Replace element in 2d list python

• Problem Formulation: Replacing One Element
• Solution Indexing
• Problem Formulation: Replacing Multiple Elements
• Method 1: For Loop
• Method 2: zip() and For Loop
• Method 3: NumPy + Slice Assignment
• Method 4: Python One-Liner Solution
• Python One-Liners Book: Master the Single Line First!

Problem Formulation: Replacing One Element

Given

• List lst
• Element x
• Index i

How to replace the element at index i in the list lst with the new element x?

Solution Indexing

You use simple indexing using the square bracket notation lst[i] = x to replace the element at index i in list lst with the new element x.

>>> lst = ['Alice', 'Bob', 'Carl', 'Dave']
>>> x = 'Chris'
>>> i = 2
>>> lst[i] = x
>>> lst
['Alice', 'Bob', 'Chris', 'Dave']

But what if you want to replace multiple elements at multiple indices?

Problem Formulation: Replacing Multiple Elements

Given

• List lst
• Elements [x_0, x_1, ..., x_n]
• Indices [i_0, i_1, ..., i_n]

How to replace the elements at indices i_0, i_1, ..., i_n in the list lst with the new elements x_0, x_1, ..., x_n in that order?

Method 1: For Loop

You can use the range() function to get the pair of the i-th index and the i-th replacement value in a for loop. Then, you replace all elements one-by-one.

lst = ['Alice', 'Bob', 'Carl', 'Dave', 'Elena', 'Frank', 'George']
repl = ['None', 'Foo', 'Bar']
indices = [0, 2, 5]


# Method 1: For Loop
for i in range(len(indices)):
    lst[indices[i]] = repl[i]

print(lst)
# ['None', 'Bob', 'Foo', 'Dave', 'Elena', 'Bar', 'George']

Method 2: zip() and For Loop

A more Pythonic approach is to zip together the indices and replacement values and then simply iterating over them in pairs using multiple assignments.

lst = ['Alice', 'Bob', 'Carl', 'Dave', 'Elena', 'Frank', 'George']
repl = ['None', 'Foo', 'Bar']
indices = [0, 2, 5]


# Method 2: zip() and for loop
for index, replacement in zip(indices, repl):
    lst[index] = replacement

print(lst)
# ['None' 'Bob' 'Foo' 'Dave' 'Elena' 'Bar' 'George']

Method 3: NumPy + Slice Assignment

Stand on the shoulders of giants! You can use NumPy’s powerful advanced indexing functionality to pass the list of indices to be replaced in the indexing scheme—and replacing those with all elements on the right of an assignment operation.

lst = ['Alice', 'Bob', 'Carl', 'Dave', 'Elena', 'Frank', 'George']
repl = ['None', 'Foo', 'Bar']
indices = [0, 2, 5]


# Method 3: NumPy + Slice Assignment
import numpy as np
lst = np.array(lst)
lst[indices] = repl
print(lst)
# ['None' 'Bob' 'Foo' 'Dave' 'Elena' 'Bar' 'George']

Method 4: Python One-Liner Solution

I love Python one-liners (that’s why I’ve written a book about them). Can we solve the multiple replacement problem in a single line? Yeah, sure!

lst = ['Alice', 'Bob', 'Carl', 'Dave', 'Elena', 'Frank', 'George']
repl = ['None', 'Foo', 'Bar']
indices = [0, 2, 5]

# Method 4: Python One-Liner
lst = [repl[indices.index(i)] if i in indices else lst[i] for i in range(len(lst))]
print(lst)
# ['None' 'Bob' 'Foo' 'Dave' 'Elena' 'Bar' 'George']

• We use list comprehension [... for i in ...] to iterate over all indices from 0 to the length of the list.
• We use the ternary operator ... if ... else ... to check whether this index is one that must be replaced.
• If the index doesn’t have to be replaced, return the original element, otherwise return the replacement element.
• We use the list.index() method to figure out the index of the element to replace the original list element.

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