问题:检查列表中的所有元素是否唯一
检查列表中所有元素是否唯一的最佳方法(与传统方法一样最佳)是什么?
我目前使用的方法Counter
是:
>>> x = [1, 1, 1, 2, 3, 4, 5, 6, 2]
>>> counter = Counter(x)
>>> for values in counter.itervalues():
if values > 1:
# do something
我可以做得更好吗?
What is the best way (best as in the conventional way) of checking whether all elements in a list are unique?
My current approach using a Counter
is:
>>> x = [1, 1, 1, 2, 3, 4, 5, 6, 2]
>>> counter = Counter(x)
>>> for values in counter.itervalues():
if values > 1:
# do something
Can I do better?
回答 0
不是最高效的,而是简单明了的:
if len(x) > len(set(x)):
pass # do something
短名单可能不会有太大的不同。
Not the most efficient, but straight forward and concise:
if len(x) > len(set(x)):
pass # do something
Probably won’t make much of a difference for short lists.
回答 1
这里有两个班轮,它们也会提前退出:
>>> def allUnique(x):
... seen = set()
... return not any(i in seen or seen.add(i) for i in x)
...
>>> allUnique("ABCDEF")
True
>>> allUnique("ABACDEF")
False
如果x的元素不可散列,那么您将不得不使用以下列表seen
:
>>> def allUnique(x):
... seen = list()
... return not any(i in seen or seen.append(i) for i in x)
...
>>> allUnique([list("ABC"), list("DEF")])
True
>>> allUnique([list("ABC"), list("DEF"), list("ABC")])
False
Here is a two-liner that will also do early exit:
>>> def allUnique(x):
... seen = set()
... return not any(i in seen or seen.add(i) for i in x)
...
>>> allUnique("ABCDEF")
True
>>> allUnique("ABACDEF")
False
If the elements of x aren’t hashable, then you’ll have to resort to using a list for seen
:
>>> def allUnique(x):
... seen = list()
... return not any(i in seen or seen.append(i) for i in x)
...
>>> allUnique([list("ABC"), list("DEF")])
True
>>> allUnique([list("ABC"), list("DEF"), list("ABC")])
False
回答 2
提前退出的解决方案可能是
def unique_values(g):
s = set()
for x in g:
if x in s: return False
s.add(x)
return True
但是对于小情况或如果提早退出并不常见,那么我希望len(x) != len(set(x))
这是最快的方法。
An early-exit solution could be
def unique_values(g):
s = set()
for x in g:
if x in s: return False
s.add(x)
return True
however for small cases or if early-exiting is not the common case then I would expect len(x) != len(set(x))
being the fastest method.
回答 3
为了速度:
import numpy as np
x = [1, 1, 1, 2, 3, 4, 5, 6, 2]
np.unique(x).size == len(x)
for speed:
import numpy as np
x = [1, 1, 1, 2, 3, 4, 5, 6, 2]
np.unique(x).size == len(x)
回答 4
如何将所有条目添加到集合中并检查其长度呢?
len(set(x)) == len(x)
How about adding all the entries to a set and checking its length?
len(set(x)) == len(x)
回答 5
替代a set
,您可以使用dict
。
len({}.fromkeys(x)) == len(x)
Alternative to a set
, you can use a dict
.
len({}.fromkeys(x)) == len(x)
回答 6
完全使用排序和分组方式的另一种方法:
from itertools import groupby
is_unique = lambda seq: all(sum(1 for _ in x[1])==1 for x in groupby(sorted(seq)))
它需要排序,但是在第一个重复值上退出。
Another approach entirely, using sorted and groupby:
from itertools import groupby
is_unique = lambda seq: all(sum(1 for _ in x[1])==1 for x in groupby(sorted(seq)))
It requires a sort, but exits on the first repeated value.
回答 7
这是一个有趣的递归O(N 2)版本:
def is_unique(lst):
if len(lst) > 1:
return is_unique(s[1:]) and (s[0] not in s[1:])
return True
Here is a recursive O(N2) version for fun:
def is_unique(lst):
if len(lst) > 1:
return is_unique(s[1:]) and (s[0] not in s[1:])
return True
回答 8
这是递归的提前退出函数:
def distinct(L):
if len(L) == 2:
return L[0] != L[1]
H = L[0]
T = L[1:]
if (H in T):
return False
else:
return distinct(T)
对于我来说,它足够快,而无需使用怪异的(慢速)转换,同时具有功能样式的方法。
Here is a recursive early-exit function:
def distinct(L):
if len(L) == 2:
return L[0] != L[1]
H = L[0]
T = L[1:]
if (H in T):
return False
else:
return distinct(T)
It’s fast enough for me without using weird(slow) conversions while having a functional-style approach.
回答 9
这个怎么样
def is_unique(lst):
if not lst:
return True
else:
return Counter(lst).most_common(1)[0][1]==1
How about this
def is_unique(lst):
if not lst:
return True
else:
return Counter(lst).most_common(1)[0][1]==1
回答 10
您可以使用Yan的语法(len(x)> len(set(x))),但可以定义一个函数来代替set(x):
def f5(seq, idfun=None):
# order preserving
if idfun is None:
def idfun(x): return x
seen = {}
result = []
for item in seq:
marker = idfun(item)
# in old Python versions:
# if seen.has_key(marker)
# but in new ones:
if marker in seen: continue
seen[marker] = 1
result.append(item)
return result
并做len(x)> len(f5(x))。这样会很快,而且还能保留订单。
此处的代码来自:http://www.peterbe.com/plog/uniqifiers-benchmark
You can use Yan’s syntax (len(x) > len(set(x))), but instead of set(x), define a function:
def f5(seq, idfun=None):
# order preserving
if idfun is None:
def idfun(x): return x
seen = {}
result = []
for item in seq:
marker = idfun(item)
# in old Python versions:
# if seen.has_key(marker)
# but in new ones:
if marker in seen: continue
seen[marker] = 1
result.append(item)
return result
and do len(x) > len(f5(x)). This will be fast and is also order preserving.
Code there is taken from: http://www.peterbe.com/plog/uniqifiers-benchmark
回答 11
在Pandas数据框中使用类似的方法来测试列的内容是否包含唯一值:
if tempDF['var1'].size == tempDF['var1'].unique().size:
print("Unique")
else:
print("Not unique")
对我来说,这在包含一百万行的日期框架中的int变量上是瞬时的。
Using a similar approach in a Pandas dataframe to test if the contents of a column contains unique values:
if tempDF['var1'].size == tempDF['var1'].unique().size:
print("Unique")
else:
print("Not unique")
For me, this is instantaneous on an int variable in a dateframe containing over a million rows.
回答 12
以上所有答案都很好,但我更喜欢使用30秒内的Pythonall_unique
示例
您需要set()
在给定列表上使用来删除重复项,并将其长度与列表的长度进行比较。
def all_unique(lst):
return len(lst) == len(set(lst))
True
如果平面列表中的所有值均为unique
,则返回,False
否则返回
x = [1,2,3,4,5,6]
y = [1,2,2,3,4,5]
all_unique(x) # True
all_unique(y) # False
all answer above are good but i prefer to use all_unique
example from 30 seconds of python
you need to use set()
on the given list to remove duplicates, compare its length with the length of the list.
def all_unique(lst):
return len(lst) == len(set(lst))
it returns True
if all the values in a flat list are unique
, False
otherwise
x = [1,2,3,4,5,6]
y = [1,2,2,3,4,5]
all_unique(x) # True
all_unique(y) # False
回答 13
对于初学者:
def AllDifferent(s):
for i in range(len(s)):
for i2 in range(len(s)):
if i != i2:
if s[i] == s[i2]:
return False
return True
For begginers:
def AllDifferent(s):
for i in range(len(s)):
for i2 in range(len(s)):
if i != i2:
if s[i] == s[i2]:
return False
return True
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