I have two strings like s1='fly,dream';s2='dream,fly'
I want the s1 equals to s2.
The code I tried is:
def Isequal(m,n):
s1=m.split(',') s2=n.split(',') s1.sort() s2.sort()
if s1 == s2:
print 'Equal'
else:
print s1,s2
Note:s1 may be equal to s2. Then
def Isequal(m,n):
s1=m.split(',')
s2=n.split(',')
if s1 == s2.reverse() || s1 == s2:
print 'Equal'
else:
print s1,s2
Is this code right? I there something to improve?
Your code splits the two strings by , (which returns a list) and calls the sort method on the list. Since the two substrings are identical, sorting the list of the substrings results in equal lists. The best way to know what is happening is printing the stuff out. See the results.
>>> s1 = 'fly,dream'
>>> s2 = 'dream,fly'
>>> s1 = s1.split(',')
>>> s1
['fly', 'dream']
>>> s2 = s2.split(',')
>>> s2
['dream', 'fly']
>>> s1.sort()
>>> s1
['dream', 'fly']
>>> s2.sort()
>>> s2
['dream', 'fly']
>>> s1 == s2
True
If you want to check that the two strings consist of the same substrings, use sets, like follows :
>>> varOne = set(s1.split(','))
>>> varTwo = set(s2.split(','))
>>> varOne == varTwo
True
Beware that sets only allow unique items, so fly,dream,fly and dream,dream,fly will result in True here.
Set would be more elegant here:
def Isequal(m, n):
s1 = set(m.split(','))
s2 = set(n.split(','))
if s1 == s2:
print 'Equal'
else:
print s1, s2
and should be more efficient too.
You probably don't want to use sort() to flip a list. The sorting that takes place entirely depends on the string (it varies on the first letter of each string). You can use .reverse to reverse a list:
def Isequal(m,n):
m = m.split(',')
m.reverse()
if m == n.split(','):
print "Equal"
else:
print m, n
If you want to sort the list in place, you can always do .reverse() instead of .sort()
If reversing the list was just an example in your question, and your strings would actually have more items when you split them, you can use sets:
def Isequal(m,n):
if not set(m.split(',')).symmetric_difference(n.split(',')):
print "Equal"
else:
print m, n
By the way, Sparse is better than dense.. The semi-colons are rather... ugly.
not set(m.split(',')).symmetric_diffeence(n.split(','))
reversed return a reverse iterator?
s1to equals2, you can just dos1 = s2. Then they will be equal. (I think you wanted to know if the order of the words, separated by commas, of each string is the inverse of the other.)