如何通过第二个值将第二低的列表查找到嵌套列表中？

Question

这里给出了一个嵌套列表：

nl = [['Harsh', 20], ['Beria', 20], ['Varun', 19], ['Kakunami', 19], ['Vikas', 21]]

现在我必须根据第二个值在嵌套列表中找到第二低的列表。并将倒数第二的列表附加到另一个列表中。

所以输出应该是：

['Harsh', 20], ['Beria', 20]

我写了下面的代码，但它不起作用：

nl = [['Harsh', 20], ['Beria', 20], ['Varun', 19], ['Kakunami', 19], ['Vikas', 21]]

result=[]

temp=max(nl, key=lambda x: x[1])

largest, larger = temp[1], temp[1]
for num in nl:
    if num[1] < largest:
        largest, larger = num[1], largest
    elif num[1] < larger:
        larger = num[1]
        result.append(larger)
print(result)

Answer 1

这是一个方便的小函数，其中包含 heapq.nlargest:

Return a list with the n largest elements from the dataset

import heapq
num = heapq.nlargest(2, [key for item, key in nl])[-1]
print [item for item in nl if item[-1] == num] #[['Harsh', 20], ['Beria', 20]]

Answer 2

您的示例可以简化为：

second_lowest = sorted(set(v[1] for v in nl))[1]
result = [v for v in nl if v[1] == second_lowest]
print(result) # [['Harsh', 20], ['Beria', 20]]

我使用您的一组值来找到第二个唯一的最低值。除了拥有此文件，在 nl 中查找与找到的值匹配的元素。

Answer 3

获取总元素的 min，使用该有效元素进行过滤，然后获取最小剩余元素并保持元素等于最小剩余元素：

from operator import itemgetter
# min of all elements
mn = min(nl, key=itemgetter(1))[1]

# remove elements equal to min
filtered = [x for x in nl if x[1] != mn]

# get min of remaining
mn_fil = min(filtered,key=itemgetter(1))[1]

# filter remaining
out = [x for x in filtered if x[1] == mn_fil]
print(out)

[['Harsh', 20], ['Beria', 20]]

适用于您的两种情况：

In [19]: nl = [['Prashant', 32], ['Pallavi', 36], ['Dheeraj', 39], ['Shivam', 40]]    
In [20]: from operator import itemgetter    
In [21]: mn = min(nl, key=itemgetter(1))[1]    
In [22]: filtered = [x for x in nl if x[1] != mn]    
In [23]: mn_fil = min(filtered,key=itemgetter(1))[1]    
In [24]: out = [x for x in filtered if x[1] == mn_fil]    
In [25]: out
Out[25]: [['Dheeraj', 36]]

使用单个 for 循环，如果我们找到一个较低的元素，我们会从临时列表中删除所有元素，如果我们找到同样较低的元素，我们会追加它：

mn = min(nl, key=itemgetter(1))[1]
temp = []
best = float("inf")
for ele in nl:
    if mn < ele[1] < best:
        best = ele[1]
        temp = []
        out.append(ele)
    elif ele[1] == best:
        temp.append(ele)
print(temp)

Answer 4

我是通过使用集合找到第二低的值，然后从列表中选择具有相同值的元素来实现的。

#ordering by value
nl.sort(key = lambda x: x[1])

values_set = set()
for value in nl:
    values_set.add(value[1])

values_list = list(values_set)
#ordering
values_list.sort()
#getting second lowest values
lowest_values = [lowest for lowest in nl if lowest[1] == values_list[1] ]

Answer 5

您可以试试下面的代码。它工作正常。

lst=[['Harry',37.21],['Berry',37.21],['Tina',37.2],['Akriti',41],['Harsh',39]]
names=[]
lowest = lst[0].__getitem__(1)
second_lowest=0
for l in lst:
    if l[1] < lowest:
        second_lowest = lowest
        lowest = l[1]
    elif l[1]<=second_lowest:
        second_lowest=l[1]
for l in lst:
    if l[1]==second_lowest:
        names.append(l[0])


print(lowest)
print(second_lowest)
print(names)

Answer 6

if __name__ == '__main__':
arr = []
for _ in range(int(input())):
    name = input()
    score = float(input())
    arr1 = [name, score]
    arr.append(arr1)
arr.sort(key=lambda x: x[1])
# print(arr)
# print(min(arr,key=lambda x:x[1]))
arr.remove(min(arr,key=lambda x:x[1]))
# print(arr)
minimum = min(arr,key=lambda x:x[1])
# print(minimum[1])
a=[]
minimum = minimum[1]
for i  in arr:
    if(i[1] == minimum):
        a.append(i[0])
a.sort()
for i in a:
    print(i)

Answer 7

我遇到过类似的问题，参考这个页面后我有了一些想法并能够解决它。

nl = [['Harsh', 20], ['Beria', 20], ['Varun', 19], ['Kakunami', 19], ['Vikas', 21]]

second = max(nl, key= lambda x: x[1])[1]
first = min(nl, key= lambda x: x[1])[1]
    
for i in range(len(nl)):
    if nl[i][1] <= second and nl[i][1] != first:
        second = nl[i][1]
    
for i in range(len(nl)):
    if second == nl[i][1]:
        print(nl[i], end=", ")

Answer 8

if __name__ == '__main__':

    n = []
    s = []
    for _ in range(int(input())):
        name = input()
        n.append(name)
        score = float(input())
        s.append(score)
        
data = [[x,y] for x,y in zip(n,s)]

min_marks = min([x[1] for x in data])

filtered_data = [d for d in data if d[1] != min_marks]

sec_min = min([x[1] for x in filtered_data])

students_with_sec_min_marks = sorted([d[0] for d in filtered_data if d[1] == sec_min])

for s in students_with_sec_min_marks:
    print(s)

Answer 9

if __name__ == '__main__':
    lst = []
    names = []
    for _ in range(int(input())):
        name = input()
        score = float(input())
        lst.append([name, score])
     
    lowest = max(lst, key=lambda x: x[1])[1]
    second_lowest=0
    for l in lst:
        if l[1] < lowest:
            second_lowest = lowest
            lowest = l[1]
        elif l[1] < second_lowest and l[1] != lowest:
            second_lowest = l[1]
    for l in lst:
        if l[1] == second_lowest:
            names.append(l[0])  
    
    names.sort()
    for nm in names:
        print(nm)

Answer 10

我使用了下面的代码，效果很好：

records = []

if __name__ == '__main__':
    for i in range(int(input())):
        name = input()
        score = float(input())
        records.append([name,score])
        records.sort(key=lambda x:x[1])
    minimum = min(records,key=lambda x:x[1])
    minimum = minimum[1]
    args = records.copy()
    
    for i in args:
        if(i[1] == minimum):
            records.remove(i)
        else:
            continue
    
    minimum = min(records,key=lambda x:x[1])
    minimum = minimum[1]
    a=[]
    for i  in records:
        if(i[1] == minimum):
            a.append(i[0])

    a.sort()
    for i in a:
        print(i)

这样，您还可以删除多个相同的最低分（如果存在）。运行它没有 copy() 并且处理原始列表会导致循环错过几个相同的最低分数。

如何通过第二个值将第二低的列表查找到嵌套列表中？

How to find the second lowest lists into a nested list by their second value?

python

list

nested-lists

python-3.x