Python set relational operation intersection, Difference DAY14

Go to weight and create a collection (unordered)

Python_1 = ['1','2','3']linux_1= ['1','2'] #python_and_linux_1= []# forP_nameinchpython_1:#ifP_nameinchlinux_1:# python_and_linux_1.append (p_name) #print (python_and_linux_1) p_s=Set(python_1) print (p_s)

Ask for Intersection

Python_1 = ['1','2','3']linux_1= ['1','2'] #python_and_linux_1= []# forP_nameinchpython_1:#ifP_nameinchlinux_1:# python_and_linux_1.append (p_name) #print (python_and_linux_1) p_s=Set(python_1) l_s=Set(linux_1) print (p_s,l_s) print (P_s.intersection (l_s)) print (p_s&l_s) #同上述函数

Seek and set

Python_1 = ['1','2','3']linux_1= ['1','2','4'] #python_and_linux_1= []# forP_nameinchpython_1:#ifP_nameinchlinux_1:# python_and_linux_1.append (p_name) #print (python_and_linux_1) p_s=Set(python_1) l_s=Set(linux_1) print (p_s,l_s) print (P_s.union (l_s)) print (p_s|l_s)

Finding the difference set

Python_1 = ['1','2','3']linux_1= ['1','2','4'] #python_and_linux_1= []# forP_nameinchpython_1:#ifP_nameinchlinux_1:# python_and_linux_1.append (p_name) #print (python_and_linux_1) p_s=Set(python_1) l_s=Set(linux_1) print (p_s,l_s)

Print (P_s.difference (l_s))

print (p_s-l_s)

Set other methods

Cross complement set-intersection

Python_1 = ['1','2','3']linux_1= ['1','2','4'] #python_and_linux_1= []# forP_nameinchpython_1:#ifP_nameinchlinux_1:# python_and_linux_1.append (p_name) #print (python_and_linux_1) p_s=Set(python_1) l_s=Set(linux_1) print (p_s,l_s) print (P_s.symmetric_difference (l_s)) print (p_s^l_s)

Difference_update

p_s=p_s-l_s

==p_s.difference_update (l_s)

Similarly intersection_update

Isdisjoint Two collections no common elements

Returns a Boolean value

Issubset,s1 contained in S2, subset

Equivalent to S1<S2

Returns a Boolean value

 python_1 = [ " 1  , "  2  , "  3   " ]linux_1  = [ " 1  , "  2  , "  4   " ]p_s  = set   ( python_1) l_s  = set   (linux_1) print (p_s,l_s ) Print (P_s.issubset (l_s))  

Issuperset, Parent set

Update updates element, add only adds one element

s1={1,2}s2={1,2,3,4}s1.update (s2) print (S1) 

Python set relational operation intersection, Difference DAY14