Question

我对python比较陌生，目前在网上找到一个学校项目。当我运行此代码时，它应该计算“Action”和“Sport”出现在阵列中的次数。

forename = ["Joe", "George", "Oliver"]
HistoryGenre=[["Action", "Action", "Action", "Action", "Sport", "Sport", "Action", "Action", "Action", "Sport"], ["Sport", "Sport", "Sport", "Sport", "Action", "Action", "Sport", "Sport", "Sport", "Action"], ["Action", "Action", "Sport", "Sport", "Action", "Action", "Action", "Sport", "Sport", "Sport"]]
rr=1
ActionCounter=0
SportCounter=0
while rr==1:
    rec=input("Who would you like to recommend games for?")
    if rec in forename:
        rr+=1
        r=forename.index(rec)
        RepeatIndex=0
        for i in HistoryGenre[r]:
            if HistoryGenre[r:RepeatIndex]=="Action":
            ActionCounter+=1
            RepeatIndex+=1
            else:
                SportCounter+=1
                RepeatIndex+=1
            if RepeatIndex==9:
                print(ActionCounter)
                print(SportCounter)

当我运行此代码时，ActionCounter打印为0，SportCounter打印为9.我不明白为什么会发生这种情况或为什么会这样，考虑到输出应该是7和3.这很可能是初学者的错误。< / p>

Answer 1

您的代码所在的问题在这里

if HistoryGenre[r:RepeatIndex]=="Action":

要引用HistoryGenre [r]中的第n项，请执行

HistoryGenre[r][n]

但是你正在使用HistoryGenre [r]进行for循环，所以你可以做到

for i in HistoryGenre[r]:
    if i == "Action":
        ...

我不知道你给出的项目是否禁止计数功能，但它会让你的生活更轻松。您可以在每个列表中找到这样的计数，而不是添加到ActionCounter和SportCounter变量：

ActionCounter = HistoryGenre[r][:9].count("Action")

[：9]获取列表中的前9个元素，这就是您想要的所有元素。此外，您可以使用字典，而不是两个并行列表。在字典中，元素具有键，因此要引用某个元素，可以使用其键。例如

 namebook = {"Joe":["Action", "Action", "Action", "Action", "Sport", "Sport", "Action", "Action", "Action", "Sport"], "George":["Sport", "Sport", "Sport", "Sport", "Action", "Action", "Sport", "Sport", "Sport", "Action"], "Oliver":["Action", "Action", "Sport", "Sport", "Action", "Action", "Action", "Sport", "Sport", "Sport"]}

要获得Joe的书籍类型，请使用

namebook["Joe"]

使用这些更改，可以缩短您的程序。

namebook = {"Joe":["Action", "Action", "Action", "Action", "Sport", "Sport", "Action", "Action", "Action", "Sport"], "George":["Sport", "Sport", "Sport", "Sport", "Action", "Action", "Sport", "Sport", "Sport", "Action"], "Oliver":["Action", "Action", "Sport", "Sport", "Action", "Action", "Action", "Sport", "Sport", "Sport"]}
inputname = input()
print(namebook[inputname][:9].count("Action"))
print(namebook[inputname][:9].count("Sport"))

Answer 2

这不能回答您的问题，但由于此处已经回答，我想出的是如果您计算每个子列表中出现"Sports"和"Action"的次数想要实现这个。

for subList in HistoryGenre:

    string = ""

    string += ' '.join(subList) + " "

    words = {}

    for word in string.split():
        try:
            words[word] += 1
        except KeyError:
            words[word] = 1

    print(words)


Out:
    {'Sport': 3, 'Action': 7}
    {'Sport': 7, 'Action': 3}
    {'Sport': 5, 'Action': 5}

word是关键

起初words[word]不存在。所以当我尝试words[word] += 1时该程序将导致KeyError，因为密钥word不存在。

因此该程序不会导致except

KeyError KeyError

在word阻止内，密钥1设置为值word。

因为string.split()每次在1中遇到该词时都会成为关键词，word会被添加到var client = sdk.getAppAuthClient('enterprise', ENTERPRISE_ID); //filter_term == admin to share the folder with client.enterprise.getUsers({filter_term: 'ken.domen@nike.com'}, function(err, users) { var userId = users.entries[0].id; client.folders.create('0', 'New Folder', function(err, newFolder) { client.collaborations.createWithUserID(userId, newFolder.id, client.collaborationRoles.VIEWER, function(err, collaboration) { console.log(err); var fileData = fs.createReadStream('/users/kdomen/Downloads/test.txt') client.files.uploadFile(newFolder.id, 'test.txt', fileData, function(err, file) { if (err){ console.log('err: ' + err); } else{ console.log('file uploaded: ' + file); } }); }); }); });的值

数组搜索不工作 - 不明白为什么？

2 个答案: