如何根据最小值获取具有某个值的元组列表 - 仅适用于 Python 的本机库

如何解决如何根据最小值获取具有某个值的元组列表 - 仅适用于 Python 的本机库

我有以下代码...但我需要获取最小值对应的文件名：

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',[4.088197921,11.135904053,11.472831274,10.691568116,9.663827636,10.659660884000001,9.392413013999999,12.586018896]),('opencountry_241',[11.366319879999999,13.314941407,2.754882813,3.998626709,9.028326501,12.145703089000001,8.675354002999999]),('waterfall23',[12.638763287,12.665527344000001,11.406341552,12.6048929,11.43774673,8.79888916]),('field26',[11.078233943,3.349212646,8.966176812,11.827669236000002,8.203674316]),('mountain_030',[10.025102839,8.78585096,11.994283939999999,7.7325744620000005]),('horse_081',[8.415467337,8.054160893999999,11.093641082000001]),('bison_052',[8.194840093,12.869559482]),('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
马_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

我需要的是获取一个元组列表，例如下面的那个，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......这就是对应的名称最小值...

您可以在下表中看到的方式：

输出

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

请只向我发送基于 Python 本地库的内容

解决方法

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)