c1_09_commonalites_in_two_dicts.py

# %% [markdown]
# # Commonalities in two dicts
# %%
# Reminder : sets
{"a", "b", "c", "a", "b"}

#%%
{1, 2, {"a": 1}}  # elements in a set must be hashable
# Similarly to dict keys, elements in a set must be hashable. To see if an element exists in a set, first a hash lookup is performed and then the __eq__ method is used to see if elements should be considered equal
# In fact, because sets and dict keys are very similar, they share some methods
# %%
aa = set(("string1", 0, 5, 8, "string2"))
bb = {"string2", "string3", 0, 6, 9}
# %%
print(aa.intersection(bb))
print(aa & bb)
# %%
print(aa.union(bb))
print(aa | bb)

# %%
print(aa.difference(bb))  # in aa but not in bb
print(aa - bb)
# %%
print(aa.symmetric_difference(bb))  # in aa but not in bb
print(aa ^ bb)
print((aa | bb) - (aa & bb) == aa ^ bb)
# %% Application to dicts
a = {"x": 1, "y": 2, "z": 3}
b = {"w": 10, "x": 11, "y": 2}

print(a.keys() | b.keys())  # returns a set
# %%
print(a.keys().intersection(b.keys()))  # it seems the only set method is isdisjoint
# %%
a.items() & b.items()  # returns a set of tuple of identical pairs
# %%
a.values() | b.values()  # not implemented for values as they are not forced to be unique
# %%
set(a.values()) | set(b.values())  # workaround
# %%
# * filter
d = {"xx": 1, "yy": 2, "zz": 3, "tt": 5}
e = {"xx": 1, "yyy": 2, "zzz": 3, "tt": 5}
{key: d[key] for key in d.keys() if key not in e.keys()}
# %%
{key: d[key] for key in d.keys() - e.keys()}
# %%