Removing Dictionary from List of Dictionaries - Python

We are given a list of dictionaries, and our task is to remove specific dictionaries based on a condition. For instance given the list: a = [{'x': 10, 'y': 20}, {'x': 30, 'y': 40}, {'x': 50, 'y': 60}], we might want to remove the dictionary where 'x' equals 30 then the output will be [{'x': 10, 'y': 20}, {'x': 50, 'y': 60}].

List Comprehension

Using list comprehension, we can filter the list while excluding dictionaries that meet the condition.

a = [{'x': 10, 'y': 20}, {'x': 30, 'y': 40}, {'x': 50, 'y': 60}]

# Removing dictionary where 'x' equals 30
a = [d for d in a if d.get('x') != 30]

print(a)

[{'x': 10, 'y': 20}, {'x': 50, 'y': 60}]

Explanation:

• We iterate through each dictionary in the list using list comprehension.
• Only dictionaries where d.get('x') != 30 are included in the new list.

Let's explore some more methods and see how we can remove dictionary from list of dictionaries.

Using filter() with Lambda

filter() function can be used to create a filtered version of the list that excludes dictionaries matching the condition.

a = [{'x': 10, 'y': 20}, {'x': 30, 'y': 40}, {'x': 50, 'y': 60}]

# Removing dictionary where 'x' equals 30
a = list(filter(lambda d: d.get('x') != 30, a))

print(a)

[{'x': 10, 'y': 20}, {'x': 50, 'y': 60}]

Explanation:

• filter() function evaluates the condition d.get('x') != 30 for each dictionary in the list.
• Dictionaries that satisfy the condition are included in the result and the result is converted back to a list using list().

Using for Loop with remove()

We can use a for loop to iterate over the list and remove dictionaries that meet the condition.

a = [{'x': 10, 'y': 20}, {'x': 30, 'y': 40}, {'x': 50, 'y': 60}]

# Removing dictionary
for d in a:
    if d.get('x') == 30:
        a.remove(d)
        break  

print(a)

[{'x': 10, 'y': 20}, {'x': 50, 'y': 60}]

Explanation:

• We loop through the list a and check each dictionary.
• If the condition d.get('x') == 30 is true, the dictionary is removed using remove.
• The break ensures we stop after the first match.

Using pop()

pop() method removes an element by index and using a reverse loop prevents index shifting issues.

a = [{'x': 10, 'y': 20}, {'x': 30, 'y': 40}, {'x': 50, 'y': 60}]

# Removing dictionary
for i in range(len(a) - 1, -1, -1):  
    if a[i].get('x') == 30:
        a.pop(i)
        break  

print(a)

[{'x': 10, 'y': 20}, {'x': 50, 'y': 60}]

Explanation:

• We iterate over the list in reverse using a range-based loop.
• When the condition a[i].get('x') == 30 is met, the dictionary is removed using pop.
• Reverse iteration avoids issues caused by removing elements while looping.