Python:使用渐进编号重命名列表中的重复项，但不对列表进行排序 [英] Python: Rename duplicates in list with progressive numbers without sorting list

问题描述

给出如下列表:

mylist = ["name", "state", "name", "city", "name", "zip", "zip"]

我想通过添加数字来重命名重复项以得到以下结果:

I would like to rename the duplicates by appending a number to get the following result:

mylist = ["name1", "state", "name2", "city", "name3", "zip1", "zip2"]

我不想更改原始列表的顺序.针对此相关的堆栈溢出问题提出的解决方案建议对该列表进行排序，而我不想这样做.

I do not want to change the order of the original list. The solutions suggested for this related Stack Overflow question sorts the list, which I do not want to do.

推荐答案

这就是我的方法.因为人们似乎喜欢这个答案，所以我把它写成了一个更通用的效用函数.

This is how I would do it. I wrote this into a more generalized utility function since people seem to like this answer.

mylist = ["name", "state", "name", "city", "name", "zip", "zip"]
check = ["name1", "state", "name2", "city", "name3", "zip1", "zip2"]
copy = mylist[:]  # so we will only mutate the copy in case of failure

from collections import Counter # Counter counts the number of occurrences of each item
from itertools import tee, count

def uniquify(seq, suffs = count(1)):
    """Make all the items unique by adding a suffix (1, 2, etc).

    `seq` is mutable sequence of strings.
    `suffs` is an optional alternative suffix iterable.
    """
    not_unique = [k for k,v in Counter(seq).items() if v>1] # so we have: ['name', 'zip']
    # suffix generator dict - e.g., {'name': <my_gen>, 'zip': <my_gen>}
    suff_gens = dict(zip(not_unique, tee(suffs, len(not_unique))))  
    for idx,s in enumerate(seq):
        try:
            suffix = str(next(suff_gens[s]))
        except KeyError:
            # s was unique
            continue
        else:
            seq[idx] += suffix

uniquify(copy)
assert copy==check  # raise an error if we failed
mylist = copy  # success

如果您想在每次计数之前添加下划线，则可以执行以下操作:

If you wanted to append an underscore before each count, you could do something like this:

>>> mylist = ["name", "state", "name", "city", "name", "zip", "zip"]
>>> uniquify(mylist, (f'_{x!s}' for x in range(1, 100)))
>>> mylist
['name_1', 'state', 'name_2', 'city', 'name_3', 'zip_1', 'zip_2']

...或者如果您想使用字母代替:

...or if you wanted to use letters instead:

>>> mylist = ["name", "state", "name", "city", "name", "zip", "zip"]
>>> import string
>>> uniquify(mylist, (f'_{x!s}' for x in string.ascii_lowercase))
>>> mylist
['name_a', 'state', 'name_b', 'city', 'name_c', 'zip_a', 'zip_b']

注意:这不是最快的算法；为此，请参考 ronakg的答案.上面函数的优点是易于理解和阅读，除非您的列表很大，否则您不会看到很多性能差异.

NOTE: this is not the fastest possible algorithm; for that, refer to the answer by ronakg. The advantage of the function above is it is easy to understand and read, and you're not going to see much of a performance difference unless you have an extremely large list.

这是我最初的回答，但是不保留顺序，而是使用.index方法，该方法非常不理想(如 queezz的答案，以获取保留订单的不错的两层式".

Here is my original answer in a one-liner, however the order is not preserved and it uses the .index method, which is extremely suboptimal (as explained in the answer by DTing). See the answer by queezz for a nice 'two-liner' that preserves order.

[s + str(suffix) if num>1 else s for s,num in Counter(mylist).items() for suffix in range(1, num+1)]
# Produces: ['zip1', 'zip2', 'city', 'state', 'name1', 'name2', 'name3']

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