Python Forum

I want to manipulate an index value on list1, but can't figure out why the manipulation('alt') appears at two locations instead of just one. list2 shows correctly what I want to do on list1.

list1 = []
list2 = [[[40], 70], [[[40], 70], 90]]


def append_values():

    amount = None

    if len(list1) == 0:
        amount = 70
        list1.append([[40],amount])
    else:
        amount = 90
        list1.append([list1[-1],amount])
    


def alter_lists():

    print('ori: ', list1) #Orginal list
    list1[1][0][1] = "alt"
    print('alt: ', list1)   #Altered list

    print('\n')

    print('ori: ', list2) #Orginal list
    list2[1][0][1] = 'alt'
    print('alt: ', list2)  #Altered list


append_values()
append_values()
alter_lists()

Output:

ori:  [[[40], 70], [[[40], 70], 90]]
alt:  [[[40], 'alt'], [[[40], 'alt'], 90]]


ori:  [[[40], 70], [[[40], 70], 90]]
alt:  [[[40], 70], [[[40], 'alt'], 90]]

There is nothing wrong, it's as it should be.

In [1]: lst = []                                                          

In [2]: lst.append([1,2])                                                 

In [3]: lst                                                               
Out[3]: [[1, 2]]

In [4]: id(lst[0])                                                        
Out[4]: 4797778952

In [5]: lst.append(lst[0])             # append same item second time                                          

In [6]: lst                                                               
Out[6]: [[1, 2], [1, 2]]

In [7]: id(lst[1])                     # both items  refer to same object                                               
Out[7]: 4797778952

In [8]: lst[0].append(3)               # changing the object                                              

In [9]: lst                                                               
Out[9]: [[1, 2, 3], [1, 2, 3]]         # changes both as the reference is to the same object

In [10]: id(lst[0]) == id(lst[1])                                         
Out[10]: True

If this behaviour is not desired you can append new object:

In [11]: lst.append([1, 2, 3])                     # append new list                                     

In [12]: lst                                                              
Out[12]: [[1, 2, 3], [1, 2, 3], [1, 2, 3]]         # not visually distinguishable from other items in list

In [13]: id(lst[2])                                                       
Out[13]: 4796667592                                # not same object (different id)

In [14]: id(lst[0]) == id(lst[2])                                         
Out[14]: False

In [15]: lst[2].append(4)                          # appending to new item                                                  

In [16]: lst                                                              
Out[16]: [[1, 2, 3], [1, 2, 3], [1, 2, 3, 4]]      # appends only to new object