Python:有条件地从列表中删除元素

类似于亚伦的回答

>>> from collections import defaultdict
>>> x = [(1,2), (3,4), (7,4), (5,4)]
>>> d = defaultdict(int)
>>> for v,k in x:
...   d[k] = max(d[k],v) 
... 
>>> y=[(k,v) for v,k in d.items()]
>>> y
[(1, 2), (7, 4)]

>>> y = [(k,v) for k,v in x if d[v]==k]
>>> y
[(1, 2), (7, 4)]

这是另一种方法。它使用更多的存储空间,但对max的调用较少,因此可能更快

>>> d = defaultdict(list)
>>> for k,v in x:
...   d[v].append(k)
... 
>>> y = [(max(k),v) for v,k in d.items()]
>>> y
[(1, 2), (7, 4)]

>>> y = [(k,v) for k,v in x if max(d[v])==k]
>>> y
[(1, 2), (7, 4)]

使用 collections.defaultdict

import collections

max_elements = collections.defaultdict(tuple)

for item in x:
    if item > max_elements[item[1]]:
        max_elements[item[1]] = item

y = max_elements.values()

如果可以假设具有相同第二个元素的元组以连续顺序出现在原始列表中 x itertools.groupby :

import itertools
import operator

def max_first_elem(x):
    groups = itertools.groupby(x, operator.itemgetter(1))
    y = [max(g[1]) for g in groups]
    return y

注意,这将保证保持组的顺序(通过第二个元组元素),如果这是输出所需的约束。

我自己的尝试,有点受阿伦斯特林的启发:

def processtuples(x):
    d = {}
    for item in x:
        if x[0] > d.get(x[1],-1):
            d[x[1]] = x[0]

    y = []
    for k in d:
        y.append((d[k],k))
    y.sort()
    return y

>>> from collections import defaultdict
>>> d = defaultdict(tuple)
>>> x = [(1,2), (3,4), (7,4), (5,4)]
>>> for a, b in x:
...     d[b] = max(d[b], (a, b))
...
>>> d.values()
[(1, 2), (7, 4)