Python: dictionaries

Creating a dictionary and what a key must be#

prices = {"tea": 120, "coffee": 250}
empty = {}                      # a dictionary, not a set
from_pairs = dict([("a", 1), ("b", 2)])
print(prices["tea"], len(prices))   # 120 2

A key can only be an immutable (hashable) value: a string, a number, a tuple, a boolean. A list cannot be a key — trying gives TypeError: unhashable type: 'list'. A value can be anything, including another dictionary.

Insertion order is preserved: since Python 3.7 this is not an implementation accident but a language guarantee. Iterating over a dictionary yields the keys in the order they were added.

Access without KeyError#

Square brackets raise KeyError when the key is missing. When a missing key is a normal situation, use get:

print(prices.get("cocoa"))        # None
print(prices.get("cocoa", 0))     # 0
print("cocoa" in prices)          # False
prices["cocoa"] = 90              # adding also uses brackets

The in operator checks keys, not values; for values you need in prices.values().

The related setdefault method returns the value for a key, and if the key is missing it stores the given default and returns that. It is handy for grouping:

groups = {}
for word in ["apple", "avocado", "banana", "blueberry", "cherry"]:
    groups.setdefault(word[0], []).append(word)
print(groups)
  # {'a': ['apple', 'avocado'], 'b': ['banana', 'blueberry'], 'c': ['cherry']}

Iteration and the three views#

for key in prices:
    print(key, prices[key])

for key, value in prices.items():
    print(key, "→", value)

print(list(prices.keys()))     # ['tea', 'coffee', 'cocoa']
print(sum(prices.values()))    # 460

keys(), values() and items() return views, not lists: they look at the live dictionary and reflect changes. If you really need a list, wrap it in list().

Sorting a dictionary gives you a list of pairs, not a dictionary:

top = sorted(prices.items(), key=lambda kv: kv[1], reverse=True)
print(top)        # [('coffee', 250), ('tea', 120), ('cocoa', 90)]
print(dict(top))  # back to a dictionary, if you need one

Changing, deleting, merging#

prices.update({"tea": 130, "juice": 100})
print(prices)   # {'tea': 130, 'coffee': 250, 'cocoa': 90, 'juice': 100}

removed = prices.pop("cocoa")
print(removed)               # 90
print(prices.pop("none", 0)) # 0 instead of KeyError
del prices["juice"]

update overwrites existing keys and adds new ones. Since Python 3.9 there is a merge operator: a | b gives a new dictionary, and a |= b changes a in place. When a key appears in both, the right-hand dictionary wins.

A dictionary comprehension is built like a list comprehension, but with a pair separated by a colon:

squares = {n: n * n for n in range(4)}
print(squares)          # {0: 0, 1: 1, 2: 4, 3: 9}
swapped = {v: k for k, v in squares.items()}
print(swapped[9])       # 3

Where it breaks#

Adding while iterating. Changing the size of a dictionary inside a loop over it raises RuntimeError: dictionary changed size during iteration. Collect the changes separately or iterate over a copy: for k in list(d).

Duplicate keys in a literal. {"a": 1, "a": 2} is not an error — you end up with {'a': 2}: the last assignment silently wins.

True and 1 are the same key. Keys are compared by value, and 1 == True, so {1: "one", True: "yes"} collapses into a single item with the value "yes".

Counting with get. A typical task is counting frequencies. Without a dictionary it grows into a dozen lines; with one it fits in three:

counts = {}
for ch in "abracadabra":
    counts[ch] = counts.get(ch, 0) + 1
print(counts)   # {'a': 5, 'b': 2, 'r': 2, 'c': 1, 'd': 1}

Try it yourself: count word frequencies in a text and print the three most common using sorted with key. For the text handling you will need Python strings, and when order matters more than keys, use Python lists.

Step-by-step plan

1. A settings dictionaryCreate a dictionary of five pairs, print a value by key and the length.
2. Safe accessCompare d["none"] with d.get("none", 0): the first call should raise KeyError.
3. Iterate with itemsPrint every pair as "key → value" using items().
4. Character frequenciesCount how many times each character appears in a string using get.
5. Sort by valueSort the pairs by value in descending order and print the first three.

Check yourself

Sources

• Dictionaries in the Python tutorialThe Dictionaries section with iteration examples
  free
• The dict type referenceEvery dictionary method and the rules for keys
  free
• collections.CounterA ready-made dictionary for counting frequencies
  free

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