从数组中提取重复值和位置的列表

Given an array a of length N, which is a list of integers, I want to extract the duplicate values, where I have a seperate list for each value containing the location of the duplicates. In pseudo-math:

给定一个长度为N的数组a，它是一个整数列表，我想提取重复值，其中我有一个包含重复项位置的每个值的单独列表。在伪数学中：

If |M| > 1:
  val -> M = { i | a[i] == val }

Example (N=11):

示例（N = 11）：

a = [0, 3, 1, 6, 8, 1, 3, 3, 2, 10, 10]

should give the following lists:

应该给出以下列表：

3  -> [1, 6, 7]
1  -> [2, 5]
10 -> [9, 10]

I added the python tag since I'm currently programming in that language (numpy and scipy are available), but I'm more interestead in a general algorithm of how to do it. Code examples are fine, though.

我添加了python标签，因为我目前正在使用该语言进行编程（numpy和scipy可用），但我更倾向于使用通用算法来完成它。但代码示例很好。

One idea, which I did not yet flesh out: Construct a list of tuples, pairing each entry of a with its index: (i, a[i]). Sort the list with the second entry as key, then check consecutive entries for which the second entry is the same.

一个想法，我还没有充实：构建一个元组列表，将a的每个条目与其索引配对：（i，a [i]）。使用第二个条目作为键对列表进行排序，然后检查第二个条目相同的连续条目。

3 个解决方案

#1

The idea is to create a dictionary mapping the values to the list of the position where it appears.

我们的想法是创建一个字典，将值映射到它出现的位置列表。

This can be done in a simple way with setdefault. This can also be done using defaultdict.

这可以通过setdefault以简单的方式完成。这也可以使用defaultdict完成。

>>> a = [0, 3, 1, 6, 8, 1, 3, 3, 2, 10, 10]
>>> dup={}
>>> for i,x in enumerate(a):
...     dup.setdefault(x,[]).append(i)
...
>>> dup
{0: [0], 1: [2, 5], 2: [8], 3: [1, 6, 7], 6: [3], 8: [4], 10: [9, 10]}

Then, actual duplicates can be extracted using set comprehension to filter out elements appearing only once.

然后，可以使用集合理解来提取实际重复项，以过滤掉仅出现一次的元素。

>>> {i:x for i,x in dup.iteritems() if len(x)>1}
{1: [2, 5], 10: [9, 10], 3: [1, 6, 7]}

#2

Here's an implementation using a python dictionary (actually a defaultdict, for convenience)

这是使用python字典的实现（为方便起见，实际上是一个defaultdict）

a = [0, 3, 1, 6, 8, 1, 3, 3, 2, 10, 10]
from collections import defaultdict
d = defaultdict(list)

for k, item in enumerate(a):
    d[item].append(k)
finalD = {key : value for key, value in d.items() if len(value) > 1}  # Filter dict for items that only occurred once.

print(finalD)    
# {1: [2, 5], 10: [9, 10], 3: [1, 6, 7]}

#3

Populate a dictionary whose keys are the values of the integers, and whose values are the lists of positions of those keys. Then go through that dictionary and remove all key/value pairs with only one position. You will be left with the ones that are duplicated.

填充字典，其字符的键是整数的值，其值是这些键的位置列表。然后浏览该字典并删除只有一个位置的所有键/值对。您将留下重复的那些。

#1