#in.py
#Demo the in operator

#strings
s = "asdfqwerty yabba dabba doo"
if 'bba' in s:    #True
    print("yes")  
else:
    print("no")
    
if 'scooby' in s:  #False
    print("yes")
else:
    print("no")

if '' in s:  #True.  every string has the empty string in it between every chararacter
    print("yes", s.find(''), s.count(''))   #index of where first found. how many
                                            #a string of length n has n+1 empty strings in it
else:
    print("no")  

if '' in '':  #True.  EVERY string has the empty string in it
    print("yes")
else:
    print("no")

#print(s.split(''))  #split a string at each empty string  CAN'T

# 'in' is boolean.    .find(substring) returns index where found, -1 if not found


#Lists
lst = [10,30,40,"hello", 3.1415]
if 30 in lst:   #True
    print("yes")
else:
    print("no")

if "hello" in lst:   #True
    print("yes")
else:
    print("no")

if "hel" in lst:   #False
    print("yes")
else:
    print("no")

if "" in lst:   #False. An empty string could be an item in the list though.
    print("yes")
else:
    print("no")

#list does not have a .find() method
# .index(val)  returns index of where val is or is Error

x = 40
if x in lst:
    print("found. at index:",lst.index(x))  #avoid the error of index when not found
x = 50
if x in lst:
    print("found. at index:",lst.index(x))  
else:
    print("not found")   #avoid the error of index when not found