----Original Message-----
From: tutor-bounces+ramit.prasad=jpmorgan....@python.org
[mailto:tutor-bounces+ramit.prasad=jpmorgan....@python.org] On Behalf Of
Prasad, Ramit
Sent: Thursday, October 13, 2011 10:09 AM
To: tutor@python.org
Subject: Re: [Tutor] problem with using set
Approach:-
>>> a='hello'
>>> set(a)
set(['h', 'e', 'l', 'o']), so i am thinking somehow if i remove 'l' and i join
the string, i will get the desired output.
2.
>>> a='microsoft'
>>> set(a)
set(['c', 'f', 'i', 'm', 'o', 's', 'r', 't'])
>>> print ''.join(set(a))
cfimosrt
When i print "Hello", i get the output as "helo"(in same sequence) but when i
write "microsoft" i get this-"cfimosrt". So, it means i can't predict the
outcome of set(a)??
=========================================================================
Are you required to use set? If you are not, I think the following will be
easier.
>>> 'hello'.replace( 'l', '' )
'heo'
>>> 'microsoft'.replace( 'o', '' )
'micrsft'
If you require set, you can do:
>>> s = set("microsoft")
>>> s
set(['c', 'f', 'i', 'm', 'o', 's', 'r', 't'])
>>> s.remove('o')
>>> string = []
>>> for letter in 'microsoft':
... if letter in s:
... string.append( letter )
...
>>> ''.join( string )
'micrsft'
==============================================
After further thought, I think you might want to keep one of each character and
not remove them all. If so, see below.
>>> string = 'microsoft'
>>> dct = {}
>>> for letter in string:
... dct[ letter ] = string.count( letter )
...
>>> for letter, count in dct.iteritems():
... reversed_string = string[::-1] # reverse string to leave first
occurrence
... reversed_string = reversed_string.replace( letter, '', count-1 )
... string = reversed_string[::-1] # return it to normal
...
>>> string
'microsft'
If you do not care which duplicate character is left you can skip the lines
with [::-1] and it will leave the last one in the string.
Ramit
Ramit Prasad | JPMorgan Chase Investment Bank | Currencies Technology
712 Main Street | Houston, TX 77002
work phone: 713 - 216 - 5423
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