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Storing Objects in a List in Python

If an object exists with variable .X
randomData.X
is the created object. If multiple objects are stored in a list and can be accessed via
randomList[3].X
Is there a way to pull all values of X from the list without looping through every object in the list as below:
for x i range(0,10)
randomList[x].X

You are probably looking for a list comprehension.
[obj.X for obj in randomList]
This produces a list with all properties X of every object in your list of objects.
Keep in mind that you can't get away from looping over the list. This is just syntactic sugar for the same loop as before.

Just in case you're looking for maximum efficiency on larger lists, an alternative to the list comprehension in this case is using map + operator.attrgetter. You can either loop over the map directly:
from operator import attrgetter
for X in map(attrgetter('X'), randomList):
which involves no temporary list (map lazily pulls items on demand in Python 3), or if you really need the list, just wrap in the list constructor or use list unpacking to run it out eagerly:
Xs = list(map(attrgetter('X'), randomList))
# or
Xs = [*map(attrgetter('X'), randomList)]
For small input lists, this will be slower than the list comprehension (it has a slightly higher setup overhead), but for medium to large inputs, it will be faster (the per-item overhead is slightly lower, as it involves no per-item bytecode execution).
To be clear, it's still going to have to loop over the list. There is no magic way to get the attributes of every item in a list without looping over it; you could go to extreme lengths to make views of the list that seamlessly read the attribute from the underlying list, but if you accessed every element of that view it would be equivalent to the loop in work required.

python - How to use map to call a method [duplicate]

In python 2, I used map to apply a function to several items, for instance, to remove all items matching a pattern:
map(os.remove,glob.glob("*.pyc"))
Of course I ignore the return code of os.remove, I just want all files to be deleted. It created a temp instance of a list for nothing, but it worked.
With Python 3, as map returns an iterator and not a list, the above code does nothing.
I found a workaround, since os.remove returns None, I use any to force iteration on the full list, without creating a list (better performance)
any(map(os.remove,glob.glob("*.pyc")))
But it seems a bit hazardous, specially when applying it to methods that return something. Another way to do that with a one-liner and not create an unnecessary list?

The change from map() (and many other functions from 2.7 to 3.x) returning a generator instead of a list is a memory saving technique. For most cases, there is no performance penalty to writing out the loop more formally (it may even be preferred for readability).
I would provide an example, but #vaultah nailed it in the comments: still a one-liner:
for x in glob.glob("*.pyc"): os.remove(x)

Set changed size during iteration

I'm new to python coming from a c++ background. I was just playing around with sets trying to calculate prime numbers and got a "Set changed size during iteration" error.
How internally does python know the set changed size during iteration?
Is it possible to do something similar in user defined objects?

The pythonic way to filter sets, lists or dicts is with list [or dict] expressions
your_filtered_set = set([elem for elem in original_set if condition(elem)])

It's trivial to do so with a user-defined object: just set a flag each time you modify the object, and have the iterator check that flag each time it tries to retrieve an item.
Generally, you should not modify a set while iterating over it, as you risk missing an item or getting the same item twice.

Sorting a list of lists with itertools imap in Python

I'm wondering what's happening when I execute this code and also if there is a better way to accomplish the same task. Is a list of lists being made in memory to preform the sort, and then bar is assigned to be an iterator of foo.values()? Or possibly foo.values() is sorted in the allotted dictionary memory space (seems unlikely)?
Imagine the first value in the list, the integer, refers to a line number in a file. I want to open the file and update only the lines referenced in the foo.values() lists with the rest of the data in the list (EG update line 1 with strings '123' and '097').
from itertools import imap
>>> foo = {'2134':[1, '123', '097'], '6543543':[3, '1'], '12315':[2, '454']}
>>> bar = imap([].sort(), foo.values())
Thanks~

First, you're passing [].sort(), which is just None, as the first argument to imap, meaning it's doing nothing at all. As the docs explain: "If function is set to None, then imap() returns the arguments as a tuple."
To pass a callable to a higher-order function like imap, you have to pass the callable itself, not call it and pass the result.
Plus, you don't want [].sort here; that's a callable with no arguments that just sorts an empty list, which is useless.
You probably wanted list.sort, the unbound method, which is a callable with one argument that will sort whatever list it's given.
So, if you did that, what would happen is that you'd creating an iterator that, if you iterated it, would generate a bunch of None values and, as a side effect, sort each list in foo.values(). No new lists would be created anywhere, because list.sort mutates the list in-place and returns None.
But since you don't ever iterate it anyway, it hardly matters what you put into imap; what it actually does is effectively nothing.
Generally, abusing map/imap/comprehensions/etc. for side-effects of the expressions is a bad idea. An iterator that generates useless values, but that you have to iterate anyway, is a recipe for confusion at best.
The simple thing to do here is to just use a loop:
for value in foo.values():
value.sort()
Or, instead of sorting in-place, generate new sorted values:
bar = imap(sorted, foo.values())
Now, as you iterate bar, each list will be sorted and given to you, so you can use it. If you iterate this, it will generate a sorted list in memory for each list… but only one will ever be alive at a time (unless you explicitly stash them somewhere).

Correct way to iterate twice over a list?

What is the correct way to perform multiple iteration over a container? From python documentation:
Iterator - A container object (such as a list) produces a fresh new
iterator each time you pass it to the iter() function or use it in a
for loop. Attempting this with an iterator will just return the same
exhausted iterator object used in the previous iteration pass, making
it appear like an empty container.
The intention of the protocol is that once an iterator’s next() method
raises StopIteration, it will continue to do so on subsequent calls.
Implementations that do not obey this property are deemed broken.
(This constraint was added in Python 2.3; in Python 2.2, various
iterators are broken according to this rule.)
If I have this code:
slist = [1,2,3,4]
rlist = reversed(slist)
list(rlist)
#[4,3,2,1]
tuple(rlist)
#()
What would be the easiest and most correct way to iterate over 'rlist' twice?

rlist = list(reversed(slist))
Then iterate as often as you want. This trick applies more generally; whenever you need to iterate over an iterator multiple times, turn it into a list. Here's a code snippet that I keep copy-pasting into different projects for exactly this purpose:
def tosequence(it):
"""Turn iterable into a sequence, avoiding a copy if possible."""
if not isinstance(it, collections.Sequence):
it = list(it)
return it
(Sequence is the abstract type of lists, tuples and many custom list-like objects.)

I wouldn't stored the list twice, if you can not combine it to iterate once, then I would
slist = [1,2,3,4]
for element in reversed(slist):
print element # do first iteration stuff
for element in reversed(slist):
print element # do second iteration stuff
Just think of the reversed() as setting up a reverse iterator on slist. The reversed is cheap. That being said, if you only ever need it reversed, I would reverse it and just have it stored like that.

What is the correct way to perform multiple iteration over a container?
Just do it twice in a row. No problem.
What would be the easiest and most correct way to iterate over 'rlist' twice?
See, the reason that isn't working for you is that rlist isn't "a container".
Notice how
list(slist) # another copy of the list
tuple(slist) # still works!
So, the simple solution is to just ensure you have an actual container of items if you need to iterate multiple times:
rlist = list(reversed(slist)) # we store the result of the first iteration
# and then that result can be iterated over multiple times.
If you really must not store the items, try itertools.tee. But note that you won't really avoid storing the items if you need to complete one full iteration before starting the next. In the general case, storage is really unavoidable under those restrictions.

Why don't you simply reverse the original list in-place (slist.reverse()), then iterate over it as many times as you wish, and finally reverse it again to obtain the original list once again?
If this doesn't work for you, the best solution for iterating over the list in reversed order is to create a new reverse iterator every time you need to iterate
for _ in xrange(as_many_times_as_i_wish_to_iterate_this_list_in_reverse_order):
for x in reversed(slist):
do_stuff(x)