将成对的匹配项目组合到匹配组（3 个以上项目）

Question

所以，我有一个带有成对匹配图像（或多或少相同）的 csv 文件，如下所示：

File1;File2
File3;File4
File5;File1
File2;File5
...

我想根据这些数据删除重复的图像。 csv 文件来自第三方程序，该程序根据内容匹配图像，但问题是它只匹配对，如果文件之前匹配过，则忽略（即三个以上重复的组没有一起列出）

关键点是上例中的第三行和第四行，其中File1也匹配File5，File2匹配File5，所以三个文件都是重复的。但如果我只是根据数据删除文件，我可能会删除所有文件（如果我删除每行的第二个）。

所以我正在尝试列出一组匹配文件，以确保我始终保留其中一个。上面的例子应该是这样的：

File1;File2;File5
File3;File4
...

这是我的代码，似乎没有这样做。 csv 文件被读入成对列表 (_data)。最后的列表是 self.__map。

    _exists = [ ]
    for _match in _data:
        _first = _match[ 0 ] in _exists
        _second = _match[ 1 ] in _exists
        if not _first and not _second:
            self.__map.append( _match )
            _exists.append( _match[ 0 ] )
            _exists.append( _match[ 1 ] )
        elif _first and not _second:
            for _node in self.__map:
                if _first in _node:
                    _node.append( _match[ 1 ] )
                    _exists.append( _match[ 1 ] )
                    break
        elif _second and not _first:
            for _node in self.__map:
                if _second in _node:
                    _node.append( _match[ 0 ] )
                    _exists.append( _match[ 0 ] )
                    break

看不出为什么它不起作用，但是当我检查 self.__map 时，它不包含 csv 文件中的所有文件。

可能有一种更简单的方法可以做到这一点，所以请随意提出更好的方法。

Answer 1

由于这两行，代码不起作用。

if _first in _node:

和

if _second in _node:

一开始_first 和_second 被评估为布尔值，表示第一个文件和第二个文件是否已经被识别。但是在这些行中，它们分别被视为第一项和第二项。将_first 更改为_match[0]，将_second 更改为_match[1]。它应该可以工作。

否则我们也可以这样实现。

files = [
    ["File1","File2"],
    ["File3","File4"],
    ["File5","File1"],
    ["File2","File5"],
]

# list of sets. each set contains duplicate/related images. set is used to avoid duplicates. we
# can use list of list here as well. in that case we need add additional logic to handle
# duplicates later
map_of_duplicates = []
for file1, file2 in files:
    new_entry = True
    for match in map_of_duplicates:
        if file1 in match and file2 in match:
            # if both files are present in the match set do nothing
            new_entry = False
            break
        elif file1 in match:
            # if only file1 is present in the match set then add file2 to match set
            new_entry = False
            match.add(file2)
            break
        elif file2 in match:
            #if only file2 is present in the match set then add file1 to the match set
            new_entry = False
            match.add(file1)
            break
    # if after the inner for loop new_entry flag is still true that means none of the files were found in the map and
    # this is indeed a new entry
    if new_entry:
        map_of_duplicates.append(set((file1, file2)))

print(map_of_duplicates)

Answer 2

您可以尝试以下方法：

data = """File1;File2
File3;File4
File5;File1
File2;File5"""

# Make sure you have a list of set (and not a list of list)
l = [set(_.split(";")) for _ in data.split("\n")]
print(l)
# [{'File1', 'File2'}, {'File4', 'File3'}, {'File1', 'File5'}, {'File5', 'File2'}]

# Add first item
out = [l[0]]
# Iterate over the list input
for pair in l[1:]:
    added = False
    # Iterate over output (already visited pairs)
    for i, pair_out in enumerate(out):
        # if any element of the pair has already been visited
        if any(file_ in pair_out for file_ in pair):
            # Update output
            out[i] = pair_out.union(pair)
            added = True
            # Exit loop
            break
    # If pair not found: append the pair
    if not added: out.append(pair)

print(out)
# [{'File1', 'File5', 'File2'}, {'File4', 'File3'}]

Answer 3

另一种方法

from collections import defaultdict

input_data = """File1;File2
File3;File4
File5;File1
File2;File5"""

data = [d.split(';') for d in input_data.split("\n")]
result = defaultdict(list)

# Create the full list of matches
for d1, d2 in data:
    result[d1].append(d2)
    result[d2].append(d1)

# Start with all the keys
keep = list(result.keys())

# Remove any keys that are repeats
for key in result:
    if key in keep:
        for repeated in result[key]:
            if repeated in keep:
                keep.remove(repeated)

# Pick only the ones we want to keep
result = {k: v for k, v in result.items() if k in keep}