Question

What I wanted to do was to search in a list and remove a value .

So I wrote the following code

for x in range(10):
   if x in list1:
      list1.remove(x)

Does this function of order ~ (n^2) since first it looks for the value and then it deletes and pushes the rest of the values backwards ??

Also is there a way to turn this in order n by using try/except

try:
  for x in range(10):
    list1.remove(x)
except ValueError:
  # make it go back to next iteration 

Answer 1

This sounds like a job for filter():

>>> filter(lambda x: not x in (4, 5, 7), xrange(10))
[0, 1, 2, 3, 6, 8, 9]

Update: one more example where I construct a list using list comprehension:

>>> filter(lambda x: not x[0] in (4, 5, 7), [[a] for a in xrange(10)])
[[0], [1], [2], [3], [6], [8], [9]]