How to Convert a Dictionary to a List in Python

To convert a dictionary to a list in Python, call list() on the part you want: list(d) gives the keys, list(d.values()) the values, and list(d.items()) a list of (key, value) tuples. All three keep the dictionary’s insertion order:

list(prices.values())   # [14.99, 9.5, 4.25]
list(prices.items())    # [('Coffee beans', 14.99), ...]

A small store price list runs through every example; I ran them on Python 3.12.5, where dicts always keep insertion order.

You’ll see keys, values and items as lists, the ‘dict_values’ subscript error, filtering and sorting while converting, flattening a dict of lists, and turning nested dicts into rows.

Convert a dictionary to a list of keys, values or items

Pick the view you need and wrap it in list(). For pairs that must be lists rather than tuples, say for JSON, use a comprehension:

prices = {"Coffee beans": 14.99, "Mug": 9.50, "Filters": 4.25}

print(list(prices))            # keys
print(list(prices.values()))   # values
print(list(prices.items()))    # (key, value) tuples
print([[k, v] for k, v in prices.items()])   # pairs as lists, e.g. for JSON

Output:

['Coffee beans', 'Mug', 'Filters']
[14.99, 9.5, 4.25]
[('Coffee beans', 14.99), ('Mug', 9.5), ('Filters', 4.25)]
[['Coffee beans', 14.99], ['Mug', 9.5], ['Filters', 4.25]]

list(d) and list(d.keys()) are the same. For more on the values case, see converting dictionary values to a list.

Why do I get ‘dict_values’ object is not subscriptable?

d.values() returns a view, not a list, so you can’t index it. Convert it first, or use next(iter(...)) for just the first value:

prices = {"Coffee beans": 14.99, "Mug": 9.50, "Filters": 4.25}
values = prices.values()

try:
    print(values[0])
except TypeError as err:
    print("TypeError:", err)

print(list(values)[0])
print(next(iter(values)))      # first value without building a list

prices["Kettle"] = 39.00
print(values)                  # a view: it sees the new item

Output:

TypeError: 'dict_values' object is not subscriptable
14.99
14.99
dict_values([14.99, 9.5, 4.25, 39.0])

Views stay live: add a key and the view shows it. Keys work the same way, as in getting the keys of a dictionary.

Filter and sort while converting a dictionary to a list

A comprehension can pick and reshape items in the same step. sorted() on items() sorts by key, or by value with a key function:

prices = {"Coffee beans": 14.99, "Mug": 9.50, "Filters": 4.25, "Kettle": 39.00, "Grinder": 89.99}

over_10 = [item for item, price in prices.items() if price > 10]
print("over $10:", over_10)

by_price = sorted(prices.items(), key=lambda kv: kv[1], reverse=True)
print("most expensive first:", by_price[:3])

labels = [f"{item}: ${price:.2f}" for item, price in prices.items()]
print(labels)

Output:

over $10: ['Coffee beans', 'Kettle', 'Grinder']
most expensive first: [('Grinder', 89.99), ('Kettle', 39.0), ('Coffee beans', 14.99)]
['Coffee beans: $14.99', 'Mug: $9.50', 'Filters: $4.25', 'Kettle: $39.00', 'Grinder: $89.99']

For the pairs as tuples in more depth, see converting a dictionary to a list of tuples.

Flatten a dictionary of lists into one list

When each value is itself a list, a nested comprehension or itertools.chain joins them into one list:

from itertools import chain

stock = {"Austin": ["Mug", "Kettle"], "Denver": ["Beans"], "Tampa": ["Mug", "Filters"]}

flat = [item for items in stock.values() for item in items]
print(flat)
print(list(chain.from_iterable(stock.values())))
print(sorted(set(flat)))

pairs = [x for kv in {"TX": 3, "CO": 1}.items() for x in kv]     # keys and values interleaved
print(pairs)

Output:

['Mug', 'Kettle', 'Beans', 'Mug', 'Filters']
['Mug', 'Kettle', 'Beans', 'Mug', 'Filters']
['Beans', 'Filters', 'Kettle', 'Mug']
['TX', 3, 'CO', 1]

Wrap the result in set() when repeats don’t matter. Building this kind of dict in the first place is covered in a Python dictionary of lists.

Convert a dictionary of dictionaries to a list of records

APIs often return records keyed by ID. Merge the key into each record to get a list you can sort, filter or write to CSV:

orders = {
    1001: {"customer": "Emily Carter", "state": "TX", "total": 129.99},
    1002: {"customer": "James Wilson", "state": "CO", "total": 54.50},
}

rows = [{"id": oid, **data} for oid, data in orders.items()]
print(rows)

totals = [o["total"] for o in orders.values()]
print(totals, f"sum ${sum(totals):.2f}")

Output:

[{'id': 1001, 'customer': 'Emily Carter', 'state': 'TX', 'total': 129.99}, {'id': 1002, 'customer': 'James Wilson', 'state': 'CO', 'total': 54.5}]
[129.99, 54.5] sum $184.49

{"id": oid, **data} builds a new dict without changing the original. The reverse direction is covered in creating a dictionary from two lists.

Dictionary to list methods compared

You want	Code
Keys	list(d)
Values	list(d.values())
(key, value) tuples	list(d.items())
[key, value] lists	[[k, v] for k, v in d.items()]
Sorted by value	sorted(d.items(), key=lambda kv: kv[1])
Flattened values	[x for v in d.values() for x in v]

More conversions between Python’s built-in types:

• Create a dictionary from a list in Python
• Compare lists, tuples, sets and dictionaries
• Convert a set to a list in Python
• Count the keys in a Python dictionary

Frequently asked questions