Question

我有下面的pandas系列，我需要选择其他行的超集。

系列：

1   [72197, 82086]
2   [72197, 82086, 194665]
3   [72230]
4   [72235, 72690, 121261]
5   [72235, 121261]
6   [72241]
7   [72251]
8   [72253, 83613]
9   [72253, 83613, 101294]
10  [72255, 122794]
11 [71962, 101646, 101663, 126351]
12 [71962, 101646, 101663, 126351, 141883]
13 [71962, 101646, 101663, 141883]
14 [72235]

输出系列：

1   [72197, 82086, 194665]
2   [72230]
3   [72235, 72690, 121261]
4   [72241]
5   [72251]
6   [72253, 83613, 101294]
7   [72255, 122794]
8   [71962, 101646, 101663, 126351, 141883]

Answer 1

如果排序不重要，则可以使用经过稍微改动的this解决方案-将内部列表首先转换为set，最后一次转换回：

s = pd.Series([[72197, 82086], [72197, 82086, 194665], [72230], [72235, 72690, 121261], [72235, 121261],
                       [72241], [72251], [72253, 83613], [72253, 83613], [72253, 83613, 101294], [72255, 122794],
                          [71962, 101646, 101663, 126351], [71962, 101646, 101663, 126351, 141883],
                          [71962, 101646, 101663, 141883], [72235]])

-

import functools,operator,collections

def is_power_of_two(n):
    """Returns True iff n is a power of two.  Assumes n > 0."""
    return (n & (n - 1)) == 0

def eliminate_subsets(sequence_of_sets):
    """Return a list of the elements of `sequence_of_sets`, removing all
    elements that are subsets of other elements.  Assumes that each
    element is a set or frozenset and that no element is repeated."""
    # The code below does not handle the case of a sequence containing
    # only the empty set, so let's just handle all easy cases now.
    if len(sequence_of_sets) <= 1:
        return list(sequence_of_sets)
    # We need an indexable sequence so that we can use a bitmap to
    # represent each set.
    if not isinstance(sequence_of_sets, collections.Sequence):
        sequence_of_sets = list(sequence_of_sets)
    # For each element, construct the list of all sets containing that
    # element.
    sets_containing_element = {}
    for i, s in enumerate(sequence_of_sets):
        for element in s:
            try:
                sets_containing_element[element] |= 1 << i
            except KeyError:
                sets_containing_element[element] = 1 << i
    # For each set, if the intersection of all of the lists in which it is
    # contained has length != 1, this set can be eliminated.
    out = [s for s in sequence_of_sets
           if s and is_power_of_two(functools.reduce(
               operator.and_, (sets_containing_element[x] for x in s)))]
    return list(map(list, out))

s = pd.Series(eliminate_subsets(list(map(set, s))))
print (s)

0                     [194665, 72197, 82086]
1                                    [72230]
2                     [72690, 72235, 121261]
3                                    [72241]
4                                    [72251]
5                     [101294, 72253, 83613]
6                            [122794, 72255]
7    [101646, 126351, 71962, 141883, 101663]
dtype: object

Answer 2

您可以尝试以下方法：

df = pd.DataFrame({'ser': [[72197, 82086], [72197, 82086, 194665], [72230], [72235, 72690, 121261], [72235, 121261],
                       [72241], [72251], [72253, 83613], [72253, 83613], [72253, 83613, 101294], [72255, 122794],
                          [71962, 101646, 101663, 126351], [71962, 101646, 101663, 126351, 141883],
                          [71962, 101646, 101663, 141883], [72235]]})
df

ser
0   [72197, 82086]
1   [72197, 82086, 194665]
2   [72230]
3   [72235, 72690, 121261]
4   [72235, 121261]
5   [72241]
6   [72251]
7   [72253, 83613]
8   [72253, 83613]
9   [72253, 83613, 101294]
10  [72255, 122794]
11  [71962, 101646, 101663, 126351]
12  [71962, 101646, 101663, 126351, 141883]
13  [71962, 101646, 101663, 141883]
14  [72235]

supersets = []
for i, x in enumerate(df['ser']):
    a = np.array([set(x).issuperset(set(row)) for row in df['ser']])
    a = np.delete(a, i)
    if any(a):
        supersets.append(x)
print(supersets)
[[72197, 82086, 194665], [72235, 72690, 121261], [72235, 121261], [72253, 83613], [72253, 83613], [72253, 83613, 101294], [71962, 101646, 101663, 126351, 141883]]

使用序列表或数据框列表效率不高

Answer 3

如果您使用普通列表，我想您可以使用set（a）

m=[[72197, 82086] ,[72197, 82086, 194665],[72230],[72235, 72690, 121261],[72235, 121261],[72241],[72251],[72253, 83613],[72253, 83613, 101294],[72255, 122794]]

del_list_idx=[]
for i in range(0,len(m)-1):
    for j in range(0,len(m)-1):
        if set(m[i])<set(m[j]): 
            del_list_idx.append(i)


for i in range (0,len(del_list_idx)):
    del m[del_list_idx[i]-i]

for i in range(0, len(m)):

    print i, m[i]

与其他行相比，如何基于列表过滤系列？

3 个答案: