Question

我有一个非唯一字符串列表：

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

我想用一个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]

只要数字是唯一标识符，该数字无关紧要。

到目前为止，我所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。我确信有更好的方法。

Answer 1

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]

另外，您不应将'list'用作变量名称，因为它会影响内置类型list。

Answer 2

以下是defaultdict的单通解决方案：

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]

这是一种伎俩但值得一提！

替代方案suggested by @user2357112 in a related question/answer，将以itertools.count递增。这允许您只在构造函数中执行此操作：

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

这可能更合适，因为default_factory方法在全局范围内不会查找seen。

Answer 3

>>> 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 4

如果你不挑剔，那么使用哈希函数：它返回一个整数。对于相同的字符串，它返回相同的哈希：

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 5

功能性方法：

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())))

您还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

5 个答案: