我想从下面的列表中获得唯一的值:

['nowplaying', 'PBS', 'PBS', 'nowplaying', 'job', 'debate', 'thenandnow']

我需要的输出是:

['nowplaying', 'PBS', 'job', 'debate', 'thenandnow']

这段代码工作:

output = []
for x in trends:
    if x not in output:
        output.append(x)
print(output)

有更好的解决方案吗?


当前回答

首先,你给出的例子不是一个有效的列表。

example_list = [u'nowplaying',u'PBS', u'PBS', u'nowplaying', u'job', u'debate',u'thenandnow']

假设以上是示例列表。然后,您可以使用下面的配方来给出itertools示例文档,该文档可以返回唯一的值,并按照您的要求保留顺序。这里的可迭代对象是example_list

from itertools import ifilterfalse

def unique_everseen(iterable, key=None):
    "List unique elements, preserving order. Remember all elements ever seen."
    # unique_everseen('AAAABBBCCDAABBB') --> A B C D
    # unique_everseen('ABBCcAD', str.lower) --> A B C D
    seen = set()
    seen_add = seen.add
    if key is None:
        for element in ifilterfalse(seen.__contains__, iterable):
            seen_add(element)
            yield element
    else:
        for element in iterable:
            k = key(element)
            if k not in seen:
                seen_add(k)
                yield element

其他回答

作为奖励,Counter是一种获得唯一值和每个值的计数的简单方法:

from collections import Counter
l = [u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']
c = Counter(l)
def setlist(lst=[]):
   return list(set(lst))

为了与我使用的类型保持一致:

mylist = list(set(mylist))

在代码开始时,只需将输出列表声明为空:output=[] 您可以使用以下代码代替您的代码trends=list(set(trends))

维持秩序:

# oneliners
# slow -> . --- 14.417 seconds ---
[x for i, x in enumerate(array) if x not in array[0:i]]

# fast -> . --- 0.0378 seconds ---
[x for i, x in enumerate(array) if array.index(x) == i]

# multiple lines
# fastest -> --- 0.012 seconds ---
uniq = []
[uniq.append(x) for x in array if x not in uniq]
uniq

顺序不重要:

# fastest-est -> --- 0.0035 seconds ---
list(set(array))