在Python中删除列表中的dupes

Question

基本上,我试图删除任何以相同值开头的列表.例如,下面的两个以数字1开头:

a = [[1,2],[1,0],[2,4],[3,5]]

因为值1存在于两个列表的开头 - 我需要删除它们以便新列表变为:

b = [[2,4],[3,5]]

我怎样才能做到这一点？

我试过下面的,但输出是: [[1, 2], [2, 4], [3, 5]]

def unique_by_first_n(n, coll):
    seen = set()
    for item in coll:
        compare = tuple(item[:n])
        print compare   # Keep only the first `n` elements in the set
        if compare not in seen:
            seen.add(compare)
            yield item

a = [[1,2],[1,0],[2,4],[3,5]]

filtered_list = list(unique_by_first_n(1, a))

Answer 1

一个有效的解决方案是创建一个Counter对象来保存第一个元素的出现,然后过滤主列表中的子列表:

from collections import Counter
counts = Counter(l[0] for l in a)
filtered = [l for l in a if counts[l[0]] == 1]
#[[2, 4], [3, 5]]

Answer 2

如果您乐意使用第三方库,可以使用Pandas:

import pandas as pd

a = [[1,2],[1,0],[2,4],[3,5]]

df = pd.DataFrame(a)
b = df.drop_duplicates(subset=[0], keep=False).values.tolist()

print(b)

[[2, 4], [3, 5]]

诀窍是keep=False论证,在文档中描述pd.DataFrame.drop_duplicates.

Answer 3

您可以使用collections.Counter列表推导来获取第一个项目仅出现一次的子列表:

from collections import Counter
c = Counter(n for n, _ in a)
b = [[x, y] for x, y in a if c[x] == 1]