How to Create a Dictionary from a List in Python

To create a dictionary from a list in Python, match the method to what the list holds: dict(pairs) for a list of key-value pairs, dict.fromkeys(keys) for a list of keys,:

dict([("NY", "New York"), ("TX", "Texas")])  # {'NY': 'New York', 'TX': 'Texas'}
dict.fromkeys(["Emily", "James"], 0.00)        # {'Emily': 0.0, 'James': 0.0}

I pasted every output below from real runs on Python 3.12.5, so you can trust them.

In this tutorial, we’ll go through each list shape, plus two traps: a shared default list and vanishing duplicate keys.

Create a dictionary from a list of tuples or pairs with dict()

When each item is already a two-element tuple or list, pass the whole list to dict(). Item one becomes the key, item two the value:

pairs = [("NY", "New York"), ("TX", "Texas"), ("CA", "California")]
states = dict(pairs)
print(states)

rows = [["SKU-101", 19.99], ["SKU-102", 4.50]]   # inner lists work too
prices = dict(rows)
print(prices)

bad = [("NY", "New York"), ("TX",)]             # one item is missing its value
try:
    dict(bad)
except ValueError as e:
    print("ValueError:", e)

Output:

{'NY': 'New York', 'TX': 'Texas', 'CA': 'California'}
{'SKU-101': 19.99, 'SKU-102': 4.5}
ValueError: dictionary update sequence element #1 has length 1; 2 is required

The error names the broken element. If your pairs live in a tuple of tuples instead, my post on converting a tuple to a dictionary in Python covers that input.

Create a dictionary from a list of keys with dict.fromkeys()

If the list only holds keys, dict.fromkeys() gives every key the same starting value. Watch out for mutable defaults:

customers = ["Emily", "James", "Olivia"]

balances = dict.fromkeys(customers, 0.00)
print(balances)

orders = dict.fromkeys(customers, [])      # trap: ONE list shared by every key
orders["Emily"].append("2026-03-14")
print(orders)

orders = {name: [] for name in customers}  # fix: a new list per key
orders["Emily"].append("2026-03-14")
print(orders)

Output:

{'Emily': 0.0, 'James': 0.0, 'Olivia': 0.0}
{'Emily': ['2026-03-14'], 'James': ['2026-03-14'], 'Olivia': ['2026-03-14']}
{'Emily': ['2026-03-14'], 'James': [], 'Olivia': []}

Use fromkeys() for numbers, strings and None, and a comprehension for lists or dicts. More starting-value patterns are in initializing a Python dictionary with keys and values.

Build dictionary values from list items with a comprehension

A dict comprehension runs an expression for each item, so values come from the items. Here each city maps to its length:

cities = ["Boston", "Denver", "Austin", "Seattle"]

lengths = {city: len(city) for city in cities}
print(lengths)

Output:

{'Boston': 6, 'Denver': 6, 'Austin': 6, 'Seattle': 7}

Comprehensions get their own deep dive in Python dictionary comprehension. When the logic needs several lines, a plain loop reads better; see creating a dictionary in Python using a for loop.

How do I convert a flat list of alternating keys and values?

Form or CSV data often arrives as [key, value, key, value]. Slice out the even and odd positions and zip them back together:

flat = ["Jan", 1250.00, "Feb", 1480.50, "Mar", 1399.99]

sales = dict(zip(flat[::2], flat[1::2]))   # even items = keys, odd items = values
print(sales)

Output:

{'Jan': 1250.0, 'Feb': 1480.5, 'Mar': 1399.99}

It’s a neat fix you’ll reuse. If your keys and values are already in separate lists, follow creating a dictionary from two lists in Python instead.

Key a list of dictionaries by ID, and watch for duplicate keys

Pick a unique field to turn records into a lookup table. A repeated key keeps only its last value:

employees = [
    {"id": 101, "name": "Megan Bowen", "dept": "Sales"},
    {"id": 102, "name": "Alex Wilber", "dept": "IT"},
    {"id": 103, "name": "Adele Vance", "dept": "Sales"},
]
by_id = {e["id"]: e for e in employees}
print(by_id[102]["name"], "works in", by_id[102]["dept"])

pairs = [("Sales", "Megan"), ("IT", "Alex"), ("Sales", "Adele")]
d = dict(pairs)
print(d)
print(len(pairs), "pairs ->", len(d), "keys")   # the later "Sales" value wins

Output:

Alex Wilber works in IT
{'Sales': 'Adele', 'IT': 'Alex'}
3 pairs -> 2 keys

Compare len() of the list and the dictionary whenever duplicates are possible; counting the keys in a Python dictionary shows the options.

Which list-to-dictionary method should you use?

List holds	Use	Result
Pairs	dict(pairs)	First item = key
Keys only	dict.fromkeys(keys, v)	Same value for every key
Items to transform	{x: f(x) for x in lst}	Computed values
Alternating	dict(zip(lst[::2], lst[1::2]))	Even = key, odd = value
Dictionaries	{r["id"]: r for r in lst}	Lookup by ID

Good next reads once your dictionary exists:

• Update values in a Python dictionary
• Sum all values in a Python dictionary
• Work with a Python dictionary of lists
• Convert a dictionary back to a list

Frequently asked questions