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Pythons generator expressions are late binding (see PEP 289 -- Generator Expressions) (what the other answers call "lazy"):

Early Binding versus Late Binding

After much discussion, it was decided that the first (outermost) for-expression [of the generator expression] should be evaluated immediately and that the remaining expressions be evaluated when the generator is executed.

[...]

However, Python takes a late binding approach to lambda expressions and has no precedent for automatic, early binding. It was felt that introducing a new paradigm would unnecessarily introduce complexity.

After exploring many possibilities, a consensus emerged that binding issues were hard to understand and that users should be strongly encouraged to use generator expressions inside functions that consume their arguments immediately. For more complex applications, full generator definitions are always superior in terms of being obvious about scope, lifetime, and binding.

That means it only evaluates the outermost for, so it actually binds the value with the name array in the "subexpression" in array. But when you iterate over the generator the if array.count call actually refers to what is currently named array.

Since it's actually a list not an array I changed the variable names in the rest of the answer to be more accurate.

So in your first case the list you iterate over and the list you count in will be different. It's as if you used:

list1 = [1, 2, 2, 4, 5]

list2 = [5, 6, 1, 2, 9]

f = (x for x in list1 if list2.count(x) == 2)

So you check for each element in list1 if it's present twice in list2.

You can easily verify this by modifying the second list:

>>> lst = [1, 2, 2, 4, 5]

>>> f = (x for x in lst if lst.count(x) == 2)

>>> lst = [1, 1, 2, 3, 4]

>>> list(f)

[1]

If it iterated over the second list the output should be [1, 1] but since it iterates over the first list (containing one 1) but checks the second list (which contains two 1s) the output is just a single 1.

Solution using a generator function

There are several possible solutions, I generally prefer not to use "generator expressions" if they aren't iterated over immediately. A simple generator function (as recommended by the PEP - see above) will suffice to make it work correctly:

def keep_only_duplicated_items(lst):

    for item in lst:

        if lst.count(item) == 2:

            yield item

And then use it like this:

lst = [1, 2, 2, 4, 5]

f = keep_only_duplicated_items(lst)

lst = [5, 6, 1, 2, 9]



>>> list(f)

[2, 2]

A better Solution using a generator function with a Counter

A better solution (avoiding the quadratic runtime behavior because you iterate over the whole array for each element in the array) would be to count (collections.Counter) the elements once and then do the lookup in constant time (resulting in linear time):

from collections import Counter



def keep_only_duplicated_items(lst):

    cnts = Counter(lst)

    for item in lst:

        if cnts[item] == 2:

            yield item

As others have mentioned Python generators are lazy. When this line is run:

f = (x for x in array if array.count(x) == 2) # Filters original

nothing actually happens yet. You've just declared how the generator function f will work. Array is not looked at yet. Then, you create a new array that replaces the first one, and finally when you call

print(list(f)) # Outputs filtered

the generator now needs the actual values and starts pulling them from the generator f. But at this point, array already refers to the second one, so you get an empty list.

If you need to reassign the list, and can't use a different variable to hold it, consider creating the list instead of a generator in the second line:

f = [x for x in array if array.count(x) == 2] # Filters original

...

print(f)

Others have already explained the root cause of the issue - the generator is binding to the name of the array local variable, rather than its value.

The most pythonic solution is definitely the list comprehension:

f = [x for x in array if array.count(x) == 2]

However, if there is some reason that you don't want to create a list, you can also force a scope close over array:

f = (lambda array=array: (x for x in array if array.count(x) == 2))()

What's happening here is that the lambda captures the reference to array at the time the line is run, ensuring that the generator sees the variable you expect, even if the variable is later redefined.

Note that this still binds to the variable (reference), not the value, so, for example, the following will print [2, 2, 4, 4]:

array = [1, 2, 2, 4, 5] # Original array



f = (lambda array=array: (x for x in array if array.count(x) == 2))() # Close over array

array.append(4)  # This *will* be captured



array = [5, 6, 1, 2, 9] # Updates original to something else



print(list(f)) # Outputs [2, 2, 4, 4]

This is a common pattern in some languages, but it's not very pythonic, so only really makes sense if there's a very good reason for not using the list comprehension (e.g., if array is very long, or is being used in a nested generator comprehension, and you're concerned about memory).

You are not using a generator correctly if this is the primary use of this code. Use a list comprehension instead of a generator comprehension. Just replace the parentheses with brackets. It evaluates to a list if you don't know.

array = [1, 2, 2, 4, 5]

f = [x for x in array if array.count(x) == 2]

array = [5, 6, 1, 2, 9]



print(f)

#[2, 2]

You are getting this response because of the nature of a generator. You're calling the generator when it't contents will evaluate to

Generators are lazy, they won't be evaluated until you iterate through them. In this case that's at the point you create the list with the generator as input, at the print.

The root cause of the problem is that generators are lazy; variables are evaluated each time:

>>> l = [1, 2, 2, 4, 5, 5, 5]

>>> filtered = (x for x in l if l.count(x) == 2)

>>> l = [1, 2, 4, 4, 5, 6, 6]

>>> list(filtered)

[4]

It iterates over the original list and evaluates the condition with the current list. In this case, 4 appeared twice in the new list, causing it to appear in the result. It only appears once in the result because it only appeared once in the original list. The 6s appear twice in the new list, but never appear in the old list and are hence never shown.

Full function introspection for the curious (the line with the comment is the important line):

>>> l = [1, 2, 2, 4, 5]

>>> filtered = (x for x in l if l.count(x) == 2)

>>> l = [1, 2, 4, 4, 5, 6, 6]

>>> list(filtered)

[4]

>>> def f(original, new, count):

    current = original

    filtered = (x for x in current if current.count(x) == count)

    current = new

    return list(filtered)



>>> from dis import dis

>>> dis(f)

  2           0 LOAD_FAST                0 (original)

              3 STORE_DEREF              1 (current)



  3           6 LOAD_CLOSURE             0 (count)

              9 LOAD_CLOSURE             1 (current)

             12 BUILD_TUPLE              2

             15 LOAD_CONST               1 (<code object <genexpr> at 0x02DD36B0, file "<pyshell#17>", line 3>)

             18 LOAD_CONST               2 ('f.<locals>.<genexpr>')

             21 MAKE_CLOSURE             0

             24 LOAD_DEREF               1 (current)

             27 GET_ITER

             28 CALL_FUNCTION            1 (1 positional, 0 keyword pair)

             31 STORE_FAST               3 (filtered)



  4          34 LOAD_FAST                1 (new)

             37 STORE_DEREF              1 (current)



  5          40 LOAD_GLOBAL              0 (list)

             43 LOAD_FAST                3 (filtered)

             46 CALL_FUNCTION            1 (1 positional, 0 keyword pair)

             49 RETURN_VALUE

>>> f.__code__.co_varnames

('original', 'new', 'count', 'filtered')

>>> f.__code__.co_cellvars

('count', 'current')

>>> f.__code__.co_consts

(None, <code object <genexpr> at 0x02DD36B0, file "<pyshell#17>", line 3>, 'f.<locals>.<genexpr>')

>>> f.__code__.co_consts[1]

<code object <genexpr> at 0x02DD36B0, file "<pyshell#17>", line 3>

>>> dis(f.__code__.co_consts[1])

  3           0 LOAD_FAST                0 (.0)

        >>    3 FOR_ITER                32 (to 38)

              6 STORE_FAST               1 (x)

              9 LOAD_DEREF               1 (current)  # This loads the current list every time, as opposed to loading a constant.

             12 LOAD_ATTR                0 (count)

             15 LOAD_FAST                1 (x)

             18 CALL_FUNCTION            1 (1 positional, 0 keyword pair)

             21 LOAD_DEREF               0 (count)

             24 COMPARE_OP               2 (==)

             27 POP_JUMP_IF_FALSE        3

             30 LOAD_FAST                1 (x)

             33 YIELD_VALUE

             34 POP_TOP

             35 JUMP_ABSOLUTE            3

        >>   38 LOAD_CONST               0 (None)

             41 RETURN_VALUE

>>> f.__code__.co_consts[1].co_consts

(None,)

To reiterate: The list to be iterated is only loaded once. Any closures in the condition or expression, however, are loaded from the enclosing scope each iteration. They are not stored in a constant.

The best solution for your problem would be to create a new variable referencing the original list and use that in your generator expression,.

Generator evaluation is "lazy" -- it doesn't get executed until you actualize it with a proper reference. With your line:

Look again at your output with the type of f: that object is a generator, not a sequence. It's waiting to be used, an iterator of sorts.

Your generator isn't evaluated until you start requiring values from it. At that point, it uses the available values at that point, not the point at which it was defined.

Code to "make it work"

That depends on what you mean by "make it work". If you want f to be a filtered list, then use a list, not a generator:

f = [x for x in array if array.count(x) == 2] # Filters original

Generators are lazy and your newly defined array is used when you exhaust your generator after redefining. Therefore, the output is correct. A quick fix is to use a list comprehension by replacing parentheses () by brackets .

Moving on to how better to write your logic, counting a value in a loop has quadratic complexity. For an algorithm that works in linear time, you can use collections.Counter to count values, and keep a copy of your original list:

from collections import Counter



array = [1, 2, 2, 4, 5]   # original array

counts = Counter(array)   # count each value in array

old_array = array.copy()  # make copy

array = [5, 6, 1, 2, 9]   # updates array



# order relevant

res = [x for x in old_array if counts[x] >= 2]

print(res)

# [2, 2]



# order irrelevant

from itertools import chain

res = list(chain.from_iterable([x]*count for x, count in counts.items() if count >= 2))

print(res)

# [2, 2]

Notice the second version doesn't even require old_array and is useful if there is no need to maintain ordering of values in your original array.