我有一个像这样的python列表,
[[12587961, 0.7777777777777778], [12587970, 0.5172413793103449], [12587979, 0.3968253968253968], [12587982, 0.88], [12587984, 0.8484848484848485], [12587992, 0.7777777777777778], [12587995, 0.8070175438596491], [12588015, 0.4358974358974359], [12588023, 0.8985507246376812], [12588037, 0.5555555555555555], [12588042, 0.9473684210526315]]
此列表的长度最多可达千个元素,如何根据子数组中的第二项获取列表中的最大值,并获取最大值的索引,该最大值是python中的子数组?
答案 0 :(得分:17)
使用max
函数及其key
参数,仅使用第二个元素来比较列表中的元素。
例如,
>>> data = [[12587961, 0.7777777777777778], [12587970, 0.5172413793103449], [12587979, 0.3968253968253968].... [12588042, 0.9473684210
526315]]
>>> max(data, key=lambda item: item[1])
[12588042, 0.9473684210526315]
现在,如果你只想要第一个元素,那么你可以简单地单独获取第一个元素,或者只是解压缩结果,就像这样
>>> index, value = max(data, key=lambda item: item[1])
>>> index
12588042
>>> value
0.9473684210526315
编辑:如果你想找到具有最大值(第二个值)的所有元素中的最大索引(第一个值),那么你可以这样做
>>> _, max_value = max(data, key=lambda item: item[1])
>>> max(index for index, value in data if value == max_value)
您可以在一次迭代中执行相同的操作,例如
max_index = float("-inf")
max_value = float("-inf")
for index, value in data:
if value > max_value:
max_value = value
max_index = index
elif value == max_value:
max_index = max(max_index, index)
答案 1 :(得分:3)
将max
与密钥一起使用。
l = [[12587961, 0.7777777777777778], [12587970, 0.5172413793103449], [12587979, 0.3968253968253968], [12587982, 0.88], [12587984, 0.8484848484848485], [12587992, 0.7777777777777778], [12587995, 0.8070175438596491], [12588015, 0.4358974358974359], [12588023, 0.8985507246376812], [12588037, 0.5555555555555555], [12588042, 0.9473684210526315]]
max_sub = max(l, key=lambda x: x[1])
max_val = max_sub[1]
max_index = max_sub[0]
答案 2 :(得分:0)
from operator import itemgetter
a = [[12587961, 0.7777777777777778], [12587970, 0.5172413793103449], [12587979, 0.3968253968253968], [12587982, 0.88], [12587984, 0.8484848484848485], [12587992, 0.7777777777777778], [12587995, 0.8070175438596491], [12588015, 0.4358974358974359], [12588023, 0.8985507246376812], [12588037, 0.5555555555555555], [12588042, 0.9473684210526315]]
max(a, key=itemgetter(1))[0]
// => 12588042
答案 3 :(得分:0)
allData = [[12587961, 0.7777777777777778], [12587970, 0.5172413793103449], [12587979, 0.3968253968253968], [12587982, 0.88], [12587984, 0.8484848484848485], [12587992, 0.7777777777777778], [12587995, 0.8070175438596491], [12588015, 0.4358974358974359], [12588023, 0.8985507246376812], [12588037, 0.5555555555555555], [12588042, 0.9473684210526315]]
listOfSecondData = [i[1] for i in allData]
result = allData[listOfSecondData.index(max(listOfSecondData))][0]
print(result)
#Output: 12588042
答案 4 :(得分:-3)
简单
list = [[12587961, 0.7777777777777778], [12587970, 0.5172413793103449], [12587979, 0.3968253968253968], [12587982, 0.88], [12587984, 0.8484848484848485], [12587992, 0.7777777777777778], [12587995, 0.8070175438596491], [12588015, 0.4358974358974359], [12588023, 0.8985507246376812], [12588037, 0.5555555555555555], [12588042, 0.9473684210526315]]
list2 = []
for x in list:
list2.append(x[1])
print "index->" + str(list[list2.index(max(list2))][0])
print "max value->" + str(list[list2.index(max(list2))][1])