如何删除python列表中的每个项目之一

Question

I have a list that looks like this:

lst = ['p','p','p','p','p','m','m','m','n','n','n','n','d','d']

I want to remove one of each item. Currently my code looks like this:

for item in lst: 
    if (lst[-1] == lst[-2]) == True: 
        del(lst[-2])

That is if the last two items of the list are the same the second from the last should be deleted, but my code does not work.

Answer 1

You can make a set of the unique characters, loop over a copy of your list, and then remove items from the set while adding to an output list:

lst = ['p','p','p','p','p','m','m','m','n','n','n','n','d','d']

chars_to_remove = set(lst)
counter = len(chars_to_remove)
output = []

for item in lst[:]:
    if item in chars_to_remove:
        chars_to_remove.remove(item)
        continue
    else:
        output.append(item)

print(output)

Result:

['p', 'p', 'p', 'p', 'm', 'm', 'n', 'n', 'n', 'd']

Note : You still need to define what happens when there is only a single instance of a string in your list. (ie Does it get deleted as well?) In the above code, it will be deleted. But that can be changed like so, by adding another condition to the loop body:

Sample input : lst = ['p','p','p','p','p','m','m','m','q','n','n','n','n','d','d']

for item in lst[:]: if lst[:].count(item) == 1: output.append(item) continue elif item in chars_to_remove: chars_to_remove.remove(item) continue else: output.append(item)

Result: ['p', 'p', 'p', 'p', 'm', 'm', 'q', 'n', 'n', 'n', 'd']

Answer 2

You can also, use sum and groupby :

from itertools import groupby

lst = ['p', 'p', 'p', 'p', 'p', 'm', 'm', 'm', 'n', 'n', 'n', 'n', 'd', 'd']
final = sum((list(g)[:-1] for _, g in groupby(lst)), [])
print(final)

Output:

['p', 'p', 'p', 'p', 'm', 'm', 'n', 'n', 'n', 'd']

Answer 3

Assuming you want to remove single occurrences in the list, can be done in one line:

[lst.remove(c) for c in set(lst)]

This does not return the answer, but modifies your list in place, so lst is now trimmed.

Or wrapped into a potentially more useful function:

def remove_first_occurence(lst):
    l = lst.copy()
    [l.remove(c) for c in set(l)]
    return l

Answer 4

You could give this a shot

result = []

for _, u in groupby(lst):
    new_u      = list(u)
    last_index = max(1, len(new_u) - 1)

    result += new_u[:last_index]

Answer 5

It is not clear what your expected result it. Your code will not work because you are iterating a list while mutating it. Instead, iterate over a copy ( lst[:] ).

for item in lst[:]: 
    if (lst[-1] == lst[-2]): 
        del(lst[-2])

lst
# ['p', 'p', 'p', 'p', 'p', 'm', 'm', 'm', 'n', 'n', 'n', 'n', 'd']

However, you code still needs more to resolve:

I want to remove one of each item

Try this instead:

import itertools as it


lst = ['p','p','p','p','p','m','m','m','n','n','n','n','d','d']
list(it.chain.from_iterable((list(g)[:-1] for _, g in it.groupby(lst))))
# ['p', 'p', 'p', 'p', 'm', 'm', 'n', 'n', 'n', 'd']