How to Count the Number of Keys in a Python Dictionary

Python’s built-in len() is all you need: to count the number of keys in a Python dictionary, pass the dictionary to len() and it returns the key total instantly, because a dict keeps that number on hand.

prices = {"laptop": 899.99, "mouse": 24.95}
len(prices)  # 2

Every result on this page came straight out of a Command Prompt window running Python 3.12.5.

In this tutorial, we’ll count keys that pass a test, total the keys inside nested dictionaries, tally keys across a list of records, and see how updates move the count.

Count dictionary keys with len()

Here’s a small store inventory. I count it with len() directly and through the keys() view, then check an empty dictionary:

inventory = {"laptop": 899.99, "monitor": 249.50, "keyboard": 49.99, "mouse": 24.95}

print(len(inventory))          # number of keys
print(len(inventory.keys()))   # same answer through the keys view

empty = {}
print(len(empty))
if not empty:
    print("No keys yet")

Output:

4
4
0
No keys yet

len(d.keys()) works but builds a view you don’t need, and an empty dict is falsy. For how len() treats other containers, see my guide to finding the length of a Python dictionary.

Count only the keys that match a condition

Sometimes you don’t want every key, just the ones that pass a test. Feed a generator of 1s to sum():

inventory = {"laptop": 899.99, "monitor": 249.50, "keyboard": 49.99, "mouse": 24.95}

over_50 = sum(1 for price in inventory.values() if price > 50.00)
print("Items over $50.00:", over_50)

m_keys = sum(1 for name in inventory if name.startswith("m"))
print("Keys starting with m:", m_keys)

Output:

Items over $50.00: 2
Keys starting with m: 2

The first test looks at values and the second at the key names themselves. Both make one pass; iterating through a Python dictionary explains keys(), values() and items() in more depth.

How do I count keys in a nested dictionary?

len() only sees the top level. To include the keys of inner dictionaries, recurse into any value that is itself a dict:

staff = {
    "Sales": {"Megan Bowen": "Manager", "Adele Vance": "Rep"},
    "IT": {"Alex Wilber": "Admin", "Lee Gu": "Developer", "Grady Archie": "Analyst"},
    "HR": {},
}

def count_keys(d):
    total = 0
    for value in d.values():
        total += 1
        if isinstance(value, dict):
            total += count_keys(value)
    return total

print("Top-level keys:", len(staff))
print("All keys, every level:", count_keys(staff))
print({dept: len(people) for dept, people in staff.items()})

Output:

Top-level keys: 3
All keys, every level: 8
{'Sales': 2, 'IT': 3, 'HR': 0}

The function counts the department key first, then adds whatever lives inside it. When you read deep values rather than count them, handling KeyError in a nested Python dictionary keeps missing levels from crashing your code.

Count how often each key appears in a list of dictionaries

API records rarely share the same fields. collections.Counter tallies every key in one line:

from collections import Counter

orders = [
    {"id": 5001, "total": 120.00, "coupon": "FALL10"},
    {"id": 5002, "total": 64.25},
    {"id": 5003, "total": 310.40, "coupon": "VIP", "gift_wrap": True},
]

key_counts = Counter(key for order in orders for key in order)
print(key_counts)
print("coupon used in", key_counts["coupon"], "of", len(orders), "orders")

unique = set().union(*orders)
print("Distinct keys:", len(unique))

Output:

Counter({'id': 3, 'total': 3, 'coupon': 2, 'gift_wrap': 1})
coupon used in 2 of 3 orders
Distinct keys: 4

set().union(*orders) answers the related question of how many distinct keys exist at all. If you need the key names themselves rather than a number, read getting the keys of a dictionary in Python.

Does updating a dictionary change its key count?

Only adding or removing keys moves the number. Assigning to a key that already exists swaps the value in place:

scores = {"Boston": 88, "Denver": 92}
print(len(scores))

scores["Boston"] = 95        # existing key: value changes, count stays
print(len(scores))

scores["Austin"] = 79        # new key: count goes up
print(len(scores))

del scores["Denver"]
print(len(scores))

try:
    scores.count()
except AttributeError as e:
    print("AttributeError:", e)

Output:

2
2
3
2
AttributeError: 'dict' object has no attribute 'count'

Note the last line: dictionaries have no count() method, so stick with len(). My post on updating values in a Python dictionary covers update() and other ways to change values safely.

Building the dictionary in a loop? creating a dictionary in Python with a for loop shows the pattern.

Which key-counting method fits your data?

You want	Use	Notes
All top-level keys	len(d)	Constant time
Keys passing a test	sum(1 for k in d if test)	One pass
Keys at every level	Recursive function	Checks isinstance(v, dict)
Key frequency across records	Counter(k for r in rows for k in r)	Missing keys show as 0
Distinct keys across records	len(set().union(*rows))	Ignores repeats

Want to keep practicing with dictionaries? Try these next:

• Sum all values in a Python dictionary
• Create a dictionary from a list in Python
• Remove multiple keys from a Python dictionary
• Check whether a Python dictionary contains a key

Frequently asked questions