这里给出一个嵌套列表：
nl = [['Harsh', 20], ['Beria', 20], ['Varun', 19], ['Kakunami', 19], ['Vikas', 21]]
现在我必须通过第二个值在嵌套列表中找到第二低的列表。并将第二低的列表附加到另一个列表中。
所以输出应该是：
['Harsh', 20], ['Beria', 20]
我编写了以下代码，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)
获取总元素的
minfrom 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，如果找到较低的元素，则从临时列表中删除所有元素，如果找到同样较低的元素，则将其附加：
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)
我通过使用集合查找第二低值，然后从列表中选择具有相同值的元素来做到这一点。
#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] ]
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)
heapq.nlargest返回包含数据集中 n 个最大元素的列表
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]]
您可以尝试下面的代码。效果很好。
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)
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)
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)
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数（如果存在）。在没有
copy()students=[]
if __name__ == '__main__':
for _ in range(int(input())):
name = input()
score = float(input())
students.append([name,score])
students.sort(key=lambda x:x[1] )
students.remove(min(students,key=lambda x:x[1] ))
names=[]
score=students[0][1]
for student in students:
if student[1]==score:
names.append(student[0])
names.sort()
for each in names:
print(each)
我做了一个简单的方法来解决这个问题，它是基本的而不是进步的
if __name__ == '__main__':
records = []
for _ in range(int(input())):
name = input()
score = float(input())
records.append([name,score])
sorted_records = sorted(records, key=lambda x: x[1])
second_lowest = 0
for item in sorted_records:
if sorted_records[0][1] < item[1]:
second_lowest = item[1]
break
second_lowest_list = [x for x in sorted_records if second_lowest==x[1]]
second_lowest_list.sort()
for element in second_lowest_list:
print(element[0])
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查找第二个唯一的最低值。 比拥有这个文件，在 nl 中查找与找到的值匹配的元素。
我遇到过类似的问题，在参考此页面后，我得到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=", ")