如何在整数列表中找到重复项并创建重复项的另一个列表?
当前回答
为了实现这个问题,我们可以使用多种不同的方法来解决它,这两种是常见的解决方案,但在实际场景中实现它们时,我们还必须考虑时间复杂性。
import random
import time
dupl_list = [random.randint(1,1000) for x in range(500)]
print("List with duplicate integers")
print (dupl_list)
#Method 1
print("******************Method 1 *************")
def Repeat_num(x):
_size = len(x)
repeated = []
for i in range(_size):
# print(i)
k = i + 1
for j in range(k, _size):
# print(j)
if x[i] == x[j] and x[i] not in repeated:
repeated.append(x[i])
return repeated
start = time.time()
print(Repeat_num(dupl_list))
end = time.time()
print("The time of execution of above program is :",(end-start) * 10**3, "ms")
print("***************Method 2****************")
#method 2 - using count()
def repeast_count(dup_list):
new = []
for a in dup_list:
# print(a)
# checking the occurrence of elements
n = dup_list.count(a)
# if the occurrence is more than
# one we add it to the output list
if n > 1:
if new.count(a) == 0: # condition to check
new.append(a)
return new
start = time.time()
print(repeast_count(dupl_list))
end = time.time()
print("The time of execution of above program is :",(end-start) * 10**3, "ms")
# #输出示例:
List with duplicate integers
[5, 45, 28, 81, 32, 98, 8, 83, 47, 95, 41, 49, 4, 1, 85, 26, 38, 82, 54, 11]
******************Method 1 *************
[]
The time of execution of above program is : 1.1069774627685547 ms
***************Method 2****************
[]
The time of execution of above program is : 0.1881122589111328 ms
对于一般的理解,方法1是好的,但是对于真正的实现,我更喜欢方法2,因为它比方法1花费的时间更少。
其他回答
一个非常简单的解决方案,但是复杂度是O(n*n)。
>>> xs = [1,2,3,4,4,5,5,6,1]
>>> set([x for x in xs if xs.count(x) > 1])
set([1, 4, 5])
你不需要计数,只需要该物品之前是否被看到过。把这个答案用在这个问题上:
def list_duplicates(seq):
seen = set()
seen_add = seen.add
# adds all elements it doesn't know yet to seen and all other to seen_twice
seen_twice = set( x for x in seq if x in seen or seen_add(x) )
# turn the set into a list (as requested)
return list( seen_twice )
a = [1,2,3,2,1,5,6,5,5,5]
list_duplicates(a) # yields [1, 2, 5]
以防速度很重要,这里有一些时间安排:
# file: test.py
import collections
def thg435(l):
return [x for x, y in collections.Counter(l).items() if y > 1]
def moooeeeep(l):
seen = set()
seen_add = seen.add
# adds all elements it doesn't know yet to seen and all other to seen_twice
seen_twice = set( x for x in l if x in seen or seen_add(x) )
# turn the set into a list (as requested)
return list( seen_twice )
def RiteshKumar(l):
return list(set([x for x in l if l.count(x) > 1]))
def JohnLaRooy(L):
seen = set()
seen2 = set()
seen_add = seen.add
seen2_add = seen2.add
for item in L:
if item in seen:
seen2_add(item)
else:
seen_add(item)
return list(seen2)
l = [1,2,3,2,1,5,6,5,5,5]*100
以下是结果:(做得好@JohnLaRooy!)
$ python -mtimeit -s 'import test' 'test.JohnLaRooy(test.l)'
10000 loops, best of 3: 74.6 usec per loop
$ python -mtimeit -s 'import test' 'test.moooeeeep(test.l)'
10000 loops, best of 3: 91.3 usec per loop
$ python -mtimeit -s 'import test' 'test.thg435(test.l)'
1000 loops, best of 3: 266 usec per loop
$ python -mtimeit -s 'import test' 'test.RiteshKumar(test.l)'
100 loops, best of 3: 8.35 msec per loop
有趣的是,除了计时本身,当使用pypy时,排名也略有变化。最有趣的是,基于counter的方法极大地受益于pypy的优化,而我建议的方法缓存方法似乎几乎没有任何效果。
$ pypy -mtimeit -s 'import test' 'test.JohnLaRooy(test.l)'
100000 loops, best of 3: 17.8 usec per loop
$ pypy -mtimeit -s 'import test' 'test.thg435(test.l)'
10000 loops, best of 3: 23 usec per loop
$ pypy -mtimeit -s 'import test' 'test.moooeeeep(test.l)'
10000 loops, best of 3: 39.3 usec per loop
显然,这种效应与输入数据的“重复性”有关。我设置了l = [random.randrange(1000000) for I in xrange(10000)],得到了这些结果:
$ pypy -mtimeit -s 'import test' 'test.moooeeeep(test.l)'
1000 loops, best of 3: 495 usec per loop
$ pypy -mtimeit -s 'import test' 'test.JohnLaRooy(test.l)'
1000 loops, best of 3: 499 usec per loop
$ pypy -mtimeit -s 'import test' 'test.thg435(test.l)'
1000 loops, best of 3: 1.68 msec per loop
raw_list = [1,2,3,3,4,5,6,6,7,2,3,4,2,3,4,1,3,4,]
clean_list = list(set(raw_list))
duplicated_items = []
for item in raw_list:
try:
clean_list.remove(item)
except ValueError:
duplicated_items.append(item)
print(duplicated_items)
# [3, 6, 2, 3, 4, 2, 3, 4, 1, 3, 4]
基本上可以通过转换为set (clean_list)来删除重复项,然后迭代raw_list,同时删除出现在raw_list中的clean列表中的每个项。如果item未找到,则捕获引发的ValueError异常,并将该item添加到duplicated_items列表中。
如果需要重复项的索引,只需枚举列表并使用索引即可。(对于index, item in enumerate(raw_list):),对于大型列表(比如上千+的元素)来说,这是更快和优化的
我们可以使用itertools。Groupby,以便找到所有有dup的项:
from itertools import groupby
myList = [2, 4, 6, 8, 4, 6, 12]
# when the list is sorted, groupby groups by consecutive elements which are similar
for x, y in groupby(sorted(myList)):
# list(y) returns all the occurences of item x
if len(list(y)) > 1:
print x
输出将是:
4
6
有点晚了,但可能对一些人有帮助。 对于一个比较大的列表,我发现这个方法很适合我。
l=[1,2,3,5,4,1,3,1]
s=set(l)
d=[]
for x in l:
if x in s:
s.remove(x)
else:
d.append(x)
d
[1,3,1]
显示正确和所有重复,并保持秩序。
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