这是我一直在努力的程序,但我遇到了一些问题。执行此程序时,它会打印滑冰者列表并有两次信息,但我不想要,但我不知道如何修复它
# THESE ARE LISTS TO STORE INDIVIDUAL SKATERS AND THERE SCORES AND MEDIUM
l_list = []
list = []
#STATMENT TO START WHILE LOOP AND TO END IT
re = True
while re == True:
#STORING THE SKATERS NAME AND ADDING OPTION TO STOP ADDING SKATERS IN TO PROGRAM
a_name = input("Please store the skaters name, if there are no more skaters enter 'stop'")
#PRINTING ALL SKATERS AND THERE SCORES AND MEDIUM AND ADDING OPTION TO ADD MORE OR END PROGRAM
if a_name == 'stop':
#were pritn(l_list was)---------------------------------------------------------------------------------------
for s in l_list:
print(*s)
exit = input("Would you like to exit? 'y/n'. ")
if exit == 'y':
break
#WHILE LOOP TO GET SCORES AND STORE THEM IN A LIST
scores = []
repeat = True
while repeat == True:
#ASKING FOR THE SCORES AND A INPUT TO END THIS PROGRAMS FUNCTION WHICH WILL STORE THE SCORES TO THE LIST(scores)
score = input("please type in your scores one at a time, and 'done' when finished.")
if score == "done":
repeat = False
else:
scores.append(score)
#SORTING SCORES, REMOVING HIGHEST AND LOWEST SCORES
scores.sort()
scores.remove(max(scores))
scores.remove(min(scores))
#STORING THE VARIABLE(a_name) AND THE LIST(scores) IN THE LIST(list)
#THEN STORING THE LIST(list) INTO THE LIST(l_list)
list.append(a_name)
list.append(scores)
l_list.append(list)
#CONVERTING THE CONTENTS OF THE LIST(scores) IN TO FLOATS SO THAT IT CAN FIND TEH MEDIUM/AVERAGE SCORE USING LEN AND SUM
#THEN STORING THE VARIABLE(med) INTO THE LIST(list)
scores = [float(i) for i in scores]
med = (sum(scores) / float(len(scores)))
list.append(med)
#RUN FUNCTION
start()
当我使用一些测试结果运行程序时,这就是它打印的内容:
[['thomas', ['2', '3', '4', '58'], 16.75, 'renee', ['2', '3', '4', '5', '6'], 4.0], ['thomas', ['2', '3', '4', '58'], 16.75, 'renee', ['2', '3', '4', '5', '6'], 4.0]]
Would you like to exit? 'y/n'.
你可以看到它打印两次,我不想要。
答案 0 :(得分:2)
每次循环都需要重新创建list
。
第一次循环时,list
和l_list
都是空的:您获得姓名和分数,将其附加到list
并将其附加到l_list
。一切都很好。
然后你再次为你的下一个选手穿过循环。您将他们的姓名和分数附加到list
已经拥有上一个滑冰运动员姓名并得分的l_list
。然后再将整个事物追加到l_list
。现在list
有两份list
,list = []
有两名选手和他们的分数。
在致电list.append()
之前添加|Company |Period |Value |
-------------------------------
|CompA |January |10000 |
|CompB |January |99999 |
|CompA |February |88888 |
可以解决问题。
答案 1 :(得分:2)
你有一个良好的开端,但我想警告你不要使用变量名list
。这是Python中类型的名称,因此您可以使用lst
或list_
。在这种情况下,我称之为skater_list
。 This question涵盖了不用于变量的所有名称。
当您在列表中添加姓名和分数时,您永远不会将其清除,因此您需要多次添加它们。
但是,我不认为列表实际上是您最有用的数据结构。我认为一个dict
,其中键是名称,值是得分最好。
首先,创建一个dict
而不是您的列表:
final_dict = {}
打印列表时,请改为打印dict
:
if a_name == 'stop':
#were pritn(l_list was)---------------------------------------------------------------------------------------
for name, scores in final_dict.items():
score, med = scores
print(name, score, med)
删除您的代码以将项目添加到列表
scores.sort()
scores.remove(max(scores))
scores.remove(min(scores))
scores = [float(i) for i in scores]
med = (sum(scores) / float(len(scores)))
相反,将它们添加到字典中。在这里,我将钥匙设为滑冰运动员的名字,并存储了所有剩余的分数和平均分数。
final_dict[a_name] = (scores, med)
注意,您仍然需要清除分数列表:
scores = []
这里的输出类似于:
Adam [7.0, 8.0] 7.5
Joe [6.0, 7.0] 6.5