num_strings = ['1','21','53','84','50','66','7','38','9']
This one is pretty much easy but not at first. It took me a good half and hour to figure out that it was so easy.
num =  # This is the new list which will contain all integers instead of the strings n = len(num_strings) # This will give the length of the list for i in range(0,n): num.insert(i,int(a[i])) print num
Generally speaking, it can't, because objects don't really have names. Essentially, assignment always binds a name to a value; The same is true of def and class statements, but in that case the value is a callable. Consider the following code:
class A: pass B = A a = B() b = a print b <__main__.A instance at 016D07CC> print a <__main__.A instance at 016D07CC>
Arguably the class has a name: even though it is bound to two names and invoked through the name B the created instance is still reported as an instance of class A. However, it is impossible to say whether the instance's name is a or b, since both names are bound to the same value.
Generally speaking it should not be necessary for your code to "know the names" of particular values. Unless you are deliberately writing introspective programs, this is usually an indication that a change of approach might be beneficial.
In comp.lang.python, Fredrik Lundh once gave an excellent analogy in answer to this question:
The same way as you get the name of that cat you found on your porch: the cat (object) itself cannot tell you its name, and it doesn't really care -- so the only way to find out what it's called is to ask all your neighbours (namespaces) if it's their cat (object)...
....and don't be surprised if you'll find that it's known by many names, or no name at all!
To convert, e.g., the number 144 to the string '144', use the built-in function str(). If you want a hexadecimal or octal representation, use the built-in functions hex() or oct(). For fancy formatting, use the % operator on strings, e.g. "%04d" % 144 yields '0144' and "%.3f" % (1/3.0) yields '0.333'. See the library reference manual for details.
You can't, because strings are immutable. If you need an object with this ability, try converting the string to a list or use the array module:
>>> s = "Hello, world" >>> a = list(s) >>>print a ['H', 'e', 'l', 'l', 'o', ',', ' ', 'w', 'o', 'r', 'l', 'd'] >>> a[7:] = list("there!") >>>''.join(a) 'Hello, there!' >>> import array >>> a = array.array('c', s) >>> print a array('c', 'Hello, world') >>> a = 'y' ; print a array('c', 'yello world') >>> a.tostring() 'yello, world'