Strings beginning with underscore not removed from lists
Reproducible as shown below:

Python 3.6.4 |Anaconda custom (64-bit)| (default, Jan 16 2018, 12:04:33) 
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> test = ["a","_a","b","_b"]
>>> for i in test: print(i)
... 
a
_a
b
_b
>>> for i in test: test.remove(i)
... 
>>> test
['_a', '_b']
>>> 

Is this a feature or a bug?
A search through the docs did not show any mention of this.

It's just the right behavior. You are modifying the list while iterating over it. It has no business of underscore.

>>> test = ['_a', 'a', '_b', 'b']
>>> for i in test: test.remove(i)
>>> test
['a', 'b']

You may refer to stackoverflow for an explanation, for example, https://stackoverflow.com/questions/6260089/strange-result-when-removing-item-from-a-list

In addition to Xiang Zhang's comments, this is neither a feature nor a bug, but a misunderstanding. This has nothing to do with strings or underscores:

py> L = [1, 2, 3, 4, 5]
py> for item in L:
...     L.remove(item)
...
py> L
[2, 4]


When you modify a list as you iterate over it, the results can be unexpected. Don't do it.

If you *must* modify a list that you are iterating over, you must do so backwards, so that you are only removing items from the end, not the beginning of the list:

py> L = [1, 2, 3, 4, 5]
py> for i in range(len(L)-1, -1, -1):
...     L.remove(L[i])
...
py> L
[]


But don't do that: it is nearly always must faster to make a copy of the list containing only the items you wish to keep, then assign back to the list using slicing. A list comprehension makes this an easy one-liner:

py> L = [1, 2, 3, 4, 5]
py> L[:] = [x for x in L if x > 4]
py> L
[5]