Python Cheatsheet 16 — Sets, Dicts, and Common Patterns

s = {1, 2, 3}
s = set([1, 2, 2, 3])      # {1, 2, 3}
empty = set()              # not {} (that's a dict!)

s.add(4)
s.remove(4)
s.discard(99)              # no error if missing
s.pop()                    # arbitrary element
1 in s                     # O(1) membership

len(s)

a | b                      # union
a & b                      # intersection
a - b                      # difference
a ^ b                      # symmetric difference

# In-place
a |= b; a &= b; a -= b; a ^= b

# Subset / superset
a < b                      # strict subset
a <= b
a > b                      # strict superset
a.issubset(b)
a.issuperset(b)
a.isdisjoint(b)

fs = frozenset([1, 2, 3])
# Can be a dict key or set element

unique = {x % 10 for x in numbers}
seen = {p.author_id for p in posts}

d = {"a": 1, "b": 2}
d["a"]                     # 1
d.get("a", 0)
d.get("missing", "default")
d.setdefault("c", 3)       # set if missing; return current
d["d"] = 4
del d["a"]
d.pop("b")
d.pop("missing", None)     # safe
"a" in d                   # O(1)

for k in d: ...
for k in d.keys(): ...
for v in d.values(): ...
for k, v in d.items(): ...

a = {"x": 1}
b = {"y": 2}

merged = a | b              # {"x": 1, "y": 2}
a |= b                      # in-place

# Pre-3.9
merged = {**a, **b}

from collections import defaultdict

by_letter = defaultdict(list)
for word in words:
    by_letter[word[0]].append(word)

# defaultdict(int) for counters
counts = defaultdict(int)
for x in items:
    counts[x] += 1

from collections import Counter

c = Counter(items)
c.most_common(5)
c.total()                  # 3.10+
c1 + c2                    # combine counts
c1 - c2
c1 & c2                    # min
c1 | c2                    # max

by_id = {u.id: u for u in users}
squared = {x: x**2 for x in range(10)}
filtered = {k: v for k, v in d.items() if v > 0}
inverted = {v: k for k, v in d.items()}

from collections import defaultdict

by_author = defaultdict(list)
for post in posts:
    by_author[post.author_id].append(post)

from itertools import groupby

items_sorted = sorted(items, key=lambda x: x.category)
for category, group in groupby(items_sorted, key=lambda x: x.category):
    items_in_cat = list(group)

sorted_by_value = dict(sorted(d.items(), key=lambda kv: kv[1]))
top_5 = dict(sorted(d.items(), key=lambda kv: kv[1], reverse=True)[:5])

# List of dicts → dict of lists
records = [{"a": 1, "b": 2}, {"a": 3, "b": 4}]
columns = {k: [r[k] for r in records] for k in records[0]}
# {"a": [1, 3], "b": [2, 4]}

from collections import Counter

most = Counter(items).most_common(1)[0]    # (value, count)

a = {"x": 1, "y": 2}
b = {"y": 3, "z": 4}

a.keys() & b.keys()        # {"y"}
a.keys() | b.keys()        # {"x", "y", "z"}
a.keys() - b.keys()        # {"x"}

d = {"a": 1, "b": 2, "c": 3}
list(d.keys())             # ["a", "b", "c"] — guaranteed

from collections import OrderedDict

od = OrderedDict(d)
od.move_to_end("a")        # move to end
od.move_to_end("a", last=False)  # move to start

keys_to_keep = {"a", "b"}
sliced = {k: v for k, v in d.items() if k in keys_to_keep}

from typing import TypedDict

class User(TypedDict):
    id: int
    name: str
    email: str | None