Note this DOES NOT WORK in older versions of Python such as 2.2

hendry@bilbo:~/datastructuresproject$ python2.3

Python 2.3a2+ (#2, Mar 21 2003, 22:13:05)

[GCC 3.2.3 20030316 (Debian prerelease)] on linux2

Type "help", "copyright", "credits" or "license" for more information.

>>> string = 'hello'

>>> if 'll' in string:

...     print string



In older versions I have used: def contained(self, pattern, x): print "Pattern: %s, x: %s" % (pattern, x) if(x.count(pattern) > 0): return 1 else: return 0


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