如何根据特定数字过滤列表字典的列表行？

Question

我有大量需要过滤的列表字典。这是其输出的示例：

d = {
    'hate': [(2310, "Experiencer: 'like hours'", 212, 222),
              (2310, "Experiencer: 'two'", 1035, 1038),
              (2310, "Experiencer: 'Anakin'", 1560, 1566),
              (2310, "Experiencer: ' '", 1619, 1620),
              (2310, "Experiencer: 'Tatooine'", 1726, 1734),
              (2310, "Experiencer: 'Anakin'", 1775, 1781),
              (2310, "Experiencer: 'Master Qui-Gon'", 1863, 1877),
              (2310, "Experiencer: 'half'", 1883, 1887),
              (2310, "Experiencer: 'One'", 2114, 2117),
              (2310, "Experiencer: 'Anakin'", 2180, 2186),
              (2310, "Stimulus: 'One'", 2484, 2487),
              (2310, "Stimulus: 'Anakin'", 2564, 2570),
              (2310, "Stimulus: 'Padme'", 2739, 2744)],
'confirmation': [(4132, "Experiencer: 'like hours'", 212, 222),
              (4132, "Experiencer: 'two'", 1035, 1038),
              (4132, "Experiencer: 'Anakin'", 1560, 1566),
              (4132, "Experiencer: ' '", 1619, 1620),
              (4132, "Experiencer: 'Tatooine'", 1726, 1734),
              (4132, "Experiencer: 'Anakin'", 1775, 1781),
              (4132, "Experiencer: 'Master Qui-Gon'", 1863, 1877),
              (4132, "Experiencer: 'half'", 1883, 1887),
              (4132, "Experiencer: 'One'", 2114, 2117),
              (4132, "Experiencer: 'Anakin'", 2180, 2186),
              (4132, "Experiencer: 'One'", 2484, 2487),
              (4132, "Experiencer: 'Anakin'", 2564, 2570),
              (4132, "Experiencer: 'Padme'", 2739, 2744),
              (4132, "Experiencer: 'Anakin'", 2782, 2788),
              (4132, "Experiencer: ' '", 2818, 2819),
              (4132, "Experiencer: 'centuries'", 3562, 3571),
              (4132, "Experiencer: 'one'", 3585, 3588),
              (4132, "Experiencer: 'Anakin'", 3679, 3685),
              (4132, "Experiencer: 'Anakin Skywalker'", 3789, 3805),
              (4132, "Experiencer: 'Obi-Wan'", 4014, 4021),
              (4132, "Experiencer: 'Qui-Gon'", 4025, 4032),
              (4132, "Experiencer: 'Qui-Gon's'", 4100, 4109),
              (4132, "Stimulus: 'Anakin'", 4281, 4287),
              (4132, "Stimulus: ' '", 4355, 4356),
              (4132, "Stimulus: 'Anakin'", 4436, 4442)]}

每个键（其中一个是 hate，如上所述）在每个列表元素的开头都有一个数字。在这里，它是：2310.

我希望能够打印出列表中的两个元素，这些元素的数字最接近该数字，次大，次小.

示例输出：

'hate': [(2310, "Experiencer: 'Anakin'", 2180, 2186),
         (2310, "Stimulus: 'One'", 2484, 2487)]

因为

(2310, "Experiencer: 'Anakin'", 2180, 2186)

有数字 2180，与 2310

相比，它是下一个最小的

并在 return 中：

(2310, "Stimulus: 'One'", 2484, 2487)

有数字 2484，与 2310

相比，它是第二大的

我猜这需要一个 for 循环？我如何遍历列表字典，将第一个自重复数字与每行的第一个数字进行比较，然后 return 最接近的数字，如上所述？

我希望我的问题足够容易理解...

提前致谢！

编辑：

目标是自动处理字典，并通过过滤更新它。

该字典的期望输出将是这样的：

 d = {
        'hate': [(2310, "Experiencer: 'Anakin'", 2180, 2186),
                 (2310, "Stimulus: 'One'", 2484, 2487)],
        'confirmation': [(4132, "Experiencer: 'Qui-Gon's'", 4100, 4109),    
                         (4132, "Stimulus: 'Anakin'", 4281, 4287)],
...}

我还编辑了上面的输出示例。这是一个列表字典

Answer 1

也许最简单的方法是

根据感兴趣的元素对列表进行排序，位置 2。
使用二进制搜索查找 2310 将出现在该列表中的位置；你应该在两个元素之间。这就是你想要的两个元素。

Answer 2

使用itertools.groupby将每个列表中的所有相似元素分组，然后对它们进行排序（基于绝对差）并获得前2个元素

>>> from itertools import groupby
>>> 
>>> f = lambda t: t[0]
>>> {key:sorted(v, key=lambda t: abs(k-t[3]))[:2] for key,lst in d.items() for k,v in groupby(sorted(lst, key=f), f)}
{'confirmation': [(4132, "Experiencer: 'Qui-Gon's'", 4100, 4109), (4132, "Experiencer: 'Qui-Gon'", 4025, 4032)], 'hate': [(2310, "Experiencer: 'Anakin'", 2180, 2186), (2310, "Stimulus: 'One'", 2484, 2487)]}

Answer 3

这是最普通、最简单的方法（我能想到）。下面有一个使用 itertools 的解决方案，它更优雅，但对于新手来说更难理解。

如果 l 是 hate 在你的字典中指向的列表：

target_num = l[0][0]
closest_smaller, closest_bigger = 0,0
closest_smaller_diff, closest_bigger_diff = float("inf"), float("-inf")
for element in l:
    for num in (l[-2],l[-1]):
        diff = target_num - num
        if diff > 0 and diff < closest_smaller_diff:
            closest_smaller = num
            closest_smaller_diff = diff
        if diff < 0 and diff > closest_bigger_diff:
            closest_bigger = num
            closest_bigger_diff = diff
print(closest_smaller, closest_bigger)

Answer 4

# let big_dict be the big list you start with
output_dict = {}
for key, value in big_dict.items(): 
    # break the list into two lists, for those with third value greater
    # and those with third value lesser/equal
    higher_tuples = [i for i in value if i[2] > i[0]]
    lower_tuples = [i for i in value if i[2] <= i[0]]
    # Get the values from that list with 
    high_closest = min(higher_tuples, key=lambda x: x[2] - x[0])
    low_closest = min(lower_tuples, key=lambda x: x[0] - x[2])
    # bind them into an output
    output_duct[key] = [high_closest, low_closest]

如果您愿意，可以将它们全部装订在一个非常大的单行纸中：

output_dict = {key: [min([i for i in value if i[2] > i[0]], key=lambda a: a[2] - a[0]), min([j for j in value if j[0] <= j[2]], key=lambda b: b[0] - b[2])] for key, value in big_dict.items()}

Answer 5

如果您的列表已经排序，我们可以使用 bisect 找到 "Experiencer" 和 "Status" 条目之间的位置：

from bisect import bisect

l=[(2310, "Experiencer: 'like hours'", 212, 222), (2310, "Experiencer: 'two'", 1035,1038), (2310, "Experiencer: 'Anakin'", 1560, 1566), (2310, "Experiencer: ' '", 1619, 1620), (2310, "Experiencer: 'Tatooine'", 1726, 1734), (2310, "Experiencer: 'Anakin'", 1775, 1781), (2310, "Experiencer: 'Master Qui-Gon'", 1863, 1877), (2310, "Experiencer: 'half'", 1883, 1887), (2310, "Experiencer: 'One'", 2114, 2117), (2310, "Experiencer: 'Anakin'", 2180, 2186), (2310, "Stimulus: 'One'", 2484, 2487), (2310, "Stimulus: 'Anakin'", 2564, 2570), (2310, "Stimulus: 'Padme'", 2739, 2744)]

right_index = bisect(l, (2310, "F"))  # "F" comes between "Experiencer" and "Status" 
lower, higher = l[right_index-1], l[right_index]

print(lower, higher, sep="\n")

# (2310, "Experiencer: 'Anakin'", 2180, 2186)
# (2310, "Stimulus: 'One'", 2484, 2487)

然后你就可以很容易地处理你的字典了

from bisect import bisect

def get_boundary(l):  # This assumes all lists in your dict have at least 2 items
    if len(l) < 2:
        return l
    right_index = bisect(l, (l[0][0], "F"))  
    return [l[right_index-1], l[right_index]]

print({key: get_boundary(value) for key, value in d.items()})

生产

{'hate': [(2310, "Experiencer: 'Anakin'", 2180, 2186), 
          (2310, "Stimulus: 'One'", 2484, 2487)], 
 'confirmation': [(4132, "Experiencer: 'Qui-Gon's'", 4100, 4109), 
                  (4132, "Stimulus: 'Anakin'", 4281, 4287)]
}

如何根据特定数字过滤列表字典的列表行？

How to filter list lines of a dictionary of lists according to a specific number?

python

dictionary

filter

closest