for line3 in myips matching too longer matches.
### CODE: elif line1.rstrip(‘\n’) in line2.strip(‘\n’): for line3 in myips: print “###” print “line1 is %s” % line1.rstrip(‘\n’) print “line2 is %s” % line2.strip(‘\n’) ### OUTPUT: line1 is 10.10.168.2 line2 is - address: 10.10.168.27 # myhost ### I think the problem is here: line1.rstrip(‘\n’) in line2.strip(‘\n’): I want it to match only 10.10.168.2 AND 10.10.168.2: NOT match 10.10.168.2[0-9] If someone out there knows a simple solution. I would love to see it. Thanks in advance. crzzy1 -- https://mail.python.org/mailman/listinfo/python-list
Python, convert an integer into an index?
(How do I make it into an index? ) Preferably something fairly easy to understand as I am new at this. results = 134523 #(Integer) Desired: results = [1, 2, 3, 4, 5, 2, 3] #(INDEX) Somehow I see ways to convert index to list to int, but not back again. Thanks, crzzy1 -- https://mail.python.org/mailman/listinfo/python-list
Re: Python, convert an integer into an index?
Fantastic! I started a course, but I have been having issues with string/index/list/integer conversions and manipulations. This variation wasn't in any of my texts or class exercises either. Your way was both simple enough to understand and very informative! Thanks. On Tue, Sep 22, 2015 at 6:21 PM, Laura Creighton wrote: > In a message of Tue, 22 Sep 2015 14:43:55 -0700, Chris Roberts writes: > > > > > >(How do I make it into an index? ) > >Preferably something fairly easy to understand as I am new at this. > > > >results = 134523 #(Integer) > > > >Desired: > >results = [1, 2, 3, 4, 5, 2, 3] #(INDEX) > > > >Somehow I see ways to convert index to list to int, but not back again. > > > >Thanks, > >crzzy1 > > You need to convert your results into a string first. > > result_int=1234523 > result_list=[] > > for digit in str(result_int): > result_list.append(int(digit)) > > digit will be assigned to successive 1 character long strings. Since > you wanted a list of integers, you have to convert it back. > > If you are learning python you may be interested in the tutor mailing > list. https://mail.python.org/mailman/listinfo/tutor > > Laura > > > -- https://mail.python.org/mailman/listinfo/python-list
Python... feeding an instance as an argument into a new instance.
Perhaps someone here could help me to get this into perspective. Somehow when we start to feed an instance as the argument in a new instance. my head explodes.. in this case... a = Foo() b = Bar(a) So... a is a 'Foo instance' with properties and methods. b is a 'Bar instance' Since b is using the "a" instance as an argument?? b=Bar(a) has what?? Is there an easier way to put into perspective how an instance of one class is used as the argument into the instance of another class? Code below: class Foo(object): bar = "Bar" # Class attribute. def __init__(self): ##^ The first variable is the class instance in methods. ## This is called "self" by convention, but could be any name you want. # ^ double underscore (dunder) methods are usually special. This one # gets called immediately after a new instance is created. self.variable = "Foo" # instance attribute. print self.variable, self.bar # <---self.bar references class attribute self.bar = " Bar is now Baz" # <---self.bar is now an instance attribute print self.variable, self.bar #variable will be a property of "a" and self is the bound variable of a. def method(self, arg1, arg2): # This method has arguments. You would call it like this: instance.method(1, 2) print "in method (args):", arg1, arg2 print "in method (attributes):", self.variable, self.bar a = Foo() # this calls __init__ (indirectly), output: #a is Foo, a has a method called method, and properties such as "variable and bar". # Foo bar # Foo Bar is now Baz print a.variable # Foo a.variable = "bar" a.method(1, 2) # output: # in method (args): 1 2 # in method (attributes): bar Bar is now Baz Foo.method(a, 1, 2) # <--- Same as a.method(1, 2). This makes it a little more explicit what the argument "self" actually is. print "##Above is all the a Foo object, and below is the b Bar object" class Bar(object): def __init__(self, arg): self.arg = arg self.Foo = Foo() #this calls and runs the Foo class again. b = Bar(a) #b is a Bar object, a is a Foo object with properties and methods. then Bar() calls the Foo class again. b.arg.variable = "something" print a.variable # something !(I don't know why "a" prints "something" here.) print b.Foo.variable # Foo OUTPUT: Foo Bar Foo Bar is now Baz Foo in method (args): 1 2 in method (attributes): bar Bar is now Baz in method (args): 1 2 in method (attributes): bar Bar is now Baz ##Above is all the a Foo object, and below is the b Bar object Foo Bar Foo Bar is now Baz something Foo Thanks in advance... crzzy1 ... -- https://mail.python.org/mailman/listinfo/python-list
Re: Python... feeding an instance as an argument into a new instance.
On Saturday, September 2, 2017 at 6:34:59 PM UTC-4, Chris Roberts wrote: > Perhaps someone here could help me to get this into perspective. > Somehow when we start to feed an instance as the argument in a new instance. > my head explodes.. > in this case... > a = Foo() > b = Bar(a) > So... > a is a 'Foo instance' with properties and methods. > b is a 'Bar instance' > Since b is using the "a" instance as an argument?? > b=Bar(a) has what?? > > Is there an easier way to put into perspective how an instance of one class > is used as the argument into the instance of another class? > > Code below: > > class Foo(object): > bar = "Bar" # Class attribute. > > def __init__(self): > ##^ The first variable is the class instance in methods. > ## This is called "self" by convention, but could be any > name you want. > # ^ double underscore (dunder) methods are usually special. This one > # gets called immediately after a new instance is created. > > self.variable = "Foo" # instance attribute. > print self.variable, self.bar # <---self.bar references class > attribute > self.bar = " Bar is now Baz" # <---self.bar is now an instance > attribute > print self.variable, self.bar #variable will be a property of "a" and > self is the bound variable of a. > > def method(self, arg1, arg2): > # This method has arguments. You would call it like this: > instance.method(1, 2) > print "in method (args):", arg1, arg2 > print "in method (attributes):", self.variable, self.bar > > > a = Foo() # this calls __init__ (indirectly), output: > #a is Foo, a has a method called method, and properties such as "variable and > bar". > # Foo bar > # Foo Bar is now Baz > print a.variable # Foo > a.variable = "bar" > a.method(1, 2) # output: > # in method (args): 1 2 > # in method (attributes): bar Bar is now Baz > Foo.method(a, 1, >2) # <--- Same as a.method(1, 2). This makes it a little more > explicit what the argument "self" actually is. > print "##Above is all the a Foo object, and below is the b Bar object" > > class Bar(object): > def __init__(self, arg): > self.arg = arg > self.Foo = Foo() #this calls and runs the Foo class again. > > > b = Bar(a) #b is a Bar object, a is a Foo object with properties and > methods. then Bar() calls the Foo class again. > b.arg.variable = "something" > print a.variable # something !(I don't know why "a" prints "something" here.) > print b.Foo.variable # Foo > > > OUTPUT: > Foo Bar > Foo Bar is now Baz > Foo > in method (args): 1 2 > in method (attributes): bar Bar is now Baz > in method (args): 1 2 > in method (attributes): bar Bar is now Baz > ##Above is all the a Foo object, and below is the b Bar object > Foo Bar > Foo Bar is now Baz > something > Foo > > Thanks in advance... crzzy1 ... Steve and Irv, I see what you mean... I picked up on this example online and then tried to work it out on my own, but after seeing your replies, and looking further into cars examples I found how much more intuitive that makes it... Anyways, now I know where to better redirect my studies and energies. My work is sending me to a semi advanced (Core Python) class, and I am not ready, so trying to work on all my weak spots. Thanks -- https://mail.python.org/mailman/listinfo/python-list
