## 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

in that order?`x_0, x_1, ..., x_n`

## 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.

Not very pretty, isn’t it? If you still want to learn how one-liners work, check out my book:

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