如何根据最小值获取具有某些值的元组列表 - 仅使用 Python 中的本机库答案

【问题标题】：How to get a list of tuples with some value based on its minimum value - Only with native libraries from Python如何根据最小值获取具有某些值的元组列表 - 仅使用 Python 中的本机库
【发布时间】：2021-07-20 07:32:32
【问题描述】：

我有以下代码...但是我需要获取对应的文件名的最小值：

import itertools

arrayFiles = []
for subset in itertools.combinations(distances, 2):
  array = []
  for k in range(K):
    array.append(min([(n[1][k]) for n in subset]))
  arrayFiles.append(array)

例如，我有以下示例数据K = 10，距离列表是：

distances = [('highway_bost174', [0.0, 8.708170812, 4.088197921, 11.366319879999999, 12.638763287, 11.078233943, 10.025102839, 8.415467337, 8.194840093, 13.455056175000001]),
             ('ibis_142', [8.708170812, 0.0, 10.518235207, 7.668395996, 10.522399903, 7.302185059, 6.417022705, 6.146172005, 10.448354985, 5.149291993]),
             ('street_par88', [4.088197921, 10.518235207, 0.0, 11.135904053, 11.472831274, 10.691568116, 9.663827636, 10.659660884000001, 9.392413013999999, 12.586018896]),
             ('opencountry_241', [11.366319879999999, 7.668395996, 11.135904053, 0.0, 13.314941407, 2.754882813, 3.998626709, 9.028326501, 12.145703089000001, 8.675354002999999]),
             ('waterfall23', [12.638763287, 10.522399903, 11.472831274, 13.314941407, 0.0, 12.665527344000001, 11.406341552, 12.6048929, 11.43774673, 8.79888916]),
             ('field26', [11.078233943, 7.302185059, 10.691568116, 2.754882813, 12.665527344000001, 0.0, 3.349212646, 8.966176812, 11.827669236000002, 8.203674316]),
             ('mountain_030', [10.025102839, 6.417022705, 9.663827636, 3.998626709, 11.406341552, 3.349212646, 0.0, 8.78585096, 11.994283939999999, 7.7325744620000005]),
             ('horse_081', [8.415467337, 6.146172005, 10.659660884000001, 9.028326501, 12.6048929, 8.966176812, 8.78585096, 0.0, 8.054160893999999, 11.093641082000001]),
             ('bison_052', [8.194840093, 10.448354985, 9.392413013999999, 12.145703089000001, 11.43774673, 11.827669236000002, 11.994283939999999, 8.054160893999999, 0.0, 12.869559482]),
             ('ibis_040', [13.455056175000001, 5.149291993, 12.586018896, 8.675354002999999, 8.79888916, 8.203674316, 7.7325744620000005, 11.093641082000001, 12.869559482, 0.0])]

	highway_bost174	ibis_142	street_par88	opencountry_241	waterfall23	field26	mountain_030	horse_081	bison_052	ibis_040
highway_bost174	0	8.708170812	4.088197921	11.366319880	12.63876329	11.07823394	10.02510284	8.415467337	8.194840093	13.45505618
ibis_142	8.708170812	0	10.518235207	7.668395996	10.5223999	7.302185059	6.417022705	6.146172005	10.44835499	5.149291993
street_par88	4.088197921	10.51823521	0.	11.135904053	11.47283127	10.69156812	9.663827636	10.65966088	9.392413014	12.5860189
opencountry_241	11.36631988	7.668395996	11.135904053	0.	13.31494141	2.754882813	3.998626709	9.028326501	12.14570309	8.675354003
waterfall23	12.63876329	10.5223999	11.472831274	13.314941407	0	12.66552734	11.40634155	12.6048929	11.43774673	8.79888916
field26	11.07823394	7.302185059	10.691568116	2.754882813	12.66552734	0	3.349212646	8.966176812	11.82766924	8.203674316
mountain_030	10.02510284	6.417022705	9.663827636	3.998626709	11.40634155	3.349212646	0	8.78585096	11.99428394	7.732574462
horse_081	8.415467337	6.146172005	10.65966088	9.028326501	12.6048929	8.966176812	8.78585096	0	8.054160894	11.09364108
bison_052	8.194840093	10.44835499	9.392413014	12.145703089	11.43774673	11.82766924	11.99428394	8.054160894	0	12.86955948
ibis_040	13.45505618	5.149291993	12.586018896	8.675354003	8.79888916	8.203674316	7.732574462	11.09364108	12.86955948	0

我需要的是获得一个元组列表，例如下面的一个，其中名称之前的字符串是列表距离的每个元组的第一个元素的相应结果......这就是对应的名称的最小值...

如下表所示：

输出

  [[("highway_bost174", 0), ("ibis_142",0),("highway_bost174", 4.088197921),("ibis_142", 7.668395996),("ibis_142", 10.5223999), ("ibis_142", 7.302185059), ( "ibis_142", 6.417022705), ("ibis_142", 6.146172005),("highway_bost174", 8.194840093),("ibis_142", 5.149291993)]...]

请仅向我发送基于 Python 本地库的内容

【问题讨论】：

标签： python list append list-comprehension min

【解决方案1】：

import itertools
arrayFiles = []
for subset in itertools.combinations(distances, 2):
    one = subset[0]
    two = subset[1]
    one_tup = [(subset[0][0], x) for x in subset[0][1]]
    two_tup = [(subset[1][0], x) for x in subset[1][1]]
    pair = [(o, two_tup[i]) for (i, o) in enumerate(one_tup)]
    res_list = []
    for e in pair:
        if e[1][1] > e[0][1]:
            res_list.append(e[0])
        else:
            res_list.append(e[1])
    arrayFiles.append(res_list)

【讨论】：

感谢您回复我，但这并不是我想要的...请查看我为您提供的输出 ' [[("highway_bost174", 0), ( "ibis_142",0),("highway_bost174", 4.088197921),("ibis_142", 7.668395996),("ibis_142", 10.5223999), ("ibis_142", 7.302185059), ("ibis_142", 6.41714), ("ibis_142", 6.41712) ", 6.146172005),("highway_bost174", 8.194840093),("ibis_142", 5.149291993)]...]'
好的，所以你想要一个按最小距离排序的元组列表（、）？我错过了什么吗，因为我认为我已经做到了。
是的，看看第二个表...那里的名称对应于其对应的每个文件...在初始选择列中
我建议运行以下命令：dist2 = distances[0:2] 后跟 sorted([(d[0], x) for d in dist2 for x in d[1]], key=lambda k: k[1])。我认为这提供了您所追求的输出。
你想要distances中每个目的地组合的列表吗？