用键替换 python 列表元素

Question

我有一个非唯一字符串列表：

list = ["a", "b", "c", "a", "a", "d", "b"]

我想用唯一标识每个字符串的整数键替换每个元素：

list = [0, 1, 2, 0, 0, 3, 1]

数字无所谓，只要是唯一标识即可。

到目前为止我能想到的就是将列表复制到一个集合中，并使用集合的索引来引用列表。不过我相信还有更好的方法。

Answer 1

>>> lst = ["a", "b", "c", "a", "a", "d", "b"]
>>> nums = [ord(x) for x in lst]
>>> print(nums)
[97, 98, 99, 97, 97, 100, 98]

Answer 2

如果你不挑剔，那就用散列函数：它returns一个整数。对于相同的字符串，它 returns 相同的散列：

li = ["a", "b", "c", "a", "a", "d", "b"]
li = map(hash, li)                # Turn list of strings into list of ints
li = [hash(item) for item in li]  # Same as above

Answer 3

这将保证唯一性，并且 ID 从 0:

开始是连续的

id_s = {c: i for i, c in enumerate(set(list))}
li = [id_s[c] for c in list]

换句话说，你不应该使用 'list' 作为变量名，因为它会隐藏内置类型 list.

Answer 4

这是 defaultdict 的单遍解决方案：

from collections import defaultdict
seen = defaultdict()
seen.default_factory = lambda: len(seen)  # you could instead bind to seen.__len__

In [11]: [seen[c] for c in list]
Out[11]: [0, 1, 2, 0, 0, 3, 1]

有点小技巧但值得一提！

另一种选择，suggested by @user2357112 in a related question/answer, is to increment with itertools.count。这允许您仅在构造函数中执行此操作：

from itertools import count
seen = defaultdict(count().__next__)  # .next in python 2

这可能更可取，因为 default_factory 方法不会在全局范围内查找 seen。

Answer 5

函数式方法：

l = ["a", "b", "c", "a", "a", "d", "b", "abc", "def", "abc"]
from itertools import count
from operator import itemgetter

mapped = itemgetter(*l)(dict(zip(l, count())))

您也可以使用简单的生成器函数：

from itertools import count

def uniq_ident(l):
    cn,d  = count(), {}
    for ele in l:
        if ele not in d:
            c = next(cn)
            d[ele] = c
            yield c
        else:
            yield d[ele]


In [35]: l = ["a", "b", "c", "a", "a", "d", "b"]

In [36]: list(uniq_ident(l))
Out[36]: [0, 1, 2, 0, 0, 3, 1]

用键替换 python 列表元素

Replacing python list elements with key

python

key

list