How to replace a element in a nested List in Python

Question

I can locate any listed content. In the example I locate ‘q’. I manually mapped its index as [1][0][1][1]. Then I replaced it with ‘z’ and it works. My question is what is the magic to get the index(q) or Object Address(q) when the if() condition get set to True?

 import ctypes
    
    lis = [['a', ['b'], 'c'], [['d', ['p', ['q']]], 'e', 'f']]
    idLis = id(lis)
    if 'q' in str(lis):
        idLisContent = ctypes.cast(idLis, ctypes.py_object).value
        print("List Content: = ", idLisContent)
        print("Index 0 = ", idLisContent[0])
        print("Index 1 = ", idLisContent[1])
        qId = id(idLisContent[1][0][1][1])
        print("Index Q = ", ctypes.cast(qId, ctypes.py_object).value)
        idLisContent[1][0][1][1] = 'z'
        print("List Content: = ", idLisContent)
    exit(1)

Output:

List Content: =  [['a', ['b'], 'c'], [['d', ['p', ['q']]], 'e', 'f']]
Index 0 =  ['a', ['b'], 'c']
Index 1 =  [['d', ['p', ['q']]], 'e', 'f']
Index Q =  ['q']
List Content: =  [['a', ['b'], 'c'], [['d', ['p', 'z']], 'e', 'f']]

Asked By: Santhosh Kumar

||

Source

Answer 1

Just convert it to string and replace ‘q’ to ‘z’

lis = [['a', ['b'], 'c'], [['d', ['p', ['q']]], 'e', 'f']]

str_lis = str(lis).replace('q', 'z');

print  str_lis;

Answered By: Soy Sin

Answer 2

I offer a true method to replace the value of a multidimensional array instead of converting it to a string first as stated in this accepted answer. Of course, these ways use a recursive function to get the index of the value in the multidimensional array and to replace that with another value using the index.

from copy import deepcopy

def getAllIndex(val, depth):
  if type(val) is str:
    allIndex[val] = deepcopy(depth)
  for i, j in enumerate(val):
    if type(val) is list:
      depth.append(i)
      getAllIndex(j, depth)
      depth.pop()
    
def getValue(val, index):
  if type(val) is str:
    return val
  else:
    return getValue(val[index[0]], index[1:]) 
    
def replaceValue(val, string, index):
  if len(index) == 1:
    val[index[0]] = string
  else:
    replaceValue(val[index[0]], string, index[1:]) 

lis = [['a', ['b'], 'c'], [['d', ['p', ['q']]], 'e', 'f']]

print("List Content: = ", lis)
print("Index 0 = ", lis[0])
print("Index 1 = ", lis[1])

allIndex = {}
getAllIndex(lis, []) # returns {'a': [0, 0], 'b': [0, 1, 0], 'c': [0, 2], 'd': [1, 0, 0], 'p': [1, 0, 1, 0], 'q': [1, 0, 1, 1, 0], 'e': [1, 1], 'f': [1, 2]}
indexQ = allIndex['q'] # get index of 'q'

print("Index Q = ", indexQ) # index of 'q' = [1][0][1][1][0] represent as [1, 0, 1, 1, 0]
print("Value of",indexQ,"=",getValue(lis, indexQ)) # get value of lis[1][0][1][1][0]

replaceValue(lis, 'z', indexQ) # replace specific value e.g. 'q' => 'z' with index of 'q'

print("List Content: = ", lis)

The output will be:

List Content: =  [['a', ['b'], 'c'], [['d', ['p', ['q']]], 'e', 'f']]
Index 0 =  ['a', ['b'], 'c']
Index 1 =  [['d', ['p', ['q']]], 'e', 'f']
Index Q =  [1, 0, 1, 1, 0]
Value of [1, 0, 1, 1, 0] = q
List Content: =  [['a', ['b'], 'c'], [['d', ['p', ['z']]], 'e', 'f']]

Answered By: Jordy

How to replace a element in a nested List in Python

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