返回列表中与另一个对象具有相同属性的所有对象？

Question

我有一个Move类，它具有三个属性：newPos，oldPos和符号。 表示法是一个字符串。 我生成了一个动作列表，我想检查这些动作的表示法是否相同。 什么是最pythonic的方式做到这一点？ 我能想到的最干净的解决方案是：

duplicateNotationMoves = []
for move in moves :
    if len([m for m in moves if m.notation == move.notation]) :
        duplicateNotationMoves.append(move)

它工作正常，但效率低下，而且不是很Python。 有没有一种更干净的方法来获得与列表中另一举动具有相同符号的所有举动？

Answer 1

这样的东西？ 一点点双重理解动作

import random

class move: # i just made a simplified version that randomly makes notations
    def __init__(self):
        self.notation = str(random.randrange(1,10))
    def __repr__(self): #so it has something to print, instead of <object@blabla>
        return self.notation


moves = [move() for x in range(20)] #spawns 20 of them in a list

dup = [[y for y in moves if x.notation == y.notation] for x in moves] #double list comprehension


>>> dup
[[4, 4, 4], [7, 7, 7, 7], [1], [2, 2], [8, 8, 8, 8, 8], [8, 8, 8, 8, 8], [4, 4,4], [7, 7, 7, 7], [3, 3], [7, 7, 7, 7], [4, 4, 4], [6, 6], [2, 2], [8, 8, 8, 8,8], [8, 8, 8, 8, 8], [9], [6, 6], [8, 8, 8, 8, 8], [3, 3], [7, 7, 7, 7]]

Answer 2

我找到了一种更简洁的方法来完成此任务，尽管这样做却牺牲了一些可读性：

duplicateNotationMoves = list(filter(lambda move : len(m for m in moves if m.notation == move.notation) > 1, moves))

Answer 3

#UNTESTED

# First, collect the data in a useful form:
notations = collections.Counter(move.notation for move in moves)

# If you want the notations that are duplicated:
duplicate_notations = [
    notation
    for notation, count in notations.items()
    if count > 1]

# Or, if you want the moves that have duplicate notations:
duplicate_moves = [
    move
    for move in moves
    if notations[move.notation] > 1]