Value error: list.remove(x) x not in list

Question

我必须找出最低分数，然后将其从列表中删除。这是我的代码：

n = int(input())
list = []
for i in range(n+1):
    name = input()
    score = float(input())
    list.append([name,score])
print(list)

min_scores = list[0][1]
for x in list:
  if (x[1]<float(min_scores)):
    min_scores = x[1]
  list.remove(min_scores)
print(list)

读取错误消息

ValueError                                Traceback (most recent call last)
<ipython-input-135-44e6d871c75e> in <module>
     11     if (x[1]<float(min_scores)):
     12         min_scores = x[1]
---> 13     list.remove(min_scores)
     14 print(list)

ValueError: list.remove(x): x not in list

Answer 1

您犯了严重错误，切勿将关键字用作变量名。希望这段代码对您有所帮助。

n = int(input())
lst = []  # never use a keyword as a variable name
for i in range(n+1):
    name = input()
    score = float(input())
    lst.append([name,score])

print(lst)

min_score = [] 
for x, y in lst:  # use two variables because lst is a 2 dimential array
    min_score.append(y)  # adding y into min_score list


min_score.sort() # sorting min_score
del(min_score[0]) #  after sorting first value of array will be minimum, so remove it.

print(min_score)

Answer 2

n = int(input())
listOfScores = []

for i in range(n + 1):
    name = input()
    score = float(input())
    listOfScores.append([name, score])

print(listOfScores)

minScoreList = listOfScores[0]

for x in listOfScores:
    if x[1] < float(minScoreList[1]):
        minScoreList = x
listOfScores.remove(minScoreList)

print(listOfScores)

Answer 3

老实说，这在程序员社区中被称为垃圾代码。如果你需要算法

找出最低分数，然后将其从列表中删除。然后这样做

scores_list = []


def get_scores():
    print('Type "stop" to stop the program')
    x = input('Enter score:')
    if x.isdigit():
        scores_list.append(x)
    if x == 'stop':
        return print('Minimum score is:', min(scores_list))

    get_scores()


get_scores()

并注意：切勿在变量名中进一步使用关键字，这可能会导致您的程序无法正常工作

Answer 4

只需将内置 min 函数与适当的键函数一起使用，并将其作为参数传递给 remove。

n = int(input())
lst = []
for i in range(n+1):
    name = input()
    score = float(input())
    lst.append([name, score])
    
lst.remove(min(lst, key=lambda t: t[1]))

也不要使用内置类型作为变量名（列表不是其他答案所说的关键字）。

Value error: list.remove(x) x not in list

Value error: list.remove(x) x not in list

python

nested-lists