Why is this code not working?

for a in massiv: for b in massiv: if len(a)> len(b): del b elif len(a)<len(b): del a

6 answers 6

The code does not work, because the del instruction does not delete the element, but decreases the number of references to it by 1, del a means that you need to reduce the number of references to the element of the list referenced by a , but since in the list there is a link to this item, it is not deleted anywhere.

    Strange logic in your script =). But still.

    During the iteration over arrays in python, the removal of elements is not allowed. To do this, either make a copy of the list - massiv [:], or iterate the old-fashioned way - by indices. In the case of indexes, you get access to the elements of the original array.

     massiv = ['12', '123', '1234'] for i in range(len(massiv) - 1): for j in range(len(massiv) - 1): if len(massiv[i]) > len(massiv[j]): del massiv[j] elif len(massiv[i]) < len(massiv[j]): del massiv[i] print massiv # => ['1234']

    maxlen= max([len(i) for i in massiv]) massiv=filter(lambda x: len(x)=maxlen, massiv)

       for a in massiv: for b in massiv: if len(a)> len(b): massiv.remove(b) elif len(a)<len(b): massiv.remove(a)

      There are more letters, but it became clear that where we are from and that, in general, this is a more correct approach ... but in general, based on the description of the topstarter, nothing is clear at all except that something doesn’t work for him ...

      PS: write more details of the problem and problems;)

      • one
        Risky code. .remove (a) removes the first occurrence of a. - alexlz 1:16 pm
      • for a in massiv and will give the first oncoming element .. the risk is zero if you do not allow semantic errors))) - Alexander Molofeev
      • for a in massiv - enumeration of the list. And here the "first comer"? And the first, and second, and third. Just for different values ​​of massiv, the sequence of deletions will be quite bizarre. Perhaps correct, but the person who will force me to understand this, I will hardly be treated as a friend. Code - delete all items with a length less than the maximum. Maybe it works, but well, it nafig. - alexlz

      To remove an item from the lst list if its index i is known:

       del lst[i]

      To remove all items except the item with the longest:

       lst[:] = [max(lst, key=len)]

      An idiomatic way to remove items from the list that satisfy a given condition, to collect the items that should be left and replace all the items in the list with them:

       lst[:] = [item for item in lst if keep(item)]
      • why not lst[:] = filter(keep, lst) - Andrio Skur
      • @AndrioSkur: if you already have a function, you can use filter() , but more often the expression is of the following type: if item and item[0] - in this case it is better to use list comprehension instead of filter + lambda. - jfs
      • First, is it a tuple or a static array?
      • Secondly, a and b are not elements, it is just an instant presentation.

      It should be like this:

       for a in massiv: for b in massiv: if len(a)> len(b): del massiv[b] elif len(a)<len(b): del massiv[a]
      • one
        Thanks to the moderators for turning the unformed kaku into a beautiful candy - Daniel Kolyasnikov
      • 2
        Unfortunately, the code from the example is incorrect. a and b are not indices of the massiv list - alexlz
      • alexlz is right: if a or b are not numbers, we will get such a nuisance as: Type Error: list indices must be int not string or what kind of data do you have there. and if there is all the same number then, we immediately get the impossibility of getting len (integer), but even if you get in some unknown way you delete the element by the index, not the value as the topic task assumes .. what this index may simply not exist in the array from here we get list index out of range read about variable variables in python - Alexander Molofeev