Set & Frozenset functions in Python (2)

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*Memo:

• My post explains set and frozenset functions (1).
• My post explains set and frozenset functions (3).
• My post explains set and frozenset functions (4).
• My post explains a set (1).
• My post explains a frozenset (1).

union() can return all the elements of the set or frozenset and *others as shown below:

*Memo:

• The 1st arguments are *others(Optional-Default:()-Type:Iterable):
  • Don't use any keywords like *others=, others=, etc.
• union() creates a new set or frozenset.
• | can do union(), supporting set and frozenset.

<Set>:

A = {10, 50}                # set
B = frozenset([10, 30, 50]) # frozenset
C = [20, 40]                # list

print(A.union(B))    # {50, 10, 30}
print(A.union(C))    # {40, 10, 50, 20}
print(A.union(B, C)) # {50, 20, 40, 10, 30}
print(A.union())     # {10, 50}

A = {10, 50}                # set
B = frozenset([10, 30, 50]) # frozenset
C = frozenset([20, 40])     # frozenset

print(A | B)     # {50, 10, 30}
print(A | C)     # {40, 10, 50, 20}
print(A | B | C) # {50, 20, 40, 10, 30}
print(A)         # {10, 50}

<Frozenset>:

A = frozenset([10, 50]) # frozenset
B = {10, 30, 50}        # set
C = [20, 40]            # list

print(A.union(B))    # frozenset({50, 10, 30})
print(A.union(C))    # frozenset({40, 10, 50, 20})
print(A.union(B, C)) # frozenset({50, 20, 40, 10, 30})
print(A.union())     # frozenset({10, 50})

A = frozenset([10, 50]) # frozenset
B = {10, 30, 50}        # set
C = {20, 40}            # set

print(A | B)     # frozenset({50, 10, 30})
print(A | C)     # frozenset({40, 10, 50, 20})
print(A | B | C) # frozenset({50, 20, 40, 10, 30})
print(A)         # frozenset({10, 50})

update() can add zero or more iterables to the set as shown below:

*Memo:

• The 1st arguments are *others(Optional-Default:()-Type:Iterable):
  • Don't use any keywords like *others=, others=, etc.
• update() doesn't create a new set.
• update() doesn't exist for a frozenset.
• |= with or without | can do update(), supporting set and frozenset.

<Set>:

A = {10, 30, 50}

print(A)
# {10, 50, 30}

A.update({20, 30, 40})            # set
A.update(frozenset([20, 30, 40])) # frozenset
A.update([20, 30, 40])            # list
A.update({20, 30, 40}, frozenset([20, 30, 40]), [20, 30, 40])
            # set            # frozenset           # list
print(A)
# {50, 20, 40, 10, 30}

A.update()

print(A)
# {50, 20, 40, 10, 30}

A = {10, 30, 50}

print(A)
# {10, 50, 30}

A |= {20, 30, 40} # set
A |= {20, 30, 40} | frozenset([20, 30, 40]) | frozenset([20, 30, 40])
        # set             # frozenset               # frozenset
print(A)
# {50, 20, 40, 10, 30}

<Frozenset>:

A = frozenset([10, 30, 50])

print(A)
# frozenset({10, 50, 30})

A |= frozenset([20, 30, 40]) # frozenset
A |= frozenset([20, 30, 40]) | {20, 30, 40} | {20, 30, 40}
           # frozenset            # set          # set
print(A)
# frozenset({50, 20, 40, 10, 30})

intersection() can return the zero or more common elements of the set or frozenset and *others as shown below:

*Memo:

• The 1st arguments are *others(Optional-Default:()-Type:Iterable):
  • Don't use any keywords like *others=, others=, etc.
• intersection() creates a new set or frozenset.
• & does intersection(), supporting set and frozenset.

<Set>:

A = {10, 20, 30, 40}        # set
B = frozenset([10, 30, 50]) # frozenset
C = [30, 40]                # list

print(A.intersection(B))    # {10, 30}
print(A.intersection(C))    # {40, 30}
print(A.intersection(B, C)) # {30}
print(A.intersection())     # {40, 10, 20, 30}

A = {10, 20, 30, 40}        # set
B = frozenset([10, 30, 50]) # frozenset
C = frozenset([30, 40])     # frozenset

print(A & B)     # {10, 30}
print(A & C)     # {40, 30}
print(A & B & C) # {30}
print(A)         # {40, 10, 20, 30}

<Frozenset>:

A = frozenset([10, 20, 30, 40]) # frozenset
B = {10, 30, 50}                # set
C = [30, 40]                    # list

print(A.intersection(B))    # frozenset({10, 30})
print(A.intersection(C))    # frozenset({40, 30})
print(A.intersection(B, C)) # frozenset({30})
print(A.intersection())     # frozenset({40, 10, 20, 30})

A = frozenset([10, 20, 30, 40]) # frozenset
B = {10, 30, 50}                # set
C = {30, 40}                    # set

print(A & B)     # frozenset({10, 30})
print(A & C)     # frozenset({40, 30})
print(A & B & C) # frozenset({30})
print(A)         # frozenset({40, 10, 20, 30})

intersection_update() can return the zero or more common elements of the set and *others as shown below:

*Memo:

• The 1st arguments are *others(Optional-Default:()-Type:Iterable):
  • Don't use any keywords like *others=, others=, etc.
• intersection_update() doesn't create a new set.
• intersection_update() doesn't exist for a frozenset.
• &= with or without & can do intersection_update(), supporting set and frozenset.

<Set>:

A = {10, 20, 30, 40}        # set
B = frozenset([10, 30, 50]) # frozenset
C = frozenset([30, 40])     # frozenset

A.intersection_update(B)
print(A)
# {10, 30}

A.intersection_update(C)
print(A)
# {30}

A.intersection_update(B, C)
print(A)
# {30}

A.intersection_update()
print(A)
# {30}

A = {10, 20, 30, 40}        # set
B = frozenset([10, 30, 50]) # frozenset
C = frozenset([30, 40])     # frozenset

A &= B
print(A)
# {10, 30}

A &= C
print(A)
# {30}

A &= B & C
print(A)
# {30}

<Frozenset>:

A = frozenset([10, 20, 30, 40]) # frozenset
B = {10, 30, 50}                # set
C = {30, 40}                    # set

A &= B
print(A)
# frozenset({10, 30})

A &= C
print(A)
# frozenset({30})

A &= B & C
print(A)
# frozenset({30})