Problem Formulation: Replacing One Element
How to replace the element at index
i in the list
lst with the new element
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
>>> 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
[x_0, x_1, ..., x_n]
[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.
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While working as a researcher in distributed systems, Dr. Christian Mayer found his love for teaching computer science students.
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His passions are writing, reading, and coding. But his greatest passion is to serve aspiring coders through Finxter and help them to boost their skills. You can join his free email academy here.