如果它们存在于单独的元组中，则保留重复的元组值，但如果它们位于同一元组中，则仅保留唯一的元组值答案

【问题标题】：Keeping duplicate tuple values if they are present inside separate tuples but only keeping unique tuple values if they are in the same tuple如果它们存在于单独的元组中，则保留重复的元组值，但如果它们位于同一元组中，则仅保留唯一的元组值
【发布时间】：2020-05-02 04:58:11
【问题描述】：

我有一个元组列表：

我想要做的是只保留每个元组中具有唯一编号的entities。

dataset = [('made of iron oxide', {'entities': [(12, 16, 'PRODUCT'), (17, 20, 'PRODUCT'), (15, 24, 'PRODUCT'), (12, 19, 'PRODUCT')]}),('made of ferric oxide', {'entities': [(10, 15, 'PRODUCT'), (17, 20, 'PRODUCT'), (624, 651, 'PRODUCT'), (30, 15, 'PRODUCT'), (1937, 1956, 'PRODUCT')]})]

从这里，预期的输出是：

[('made of iron oxide', {'entities': [(12, 16, 'PRODUCT'), (17, 20, 'PRODUCT'), (15, 24, 'PRODUCT')]}), ('made of ferric oxide', {'entities': [(10, 15, 'PRODUCT'), (17, 20, 'PRODUCT'), (624, 651, 'PRODUCT'), (1937, 1956, 'PRODUCT')]})]

到目前为止的代码：

dataset = [('made of iron oxide', {'entities': [(12, 16, 'PRODUCT'), (17, 20, 'PRODUCT'), (15, 24, 'PRODUCT'), (12, 19, 'PRODUCT')]}),('made of ferric oxide', {'entities': [(10, 15, 'PRODUCT'), (17, 20, 'PRODUCT'), (624, 651, 'PRODUCT'), (30, 15, 'PRODUCT'), (1937, 1956, 'PRODUCT')]})]

seen_values = []
clean_data = []

# loop through each sentence and dict of values
for sentence, values in dataset:
    for value in values['entities']:

        if value[0] in seen_values:
            # remove if we have seen this before
            values['entities'].remove(value)
        else:
            # add to list if we have not seen this before
            seen_values.append(value[0])
    clean_data.append((sentence, values))

print(clean_data)

这给了[('made of iron oxide', {'entities': [(12, 16, 'PRODUCT'), (17, 20, 'PRODUCT'), (15, 24, 'PRODUCT')]}), ('made of ferric oxide', {'entities': [(10, 15, 'PRODUCT'), (624, 651, 'PRODUCT'), (30, 15, 'PRODUCT'), (1937, 1956, 'PRODUCT')]})]

谁能帮我解决这个问题

【问题讨论】：

标签： python python-3.x list dictionary tuples

【解决方案1】：

试试这个：

data = dataset = [('made of iron oxide', {'entities': [(12, 16, 'PRODUCT'), (17, 20, 'PRODUCT'), (15, 24, 'PRODUCT'), (12, 19, 'PRODUCT')]}),('made of ferric oxide', {'entities': [(10, 15, 'PRODUCT'), (17, 20, 'PRODUCT'), (624, 651, 'PRODUCT'), (30, 15, 'PRODUCT'), (1937, 1956, 'PRODUCT')]})]
res = []
for x in data:
    t, flag = [], set()
    for d in x[1]['entities']:
        if d[0] not in flag and d[1] not in flag:
                t.append(d)
        flag.update(d[:2])
    res.append((x[0], {'entities': t}))

print(res)

输出：

[('made of iron oxide', {'entities': [(12, 16, 'PRODUCT'), (17, 20, 'PRODUCT'), (15, 24, 'PRODUCT')]}), ('made of ferric oxide', {'entities': [(10, 15, 'PRODUCT'), (17, 20, 'PRODUCT'), (624, 651, 'PRODUCT'), (1937, 1956, 'PRODUCT')]})]

【讨论】：