Question

我的程序生成如下列表：

mydata = ["foo", "bar", "baz", "quux", "quid", "quo"]

我从其他数据中知道这些可以分组成几对（这里是一个元组列表，但可以改为任何东西）：

static_mapping = [("foo", "quo"), ("baz", "quux"), ("quid", "bar")]

夫妇没有订购。

现在解决问题：我的程序生成mydata，我需要将数据分组，但保留单独的不匹配项列表。原因是任何时候mydata可能不包含所有属于夫妻的项目。

关于这种假设函数的预期结果：

 mydata = ["foo", "bar", "quo", "baz"]
 couples, remainder = group_and_split(mydata, static_mapping)
 print(couples)
 [("foo", "quo")]
 print(remainder)
 ["bar", "baz"]

编辑：我尝试过的例子（但他们停止寻找耦合）：

found_pairs = list()
for coupling in static_mapping:
    pairs = set(mydata).intersect(set(coupling))
    if not pairs or len(pairs) != 2:
        continue
    found_pairs.append(pairs)

我一直在寻找一种可靠的方法来获取提醒。

Answer 1

你可以试试这个：

import copy

def group_and_split(mydata, static_mapping):
    remainder = copy.deepcopy(mydata)
    couples = []
    for couple in static_mapping:
        if couple[0] in mydata and couple[1] in mydata:
            remainder.remove(couple[0])
            remainder.remove(couple[1])
            couples.append(couple)
    return [couples, remainder]

Answer 2

如果值很大，Set会为您提供更快的运行时间，但需要内存，而deepcopy会保持原始数据的完整性。

假设函数的一个实现可能是： -

from copy import deepcopy

def group_and_split(mydata, static_mapping):
        temp  = set(mydata)
        couples = []
        remainder = deepcopy(mydata)
        for value1,value2 in static_mapping:
                if value1 in temp and value2 in temp:
                        couples.append((value1,value2))
                        remainder.remove(value1)
                        remainder.remove(value2)
        return couples, remainder

Answer 3

您可以使用sets函数

来使用symmetric_difference进行操作

>>> full =  ["foo", "quo", "baz", "quux", "quid", "bar", 'newone', 'anotherone']
>>> couples = [("foo", "quo"), ("baz", "quux"), ("quid", "bar")]

## now for the remainder. Note you have to flatten your couples:
>>> set(full).symmetric_difference([item for sublist in couples for item in sublist])
set(['anotherone', 'newone'])
>>>

Python：根据项目对

3 个答案: