Question

如何对模糊逻辑匹配80％的数组中的值进行分组

combined_list = ['magic', 'simple power', 'matrix', 'simple aa', 'madness', 'magics', 'mgcsa', 'simple pws', 'seek', 'dour', 'softy']

的产率：

['magic, magics'], ['simple pws', 'simple aa'], ['simple power'], [matrix]

这是我所取得的成就，但与我的目标截然不同。另外它只支持很少的值，我打算用大约50,000条记录来运行它

from difflib import SequenceMatcher as sm

combined_list = ['magic', 'simple power', 'matrix', 'madness', 'magics', 'mgcsa', 'simple pws', 'seek', 'sour', 'soft']
result = list()
result_group = list()

for x in combined_list:

    for name in combined_list:
        if(sm(None, x, name).ratio() >= 0.80):
            result_group.append(name)
        else:
            pass

    result.append(result_group)
    print(result)
    del result_group[:]


print(result)

循环外的打印结果为空，但循环内的结果包含我需要的值。虽然输出与我需要的不同

['magic', 'magics']]
[['simple power', 'simple pws'], ['simple power', 'simple pws']]
[['matrix'], ['matrix'], ['matrix']]
[['madness'], ['madness'], ['madness'], ['madness']]
[['magic', 'magics'], ['magic', 'magics'], ['magic', 'magics'], ['magic', 'magics'], ['magic', 'magics']]
[['mgcsa'], ['mgcsa'], ['mgcsa'], ['mgcsa'], ['mgcsa'], ['mgcsa']]
[['simple power', 'simple pws'], ['simple power', 'simple pws'], ['simple power', 'simple pws'], ['simple power', 'simple pws'], ['simple power', 'simple pws'], ['simple power', 'simple pws'], ['simple power', 'simple pws']]
[['seek'], ['seek'], ['seek'], ['seek'], ['seek'], ['seek'], ['seek'], ['seek']]
[['sour'], ['sour'], ['sour'], ['sour'], ['sour'], ['sour'], ['sour'], ['sour'], ['sour']]
[['soft'], ['soft'], ['soft'], ['soft'], ['soft'], ['soft'], ['soft'], ['soft'], ['soft'], ['soft']]
[['simple aa'], ['simple aa'], ['simple aa'], ['simple aa'], ['simple aa'], ['simple aa'], ['simple aa'], ['simple aa'], ['simple aa'], ['simple aa'], ['simple aa']]

[[], [], [], [], [], [], [], [], [], [], []]

Answer 1

问题在于：

result.append(result_group)
print(result)
del result_group[:]

您可以在结果中附加一个列表，但由于列表是可变类型，因此只存储一个引用。因此，当您更改原始列表（result_group）时，您也会更改result中的引用，在您的情况下删除所有元素。相反，像这样复制它：

result.append(result_group[:])
print(result)
del result_group[:]

或者不要删除列表元素，但为每次迭代创建一个新列表：

for x in combined_list:
    result_group = []
    for name in combined_list:
        ...

result.append(result_group)

修改：如果您想摆脱重复项，请尝试使用集合而不是列表：

# result = list()
result = set([])

...
# result.append(result_group)
result.add(tuple(result_group))

集总是包含唯一成员，但是，由于列表是不可清除的，因此您需要先将它们转换为元组。

Edit2 ：将所有内容放在一起并检查2个以上成员的实际群组：

from difflib import SequenceMatcher as sm

combined_list = ['magic', 'simple power', 'matrix', 'madness',
                 'magics', 'mgcsa', 'simple pws', 'seek', 'sour', 'soft']

# using a set ensures there are no duplicates
result = set([])

for x in combined_list:
    result_group = []
    for name in combined_list:
        if(sm(None, x, name).ratio() >= 0.80):
            result_group.append(name)

    if len(result_group) > 1: # this gets rid of single-word groups
        result.add(tuple(result_group))

print(result)

Answer 2

from difflib import SequenceMatcher as sm

combined_list = ['magic', 'simple power', 'matrix', 'madness', 'magics', 
'mgcsa', 'simple pws', 'seek', 'sour', 'soft']
result = list()
result_group = list()
usedElements = list()
skip = False

for firstName in combined_list:
    skip = False

    for x in usedElements:
        if x == firstName:
            skip = True
    if skip == True:
        continue

    for secondName in combined_list:

        if(sm(None, firstName, secondName).ratio() >= 0.80):            

            result_group.append(secondName)
            usedElements.append(secondName)
        else:
            pass

    result.append(result_group[:])
    del result_group[:]

print(result)

我添加了一种方法来删除重复项，方法是将列表中已经放入组中的元素扔到usedElements列表中。

它确实保留了一个组，但是如果你不希望组中没有元素，你可以将最后一段代码更改为：

    if len(result_group) > 1:
        result.append(result_group[:])
        del result_group[:]
    del result_group[:]

print(result)

希望这有帮助。

Answer 3

from difflib import SequenceMatcher as sm

combined_list = ['magic', 'simple power', 'matrix', 'madness', 'magics', 'mgcsa', 'simple pws', 'seek', 'sour', 'soft']
combined_list.sort()


def getPairs(combined_list):
    results = list()
    grouped = set()
    for x in combined_list:
        result_group = list()
        if(grouped.__contains__(x)):
            continue
        for name in combined_list:
            if(sm(None, x, name).ratio() >= 0.80):
                result_group.append(name)
                grouped.add(name);
            else:
                pass;

        results.append(result_group)
    return results;

print(getPairs(combined_list))

Python：分组列表中的类似文本

3 个答案: