1) Almost-English style:

Test for presence using the in operator, then apply the remove method.

if thing in some_list: some_list.remove(thing)

The removemethod will remove only the first occurrence of thing, in order to remove all occurrences you can use while instead of if.

while thing in some_list: some_list.remove(thing)    

• Simple enough, probably my choice.for small lists (can't resist one-liners)

2) Duck-typed, EAFP style:

This shoot-first-ask-questions-last attitude is common in Python. Instead of testing in advance if the object is suitable, just carry out the operation and catch relevant Exceptions:

try:
    some_list.remove(thing)
except ValueError:
    pass # or scream: thing not in some_list!
except AttributeError:
    call_security("some_list not quacking like a list!")

Off course the second except clause in the example above is not only of questionable humor but totally unnecessary (the point was to illustrate duck-typing for people not familiar with the concept).

If you expect multiple occurrences of thing:

while True:
    try:
        some_list.remove(thing)
    except ValueError:
        break

• a little verbose for this specific use case, but very idiomatic in Python.
• this performs better than #1
• PEP 463 proposed a shorter syntax for try/except simple usage that would be handy here, but it was not approved.

However, with contextlib's suppress() contextmanager (introduced in python 3.4) the above code can be simplified to this:

with suppress(ValueError, AttributeError):
    some_list.remove(thing)

Again, if you expect multiple occurrences of thing:

with suppress(ValueError):
    while True:
        some_list.remove(thing)

3) Functional style:

Around 1993, Python got lambda, reduce(), filter() and map(), courtesy of a Lisp hacker who missed them and submitted working patches*. You can use filter to remove elements from the list:

is_not_thing = lambda x: x is not thing
cleaned_list = filter(is_not_thing, some_list)

There is a shortcut that may be useful for your case: if you want to filter out empty items (in fact items where bool(item) == False, like None, zero, empty strings or other empty collections), you can pass None as the first argument:

cleaned_list = filter(None, some_list)

• [update]: in Python 2.x, filter(function, iterable) used to be equivalent to [item for item in iterable if function(item)] (or [item for item in iterable if item] if the first argument is None); in Python 3.x, it is now equivalent to (item for item in iterable if function(item)). The subtle difference is that filter used to return a list, now it works like a generator expression - this is OK if you are only iterating over the cleaned list and discarding it, but if you really need a list, you have to enclose the filter() call with the list() constructor.
• *These Lispy flavored constructs are considered a little alien in Python. Around 2005, Guido was even talking about dropping filter - along with companions map and reduce (they are not gone yet but reduce was moved into the functools module, which is worth a look if you like high order functions).

4) Mathematical style:

List comprehensions became the preferred style for list manipulation in Python since introduced in version 2.0 by PEP 202. The rationale behind it is that List comprehensions provide a more concise way to create lists in situations where map() and filter() and/or nested loops would currently be used.

cleaned_list = [ x for x in some_list if x is not thing ]

Generator expressions were introduced in version 2.4 by PEP 289. A generator expression is better for situations where you don't really need (or want) to have a full list created in memory - like when you just want to iterate over the elements one at a time. If you are only iterating over the list, you can think of a generator expression as a lazy evaluated list comprehension:

for item in (x for x in some_list if x is not thing):
    do_your_thing_with(item)

• See this Python history blog post by GvR.
• This syntax is inspired by the set-builder notation in math.
• Python 3 has also set and dict comprehensions.

Notes

1. you may want to use the inequality operator != instead of is not (the difference is important)
2. for critics of methods implying a list copy: contrary to popular belief, generator expressions are not always more efficient than list comprehensions - please profile before complaining

if [items] in list checks that: a list containing the list of items is an element of the list. That is, you are asking: is [[1, 2, 3]] a member of the list? Probably not.

What you want to do is iterate over the element of items_to_remove and do what you did

for item in items_to_remove:
    if item in list:
        list.remove(item)

To remove the first occurrence of an element, use list.remove:

>>> xs = ['a', 'b', 'c', 'd']
>>> xs.remove('b')
>>> print(xs)
['a', 'c', 'd']

To remove all occurrences of an element, use a list comprehension:

>>> xs = ['a', 'b', 'c', 'd', 'b', 'b', 'b', 'b']
>>> xs = [x for x in xs if x != 'b']
>>> print(xs)
['a', 'c', 'd']

If new_dict is "definitely" in my_list, then my_list.remove(new_dict) should do the trick (i.e., no need for the if new_dict in my_list, that just slows it down).