range(4)]

It's quite possible that you created the list like this:
mySolution = [0]*4
mySolution = [mySolution]*4
Or equivalently:
mySolution = [[0]*4]*4
Either of the above snippets will create a list with four sublists which are copies of the exact, same sublist, so any modification on one sublist will be reflected on the others - they're one and the same. The solution is to create four different sublists:
mySolution = [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
Or a bit shorter:
mySolution = [[0]*4 for _ in xrange(4)]

Because all the contained lists are actually the same list. When you do:
l = [0, 0, 0, 0]
my_solution = [l, l, l]
Then, my_solution[0], my_solution[1], and my_solution[2] are references to the same object (l).
If you modify the list in one location, it changes everywhere. That is because lists are mutable objects.
Instead, use multiple lists:
l1 = [0, 0, 0, 0]
l2 = [0, 0, 0, 0]
l3 = [0, 0, 0, 0]
my_solution = [l1, l2, l3]
Which will work as intended.

please note that this is doing fine:
mySolution = [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
mySolution[0][0] = 1
print mySolution
>>>
[[1, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
it all depends on how you initialized your solution. this
mySolution = [[0, 0, 0, 0]]*4
mySolution[0][0] = 1
print mySolution
gives
>>>
[[1, 0, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0]]
>>>
because here each array [0, 0, 0, 0] in mySolution is a copy of initialization array [0, 0, 0, 0] in [[0, 0, 0, 0]]*4. if you change first element of first array, copy of it also change.
with this initialization mySolution = [[0, 0, 0, 0] for x in range(4)] you are not copying the array but appending [0,0,0,0] four times, giving the result that you are expecting.

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