Here's a pure lambda implementation, which takes w, x, and y as arguments to a top-level lambda:

>>> (lambda w,x,y: (lambda s: map(lambda v: 0 if v in s else v, y))(set(range(w*w,x,w))))(2,10,range(10))
[0, 1, 2, 3, 0, 5, 0, 7, 0, 9]
>>> 

Note that this avoids the use of __setitem__.

Since a for loop is a statement (as is print, in Python 2.x), you cannot include it in a lambda expression. Instead, you need to use the write method on sys.stdout along with the join method.

x = lambda x: sys.stdout.write("\n".join(x) + "\n")

This question touches a very stinking part of the "famous" and "obvious" Python syntax - what takes precedence, the lambda, or the for of list comprehension.

I don't think the purpose of the OP was to generate a list of squares from 0 to 9. If that was the case, we could give even more solutions:

squares = []
for x in range(10): squares.append(x*x)

• this is the good ol' way of imperative syntax.

But it's not the point. The point is W(hy)TF is this ambiguous expression so counter-intuitive? And I have an idiotic case for you at the end, so don't dismiss my answer too early (I had it on a job interview).

So, the OP's comprehension returned a list of lambdas:

[(lambda x: x*x) for x in range(10)]

This is of course just 10 different copies of the squaring function, see:

>>> [lambda x: x*x for _ in range(3)]
[<function <lambda> at 0x00000000023AD438>, <function <lambda> at 0x00000000023AD4A8>, <function <lambda> at 0x00000000023AD3C8>]

Note the memory addresses of the lambdas - they are all different!

You could of course have a more "optimal" (haha) version of this expression:

>>> [lambda x: x*x] * 3
[<function <lambda> at 0x00000000023AD2E8>, <function <lambda> at 0x00000000023AD2E8>, <function <lambda> at 0x00000000023AD2E8>]

See? 3 time the same lambda.

Please note, that I used _ as the for variable. It has nothing to do with the x in the lambda (it is overshadowed lexically!). Get it?

I'm leaving out the discussion, why the syntax precedence is not so, that it all meant:

[lambda x: (x*x for x in range(10))]

which could be: [[0, 1, 4, ..., 81]], or [(0, 1, 4, ..., 81)], or which I find most logical, this would be a list of 1 element - a generator returning the values. It is just not the case, the language doesn't work this way.

BUT What, If...

What if you DON'T overshadow the for variable, AND use it in your lambdas???

Well, then crap happens. Look at this:

[lambda x: x * i for i in range(4)]

this means of course:

[(lambda x: x * i) for i in range(4)]

BUT it DOESN'T mean:

[(lambda x: x * 0), (lambda x: x * 1), ... (lambda x: x * 3)]

This is just crazy!

The lambdas in the list comprehension are a closure over the scope of this comprehension. A lexical closure, so they refer to the i via reference, and not its value when they were evaluated!

So, this expression:

[(lambda x: x * i) for i in range(4)]

IS roughly EQUIVALENT to:

[(lambda x: x * 3), (lambda x: x * 3), ... (lambda x: x * 3)]

I'm sure we could see more here using a python decompiler (by which I mean e.g. the dis module), but for Python-VM-agnostic discussion this is enough. So much for the job interview question.

Now, how to make a list of multiplier lambdas, which really multiply by consecutive integers? Well, similarly to the accepted answer, we need to break the direct tie to i by wrapping it in another lambda, which is getting called inside the list comprehension expression:

Before:

>>> a = [(lambda x: x * i) for i in (1, 2)]
>>> a1
2
>>> a0
2

After:

>>> a = [(lambda y: (lambda x: y * x))(i) for i in (1, 2)]
>>> a1
2
>>> a0
1

(I had the outer lambda variable also = i, but I decided this is the clearer solution - I introduced y so that we can all see which witch is which).

Edit 2019-08-30:

Following a suggestion by @josoler, which is also present in an answer by @sheridp - the value of the list comprehension "loop variable" can be "embedded" inside an object - the key is for it to be accessed at the right time. The section "After" above does it by wrapping it in another lambda and calling it immediately with the current value of i. Another way (a little bit easier to read - it produces no 'WAT' effect) is to store the value of i inside a partial object, and have the "inner" (original) lambda take it as an argument (passed supplied by the partial object at the time of the call), i.e.:

After 2:

>>> from functools import partial
>>> a = [partial(lambda y, x: y * x, i) for i in (1, 2)]
>>> a0, a1
(2, 4)

Great, but there is still a little twist for you! Let's say we wan't to make it easier on the code reader, and pass the factor by name (as a keyword argument to partial). Let's do some renaming:

After 2.5:

>>> a = [partial(lambda coef, x: coef * x, coef=i) for i in (1, 2)]
>>> a0
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: <lambda>() got multiple values for argument 'coef'

WAT?

>>> a[0]()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: <lambda>() missing 1 required positional argument: 'x'

Wait... We're changing the number of arguments by 1, and going from "too many" to "too few"?

Well, it's not a real WAT, when we pass coef to partial in this way, it becomes a keyword argument, so it must come after the positional x argument, like so:

After 3:

>>> a = [partial(lambda x, coef: coef * x, coef=i) for i in (1, 2)]
>>> a0, a1
(2, 4)

I would prefer the last version over the nested lambda, but to each their own...

Edit 2020-08-18:

Thanks to commenter dasWesen, I found out that this stuff is covered in the Python documentation: https://docs.python.org/3.4/faq/programming.html#why-do-lambdas-defined-in-a-loop-with-different-values-all-return-the-same-result - it deals with loops instead of list comprehensions, but the idea is the same - global or nonlocal variable access in the lambda function. There's even a solution - using default argument values (like for any function):

>>> a = [lambda x, coef=i: coef * x for i in (1, 2)]
>>> a0, a1
(2, 4)

This way the coef value is bound to the value of i at the time of function definition (see James Powell's talk "Top To Down, Left To Right", which also explains why mutable default values are shunned).

There's no need to use a lambda in this case, a simple for loop will do:

my_test  = 'test_name_dup'  
testlist = ['test_name', 'test_name_dup','test_name_dup_1', 'test_name_dup_3']

for i in xrange(1, len(testlist)):
    if my_test + '_' + str(i) not in testlist:
        break

print my_test + '_' + str(i)
> test_name_dup_2

If you really, really want to use a lambda for this problem, you'll also have to learn about itertools, iterators, filters, etc. I'm gonna build on thg435's answer, writing it in a more idiomatic fashion and explaining it:

import itertools as it

iterator = it.dropwhile(
    lambda n: '{0}_{1}'.format(my_test, n) in testlist,
    it.count(1))

print my_test + '_' + str(iterator.next())
> test_name_dup_2

The key to understanding the above solution lies in the dropwhile() procedure. It takes two parameters: a predicate and an iterable, and returns an iterator that drops elements from the iterable as long as the predicate is true; afterwards, returns every element.

For the iterable, I'm passing count(1), an iterator that produces an infinite number of integers starting from 1.

Then dropwhile() starts to consume the integers until the predicate is false; this is a good opportunity for passing an in-line defined function - and here's our lambda. It receives each generated integer in turn, checking to see if the string test_name_dup_# is present in the list.

When the predicate returns false, dropwhile() returns and we can retrieve the value that made it stop by calling next() on it.

This is due to the point at which d is being bound. The lambda functions all point at the variable d rather than the current value of it, so when you update d in the next iteration, this update is seen across all your functions.

For a simpler example:

funcs = []
for x in [1,2,3]:
  funcs.append(lambda: x)

for f in funcs:
  print f()

# output:
3
3
3

You can get around this by adding an additional function, like so:

def makeFunc(x):
  return lambda: x

funcs = []
for x in [1,2,3]:
  funcs.append(makeFunc(x))

for f in funcs:
  print f()

# output:
1
2
3

You can also fix the scoping inside the lambda expression

lambda bound_x=x: bound_x

However in general this is not good practice as you have changed the signature of your function.