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Super Kai (Kazuya Ito)
Super Kai (Kazuya Ito)

Posted on • Edited on

Set in Python (2)

#python #set #curlybraces #function

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*Memo:

  • My post explains a set and the set with copy.
  • My post explains set functions (2).

add() can add a value to the set as shown below:

*Memo:

  • The 1st argument is elem(Required-Type:Any):
    • Don't use elem=. 
A = {10, 30, 50} print(A) # {10, 50, 30}  A.add(20) print(A) # {10, 50, 20, 30}  A.add(30) print(A) # {10, 50, 20, 30}  A.add(40) print(A) # {40, 10, 50, 20, 30}  A.add(50) print(A) # {40, 10, 50, 20, 30}
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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.
  • |= with or without | also does update() but |= and | don't support the combination of the set and list or tuple. 
A = {10, 30, 50} print(A) # {10, 50, 30}  A.update({20, 30, 40}) A.update([20, 30, 40]) A.update((20, 30, 40)) A.update({20, 30, 40}, [20, 30, 40], (20, 30, 40)) print(A) # {50, 20, 40, 10, 30}  A.update() print(A) # {50, 20, 40, 10, 30}
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A = {10, 30, 50} print(A) # {10, 50, 30}  A |= {20, 30, 40} A |= {20, 30, 40} | {20, 30, 40} | {20, 30, 40} print(A) # {50, 20, 40, 10, 30}
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remove() can remove the element matched to elem from the set, getting error if the element doesn't exist in the set as shown below:

*Memo:

  • The 1st argument is elem(Required-Type:Any):
    • Don't use elem=. 
A = {10, 30, 50, 70} print(A) # {10, 50, 70, 30}  A.remove(50) print(A) # {10, 70, 30}  A.remove(70) print(A) # {10, 30}  A.remove(20) # KeyError: 20
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discard() can remove the element matched to elem from the set, not getting error even if the element doesn't exist in the set as shown below:

*Memo:

  • The 1st argument is elem(Required-Type:Any):
    • Don't use elem=. 
A = {10, 30, 50, 70} print(A) # {10, 50, 70, 30}  A.discard(50) print(A) # {10, 70, 30}  A.discard(70) print(A) # {10, 30}  A.discard(20) print(A) # {10, 30}
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pop() can remove and throw an arbitrary element from the set as shown below:

*Memo:

  • It has no arguments.
  • Error occurs if the set is empty. 
A = {10, 30, 50} print(A) # {10, 50, 30}  print(A.pop()) # 10 print(A) # {50, 30}  print(A.pop()) # 50 print(A) # {30}  print(A.pop()) # 30 print(A) # set()  print(A.pop()) # KeyError: 'pop from an empty set'
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clear() can remove all elements from the set as shown below:

*Memo:

  • It has no arguments: 
A = {10, 30, 50} print(A) # {10, 50, 30}  A.clear() print(A) # set()
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union() can return all the 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.
  • | also does union() but | doesn't support the combination of a set and list or tuple. 
A = {10, 50} B = {10, 30, 50} C = {20, 40} 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}  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}
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A = {10, 50} # set B = [10, 30, 50] # list C = (20, 40) # tuple  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}
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intersection() 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.
  • & also does intersection() but & doesn't support the combination of a set and list or tuple.
  • intersection() creates a copy while intersection_update() doesn't. 
A = {10, 20, 30, 40} B = {10, 30, 50} C = {30, 40} 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}  print(A & B) # {10, 30} print(A & C) # {40, 30} print(A & B & C) # {30} print(A) # {40, 10, 20, 30}
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A = {10, 20, 30, 40} # set B = [10, 30, 50] # list C = (30, 40) # tuple  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}
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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.
  • &= with or without & does intersection_update() but &= and & doesn't support the combination of a set and list or tuple.
  • intersection_update() doesn't create a copy while intersection() does. 
A = {10, 20, 30, 40} B = {10, 30, 50} C = {30, 40} 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}
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A = {10, 20, 30, 40} # set B = [10, 30, 50] # list C = (30, 40) # tuple  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}
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A = {10, 20, 30, 40} B = {10, 30, 50} C = {30, 40} A &= B print(A) # {10, 30}  A &= C print(A) # {30}  A &= B & C print(A) # {30}
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