Removing items from unnamed lists in Python

Question

Why doesn't the following statement return a list without 'item' in Python?

 list(something_convertible_to_list).remove('item') 

?

I would like to use the above construction to avoid explicitly specifying a list for the sole purpose of passing it to a function, that is:

 operate_on_list(list(something_convertible_to_list).remove('item')) def operate_on_list(my_list): # do_something print my_list return 

Is this possible in Python?

Answer 1

In python, inline methods that work in place return None to make it absolutely transparent that they mutated the object.

Of course, you can ignore this convention and write your own wrapper:

 def my_remove(lst,what): lst.remove(what) return lst 

But I would not recommend it.

As a side note, if you want to do something like:

 list(something_convertible_to_list).remove('item') 

but return the list, the following may be useful enough to be useful:

 [x for x in something_iterable if x != 'item'] 

And this returns the list, but where list.remove removes only 1 'item' , this will lead to the creation of a new list without the 'item' in it.

Answer 2

You can try something like:

 my_list[:my_list.index('item')]+my_list[my_list.index('item')+1:] 

although you have two requests.

or

 [item for item in my_sequence if item != 'item'] 

The first one will remove the first 'item' from the list, and the second one will delete all the 'item' .