给出学生的物理班级中每个学生的姓名和成绩，将它们存储在嵌套列表中，并打印出成绩第二低的任何学生的姓名。注意：如果有多个同年级的学生，请按字母顺序排列其姓名，并将每个姓名打印在新行上。
输入：
5
Harry
37.21
Berry
37.21
Tina
37.2
Akriti
41
Harsh
39
输出：
>>Berry
>>Harry
我做到了
if __name__ == '__main__':
l2=[]
l1=[]
for i in range(int(input())):
name = input()
l1.append(name)
score = float(input())
l1.append(score)
l2.append(l1)
print(l2)
但它输出：
>>[['Harry', 37.21], ['Berry', 37.21], ['Tina', 37.2], ['Akriti', 41], ['Harsh', 39]]
答案 0 :(得分：2)
if __name__ == '__main__':
n = []
s = []
for _ in range(int(input())):
name = input()
n.append(name)
score = float(input())
s.append(score)
a=max(s)
b=min(s)
for i in range(len(n)):
if s[i]<a and s[i]>b:
a=s[i]
for j in range(len(n)):
if s[j]==a:
print(n[j])
答案 1 :(得分：2)
您可以在Linux或Windows Shell的命令行中将以下内容作为脚本运行。
它希望您通过首先输入一个整数来给出要遵循的记录数，从而输入数据。
该代码适用于以上问题中给出的示例以及其他情况。它检查最重要的与数据相关的边缘情况，这可能会带来麻烦。
if __name__ == '__main__':
# Note, Python evaluates expressions from left to right, which helps us
# to do the following. After reading the integer used within the range
# it will read [str, float] combinations as many times as the
# range requests
inputs = [[str(input()), float(input())] for dummy in range(int(input()))]
# Given the inputs, find the unique score values first. Then we take
# the second lowest from it and search for the names of the students
# with this score.
#
# Additionally, we validate the inputs to check for edge cases where
# the question cannot be answered at all. This depends on the data
# given.
#
# To get the unique score values we use the trick to convert to a
# set and back to a list
#
unique_scores = list(set([score for name, score in inputs]))
# Let's deal with the edge cases. For these we return simply None.
students = None
if len(unique_scores) == 0:
# It might be that no data has been provided. We could handle this earlier
# but let's do it here, with the other cases.
# Let's return None in this case
#
print('No data provided. Cannot answer your question.')
elif len(unique_scores) == 1:
# Suppose, you have quiete some students, but all have the same grade.
# So there is no second lowest. Let's be strict now and return None
# in this case too.
#
print('Found only one distinct score. Cannot answer your question.')
else:
# Finally, after all checks passed we return the names of those students,
# which have the second lowest case.
unique_scores.sort()
second_lowest_score = unique_scores[1]
students = [name for name, score in inputs if score == second_lowest_score]
# Finally, present the result
print(students)
答案 2 :(得分：1)
if __name__ == '__main__':
score_list={}
for _ in range(int(input())):
name = input()
score = float(input())
score_list[name]=score
second_val=sorted(set(score_list.values()))[1]
print('\n'.join(sorted ([name for name,val in score_list.items() if val==second_val])))
说明：
我用过字典。我首先获取输入，然后通过创建值列表然后对其进行排序以获取值来找出第二个最小值。
然后在print语句的下一行中，我从dict中取出键为name，值为val以使用值来验证条件是否满足条件以匹配第二个最小值然后命名排序（如果不止一个）
并通过 '\n' 加入 next 并打印。
答案 3 :(得分：1)
这是我为解决问题所做的诚实努力。
# Run it in Python 3
students = int(input()) # taking the inputs
parentlist = [] # for saving both the students and the marks in an array
scores = list() # seperate list for the scores
newlist = []
sec_low = []
# saving elements in the array
for i in range(0, students):
name = input()
score = float(input())
parentlist.append([name, score])
scores.append(score)
scores.sort() # sorting score in ascending order
res = []
[res.append(x) for x in scores if x not in res] # with the help of res checking for repetation of marks in the scores list
# taking the marks in the 1st position and comparing with the 2nd lowest in the parentlist.
for i in parentlist:
if (res[1] == i[1]):
newlist.append(i[0]) # appending the names of students having 2nd lowest marks to newlist
# sorting the newlist alphabetically and printing values
newlist.sort()
for i in newlist:
print(i)
答案 4 :(得分：1)
if __name__ == '__main__':
total = []
for i in range(int(input())):
name = input()
score = float(input())
name_score = list((score, name))
total.append(name_score)
total.sort()
min_mark = total[0][0]
count = 0
for i in range(len(total)):
if min_mark == total[i][0]:
count = count+1
if count >= 1:
for j in range(count):
total.pop(0)
nd_min_mark = total[0][0]
for i in range(len(total)):
if nd_min_mark == total[i][0]:
print(total[i][1])
答案 5 :(得分：1)
if __name__ == '__main__':n=int(input())
names=[]
scores=[]
for i in range(n):
name=input()
names.append(name)
score=float(input())
scores.append(score)
dic=dict(sorted(zip(names,scores)))
mini=(min(scores))
cnt=scores.count(mini)
for i in range(cnt):
scores.remove(mini)
minim=min(scores)
for i,j in dic.items():
if j==minim:
print(i)
答案 6 :(得分：1)
这个问题肯定来自Hacker等级。下面是简单的解决方案。
if __name__ == '__main__':
student = []
for _ in range(int(input())):
name = input()
score = float(input())
student.append([score,name])
# Sort list in Ascending order
student.sort()
# Assign the first score to a list which will be the minimum
grade = student[0][0]
# Loop through the list by excluding first item in the list and find the second lowest
for i in range(1,len(student)):
if grade < student[i][0]:
grade = student[i][0]
break
# Loop through the list and print names based on the second lowest score
for i in range(1,len(student)):
if student[i][0] == grade:
print(student[i][1])
答案 7 :(得分：1)
Ctrl+D答案 8 :(得分：1)
这是上述问题的解决方法
temp_list = []
n=int(raw_input())
if 2<=n<=5:
for _ in range(n):
name = raw_input()
score = float(raw_input())
temp_list.append((name,score))
required_dict = dict(temp_list)
scores_list = [x[1] for x in temp_list]
second_last_score = sorted(list(set(scores_list)))[1]
temp_list = []
for key,val in required_dict.items():
if val == second_last_score:
temp_list.append(key)
for _ in sorted(temp_list):
print(_)
else:
print("Please provide n >= 2 and n<=5")
答案 9 :(得分：0)
请注意，最后我们需要排序以按字母顺序显示。
name_and_scores = []
# Store name and scores
for _ in range(int(input())):
name_and_scores.append([input(), float(input())])
# Finding second minimum score
second_min_score = sorted(list(set([marks for name, marks in name_and_scores])))[1]
# Printing second minimum names
print('\n'.join([name for name,marks in sorted(name_and_scores) if marks == second_min_score]))
答案 10 :(得分：0)
我这样做很复杂
n=int(input())
list1 = []
for j in range(0,n):
name = input()
score = float(input())
list1.append([name,score])
k=min([list1[i][1] for i in range(0,n)])
for i in range(0,len(list1)-1):
if(list1[i][1]==k):
list1.remove(list1[i])
k=min([list1[i][1] for i in range(0,len(list1))])
list2=[]
for i in range(0,len(list1)):
if(list1[i][1]==k):
list2.append(list1[i])
list2.sort()
for i in range(0,len(list2)):
print(list2[i][0])
答案 11 :(得分：0)
简单有效
if __name__ == '__main__':
ls1=[]
ls2=[]
ls4=[]
for _ in range(int(input())):
name = input()
score = float(input())
ls1.append([name, score])
for i in ls1:
ls2.append(i[-1])
ls3 = list(set(ls2))
for j in ls1:
if j[-1]==ls3[-2]:
ls4.append(j[0])
ls4.sort()
for k in ls4:
print(k)
答案 12 :(得分：0)
if __name__ == '__main__':
lst=[]
for _ in range(int(input())):
name = input()
score = float(input())
data=tuple((name,score))
lst.append(data)
lst.sort(key = lambda x: (x[1],x[0]))
print(lst)
答案 13 :(得分：0)
对于初学者，此代码可能会有所帮助：
if __name__ == '__main__':
N = int(input())
l2 = []
if 2 <= N <= 5:
for i in range(N):
l1 = []
name = input()
l1.append(name)
score = float(input())
l1.append(score)
l2.append(l1)
scores = set()
for i in l2:
scores.add(i[1])
names = []
for i in l2:
if i[1] == sorted(list(scores))[1]:
names.append(i[0])
for i in sorted(names):
print(i)
答案 14 :(得分：0)
if __name__ == '__main__':
students=[[input(),float(input())] for _ in range(0,int(input()))] #read the students names and marks into an array
students.sort(key=lambda x: (x[1],x[0]),reverse=True) #sort the elements(priority to marks then, names)
marks = [i[1] for i in students] # get marks from sorted array
for i in sorted(students[marks.index(marks[marks.index(min(marks))-1]):marks.index(marks[marks.index(min(marks))])]): #get index of lowest grades and second lowest grades then print the elements between these two points.
print(i[0]) #print the names from the list
students=[[input(),float(input())] for _ in range(0,int(input()))]
上面这行是把所有的输入数据放到一个数组中（students）。
首先我们不给学生。这将由 int(input()) 行接收。
然后 for 循环运行，剩余的输入按顺序转到 [input(),float(input())]（名称在先，然后是标记）
输入示例：
5
Harry
37.21
Berry
37.21
Tina
37.2
Akriti
41
Harsh
39
数组内部的数据如下：
[['Harry', 37.21],
['Berry', 37.21],
['Tina', 37.2],
['Akriti', 41],
['Harsh', 39]]
这一行是按标记和名称对数组进行排序（标记优先，名称第二）
students.sort(key=lambda x: (x[1],x[0]),reverse=True)
排序后数组里面的数据是这样的：
[['Akriti', 41],
['Harsh', 39],
['Harry', 37.21],
['Berry', 37.21],
['Tina', 37.2]]
这一行将从学生数组中读取排序的标记
marks = [i[1] for i in students]
在数组里面，数据会是这样的：
[41, 39, 37.21, 37.21, 37.2]
这些行可能有点混乱，但它的工作很简单。
for i in sorted(students[marks.index(marks[marks.index(min(marks))-1]):marks.index(marks[marks.index(min(marks))])]): #get index of lowest grades and second lowest grades then print the elements between these two points.
print(i[0]) #print the names from the list
为了便于理解，截取一部分
(students[marks.index(marks[marks.index(min(marks))-1]):marks.index(marks[marks.index(min(marks))])])
上面这行代码是找出次低标记的索引范围。 如果我们将上面的示例输入提供给此代码，则索引位置将在 2-3 之间 所以（哈利的标记的位置是索引 2，浆果的标记是索引 3），下面的代码和上面的代码一样（如果我们给出上面提到的示例输入）
students[2:4]
这些行的输出类似于
[['Harry', 37.21], ['Berry', 37.21]]
我们只想要这个列表中的名字，我们还想按字母顺序对名字进行排序。
for i in sorted(2:4):
print(i[0])
如果我们提供与上面提到的相同的输入，上面和下面几行的工作方式将相同。
for i in sorted(students[marks.index(marks[marks.index(min(marks))-1]):marks.index(marks[marks.index(min(marks))])]):
print(i[0])
最终输出是
Berry
Harry
答案 15 :(得分：0)
# Uncomment this line to define a global platform for your project
# platform :ios, '9.0'
答案 16 :(得分：0)
def minl(l):
mx=0
ans=[]
for [i,j] in l:
if j>mx:
mx=j
for [i,j] in l:
if mx>j:
mx=j
for [i,j] in l:
if j==mx:
ans+=[[i,j]]
return ans
if __name__ == '__main__':
l=[]
m=[]
for _ in range(int(input())):
name = input()
score = float(input())
r=[name,score]
l.append(r)
for i in minl(l):
l.remove(i)
for i in minl(l):
m+=[i[0]]
for i in sorted(m):
print(i)
答案 17 :(得分：0)
n = int(input())
mainlist = []
marks_list = []
for _ in range(n):
name = input()
marks = float(input())
marks_list.append(marks)
list1 = []
list1.append(name)
list1.append(marks)
mainlist.append(list1)
marks_list_set = sorted(marks_list)
nameslist = []
length = len(mainlist)
i = 0
while i < length:
if(mainlist[i][1] == marks_list_set[1]):
nameslist.append(mainlist[i][0])
i += 1
nameslist=sorted(nameslist)
for _ in nameslist:
print (_)
答案 18 :(得分：0)
students_scores = []
for _ in range(int(input())):
name = input()
score = float(input())
students_scores.append([name, score]) # create input array
unique_scores_set = set((x[1] for x in students_scores)) # calculate unique scores
second_lowest_score = sorted(unique_scores_set)[1] # calculate second lowest score
output = [y[0] for y in students_scores if y[1] == second_lowest_score] # create output array containing names of students with the second lowest score
output.sort() # sort the names aplphabetically in place
print (*output, sep="\n") #print output list one line at a time
答案 19 :(得分：0)
我的回答很简单
n = int(input())
l = [[input(),float(input())] for _ in range(0,n)]
s = sorted(l,key=lambda k: (k[1],k[0]))
minvalue = min(s,key=lambda k : k[1])
while s[0][1] == minvalue[1] :
s.remove(s[0])
minvalue = min(s,key=lambda k : k[1])
while s and s[0][1] == minvalue[1]:
print(s[0][0])
s.remove(s[0])
答案 20 :(得分：0)
if __name__ == '__main__':
student={}
for _ in range(int(input())):
name = input()
score = float(input())
#tmp=[name]
if score in student:
temp=[]
temp=student[score]
temp.append(name)
student[score]=temp
else:
temp=[]
temp.append(name)
student[score]=temp
key=list(student.keys());
key.sort()
temp=list(student[key[1]])
temp.sort()
for i in (temp):
print(i)
#print(student)
#print(key)
这是所有后续措施和易于理解的代码
答案 21 :(得分：0)
要回答您的问题，您可以将输入作为嵌套列表
$('#c_Roles :checked').map(function () {
return $(this).val();
});
我相信这是一个黑客等级问题 这是解决方案
n=int(input())
arr=[[input(),float(input())] for _ in range(0,n)]
答案 22 :(得分：0)
我晚参加聚会吗？ 这是我的解决方案。尽管它被划分为仅执行一件事的功能，但并不是最有效的（我认为托马斯的效果最好）：
def get_second_lowest_value(sco):
s = list(set(sco))
if len(s) == 1:
return s[0]
else:
s.sort()
return s[1]
def get_all_indexes_of_an_element(arr, element):
answer = []
for i, a in enumerate(arr):
if a == element:
answer.append(i)
return answer
if __name__ == '__main__':
names = []
scores = []
for _ in range(int(input())):
names.append(input())
scores.append(float(input()))
indexes_of_second_lowest_score = get_all_indexes_of_an_element(
scores, get_second_lowest_value(scores)
)
second_lowest_names = [names[i] for i in indexes_of_second_lowest_score]
second_lowest_names.sort()
for p in second_lowest_names:
print(p)
关于托马斯答案的进一步阐述，使其更加简洁：
if __name__ == '__main__':
names = []
scores = []
for _ in range(int(input())):
names.append(input())
scores.append(float(input()))
sorted_scores = sorted(list(set(scores)))
s2 = sorted_scores[1] if len(sorted_scores) > 1 else sorted_scores[0]
stu = sorted([n for n, s in zip(names, scores) if s == s2])
print(stu)
答案 23 :(得分：0)
Student=[]
Grade=[]
scorel=[]
namel=[]
for i in range(int(input())):
name = input()
Student.append(name)
score = float(input())
Grade.append(score)
scorel.append(score)
scorel.sort()
for j in range (len(Grade)):
if scorel[-2]==scorel[0]:
if scorel[-1]==Grade[j]:
namel.append(Student[j])
else:
if scorel[0]==scorel[1]:
if scorel[2]==Grade[j]:
namel.append(Student[j])
if scorel[0]!=scorel[1]:
if scorel[1]==Grade[j]:
namel.append(Student[j])
namel.sort()
for k in range (len(namel)):
print(namel[k]).
我不知道这是否是一种很好的编码方式，但是它可以工作。
答案 24 :(得分：0)
这可能是解决此问题的最有效方法。
'''
This solution is highly efficient because it uses
heapsort and binary search. It sorts only one time
and then uses binary search to quickly find the
boundaries of all of the values that have the same
rank. Also, since it is sorted, it breaks out of the
loop once it collects all of the values for the rank.
'''
from bisect import bisect, bisect_left
from heapq import heappush, heappop
n = 2
student_scores = []
scores_sorted = []
student_scores_sorted = []
ordinal_map = {}
seen = set()
enum = 1
if __name__ == '__main__':
for _ in range(int(input())):
name = input()
score = float(input())
heappush(student_scores, (score, name))
for i in range(len(student_scores)):
val_to_append = heappop(student_scores)
scores_sorted.append(val_to_append[0])
student_scores_sorted.append(val_to_append)
if val_to_append[0] not in seen:
if enum > n:
break
seen.add(val_to_append[0])
ordinal_map[enum] = val_to_append[0]
enum += 1
low = bisect_left(scores_sorted, ordinal_map[n])
high = bisect(scores_sorted, ordinal_map[n])
for i in range(low, high):
print(student_scores_sorted[i][1])
答案 25 :(得分：0)
n=int(input())
stud=[]
grad=[]
for i in range(n):
stu=input()
stud.append(stu)
grd=float(input())
grad.append(grd)
dic=dict(sorted(zip(stud,grad)))
mini=(min(grad))
cnt=grad.count(mini)
for i in range(cnt):
grad.remove(mini)
minf=min(grad)
for i,j in dic.items():
if j==minf:
print(i)
答案 26 :(得分：0)
from operator import itemgetter
if __name__ == '__main__':
listt = [] #list will contain name and scores, to know the relation between then
names = [] #list will contain output names
n = int(input())
for _ in range(n):
name = input()
score = float(input())
nested_list = [name,score]
listt.append( nested_list )
listt.sort( key = itemgetter(1) )
minimum = listt[0][1] #contain the minimum score
i = 0
for i in range(0,n):
# if an element its highest than the minimum, its the second lowest
if(listt[i][1] > minimum):
# the second lowest
minimum=listt[i][1]
names.clear() # clear the names list if we before inserted names in else block
while(listt[i][1] == minimum): # of all scores equals to the minimum
names.append( listt[i][0] ) #Save their names
if i!=n-1: # if is not the last score
i+=1
else:
break
break # once we obtain all equals scored, break loop
else: # its possible that all students have same score
if (listt[i][1] == minimum): # if all score are equal to the minimum
names.append( listt[i][0] )
names.sort()
for name in names:
print(name)
答案 27 :(得分：0)
对最低（f）和第二最低（s）使用两个指针，并一次两次地比较记录。这应该是O（N）复杂度。
def get_second_lowest_grade_students(student_recs):
f, s = float('inf'), float('inf')
fl_names, sl_names = [], []
for (score, names) in student_recs.items():
if score < f:
s = f
f = score
sl_names = fl_names
fl_names = names
elif score < s and score != f:
s = score
sl_names = names
sl_names.sort()
return sl_names
if __name__ == '__main__':
student_recs = {}
for _ in range(int(input())):
name = input()
score = float(input())
if score not in student_recs:
student_recs[score] = []
student_recs[score].append(name)
print("\n".join(get_second_lowest_grade_students(student_recs)))
答案 28 :(得分：0)
# function for removing duplicate elements
def Remove(scorelist):# scorelist is sorted list with duplicate element
uniquelist=[] #uniquelist is list without duplicate element
for num in scorelist:
if num not in uniquelist:
uniquelist.append(num)
# print("this is unique list",uniquelist)
return uniquelist
if __name__ == '__main__':
newlist = [] #list contain name and number both
scorelist=[] #list contain number only
finallist=[]
n =int(input())
for i in range(0,n):
name = input()
score = float(input())
scorelist.append(score)
lst=[name,score]
newlist.append(lst)
scorelist.sort()
uniquelist=[]
uniquelist=Remove(scorelist) #sorted and unique list
seclow=uniquelist[1]
for i in range(len(scorelist)):
if newlist[i][1]==seclow :
finallist.append(newlist[i][0])
finallist.sort()
print(*finallist, sep="\n")
答案 29 :(得分：0)
您可以使用：
start=int(input())
for _ in range(start):
print("enter Name: ")
name = input()
print("enter Score: ")
score = float(input())
d[name]=score
l = sorted(d.items(),key=lambda x: x[0])
m=min(l,key=lambda item:item[1])
#print("minumul este: ")
#print(m)
l.remove(m)
print("lista este: ")
for i in l:
if i[1]==m[1]:
#print(rf"elementul {i[1]} a fost exclus")
l.remove(i)
print(l)
n=min(l,key=lambda item:item[1])
#print(n)
nn=[]
for i in l:
if i[1]==n[1]:
nn.append(i[0])
for i in sorted(nn):
print(i)
答案 30 :(得分：0)
这是我的解决方案。
students = [[input(), float(input())] for _ in range(int(input()))]
students.sort(key=lambda x: (x[1], x[0]))
first_min_score = students[0]
second_min_score = min(students, key=lambda x: (x[1]==first_min_score[1], x[1]))
for student in students:
if student[1] == second_min_score[1]:
print(student[0])
答案 31 :(得分：0)
一个老问题，但添加了一个答案，以便可以获取帮助
这确实是hackerrank的采访问题。
下面是我对问题的简单解决方案-我只是遍历嵌套的成绩单。并且在每次迭代中，都会将新分数与lowest分数和第二低分数（slowest）进行比较，同时维护names那些分数等于第二低分数的列表。
注意：names始终与slowest得分相关联，如果slowest发生变化，我们也会更改names。如果新分数等于slowest分数–请添加新名称。评论将帮助您进一步...
def second_lowests(grades):
"""
returns list of names of students with second lowest score
"""
# intialize the `lowest` and second lowest `slowest` score and names
grades = iter(grades)
lname, lowest = next(grades)
slname, slowest = next(grades)
if slowest < lowest:
lowest, slowest = slowest, lowest
names = [lname]
elif slowest == lowest: # we don't know, if lowest can be second lowest!
names = [lname, slname]
else:
names = [slname]
for name, score in grades:
if score == slowest:
names.append(name)
continue
if score == lowest:
continue
if score < lowest:
if slowest == lowest:
pass
else:
names = [lname]
lowest, slowest = score, lowest
lname = name
elif score < slowest:
slowest = score
names = [name]
elif score > slowest and slowest == lowest:
slowest = score
names = [name]
if slowest == lowest: # all have same score
return []
names.sort()
return names
if __name__ == '__main__':
nested_list = []
for _ in range(int(raw_input())):
name = raw_input()
score = float(raw_input())
nested_list.append([name, score])
assert 2 <= len(nested_list) <= 5
print ("\n".join(second_lowests(nested_list)))
我不知道这是否是更好的解决方案，但它可以通过所有测试用例。
答案 32 :(得分：0)
这是我此Hackerrank练习的解决方案：
if __name__ == '__main__':
students = []
for _ in range(int(input())):
name = input()
score = float(input())
new = [name, score]
students.append(new)
def removeMinimum(oldlist):
oldlist = sorted(oldlist, key=lambda x: x[1])
min_ = min(students, key=lambda x: x[1])
newlist = []
for a in range(0, len(oldlist)):
if min_[1] != oldlist[a][1]:
newlist.append(oldlist[a])
return newlist
students = removeMinimum(students);
# find the second minimum value
min_ = min(students, key=lambda x: x[1])
# sort alphabetic order
students = sorted(students, key=lambda x: x[0])
for a in range(0, len(students)):
if min_[1] == students[a][1]:
print(students[a][0])
答案 33 :(得分：-1)
res={}
for _ in range(int(input())):
name = input()
score = float(input())
res[name]=score
new_list = list(dict.fromkeys(list(res.values())))
new_list.sort()
final_list = []
for key, value in res.items():
if value == new_list[1]:
final_list.append(key)
final_list.sort()
print(*final_list, sep = "\n")
答案 34 :(得分：-1)
如果名称 =='主要'：
names = []
scores = []
for _ in range(int(input())):
name = input()
score = float(input())
names.append(name)
scores.append(score)
##在本节中，创建一个循环以查找倒数第三或最高的那个，依此类推
print(names)
print(scores)
min_score = min(scores)
len_scores = len(scores)
print(min_score)
i = 0
while i < len_scores:
print(i)
if scores[i] == min_score:
print(i)
scores.pop(i)
names.pop(i)
len_scores = len_scores-1
else:
i = i+1
print(names)
print(scores)
min_score = min(scores)
print(min_score)
final_names_list = []
for j in range(len(scores)):
if scores[j] == min_score:
print(j)
final_name = names[j]
final_names_list.append(final_name)
print(final_names_list)
final_names_list.sort()
for k in range(len(final_names_list)):
print( final_names_list[k])