[英]Making unique list in python
What is the most pythonic way of making a list unique using custom equality operator? 使用自定义相等运算符使列表唯一的最有效的方法是什么?
For instance you have a list of dicts L
, and you want a new list M
such that for all dicts d
, e
in M
and one specific x
例如,您有一个字典
L
的列表,并且您想要一个新列表M
,以便对所有字典d
, M
e
和一个特定的x
d[x] != e[x]
How can this be done? 如何才能做到这一点?
In your case (and all cases where equivalence boils down to the equivalence of some kind of key), you can simply construct a dictionary, where the keys are the values you want to compare: 在您的情况下(以及所有等效性归结为某种键的等效性的所有情况),您可以简单地构造一个字典,其中的键是您要比较的值:
L = [{'key': 'foo', 'v': 42}, {'key': 'bar', 'v': 43}, {'key': 'foo', 'v': 44}]
x = 'key'
M = {d[x]:d for d in L}.values()
# In old Python versions: dict((d[x],d for d in L)).values()
Note that the result is not deterministic, both 请注意,结果不是确定性的,两者
[{'key': 'foo', 'v': 44}, {'key': 'bar', 'v': 43}]
and 和
[{'key': 'foo', 'v': 42}, {'key': 'bar', 'v': 43}]
are valid results. 是有效的结果。
In the general case, simply check all accepted values: 在一般情况下,只需检查所有接受的值:
def unique(iterable, is_eq):
tmp = []
for el in iterable:
if not any(is_eq(inTmp, el) for inTmp in tmp):
tmp.append(is_eq)
return tmp
Note that this means that your comparison function will be called O(n²)
times instead of n
times. 请注意,这意味着您的比较函数将被称为
O(n²)
次而不是n
次。
Based on FUD's comment to phihag. 基于FUD对phihag的评论。 Note that
key
function must return a hashable value. 请注意,
key
函数必须返回可哈希值。
def unique(iterable, key=lambda x : x):
seen = set()
res = []
for item in iterable:
k = key(item)
if k not in seen:
res.append(item)
seen.add(k)
return res
from operator import itemgetter
L = [{'key': 'foo', 'v': 42}, {'key': 'bar', 'v': 43}, {'key': 'foo', 'v': 44}]
print unique(L, key=itemgetter('key'))
#[{'key': 'foo', 'v': 42}, {'key': 'bar', 'v': 43}]
I'm not sure this sort of thing admits a one-liner, but it seems to me that the set
class is the key to what you want. 我不确定这种事情是否允许单行,但是在我看来,
set
类是您想要的关键。
M = []
uniques = set(d[x] for d in L)
for d in L:
if d[x] in uniques:
uniques.remove(d[x])
M.append(d)
Note: phihag's answer seems more Pythonic, but this might be a bit more self-documenting 注意:phihag的答案似乎更像Python,但是这可能更多是自我记录
Using dictionary comprehension: 使用字典理解:
def unique(itrable,key):
return {key(x):x for x in itrable}.values()
>>> unique('abcdbbcdab', lambda x: x)
['a', 'c', 'b', 'd']
>>> unique([10, -20, 20, 30], lambda x: abs(x))
[10, 20, 30]
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