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))
