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如何修复TypeError:“ int”对象不可迭代?(int对象不可调用怎么解决)

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针对如何修复TypeError:“int”对象不可迭代?和int对象不可调用怎么解决这两个问题,本篇文章进行了详细的解答,同时本文还将给你拓展PythonFlask,TypeError:“dict”对

针对如何修复TypeError:“ int”对象不可迭代?int对象不可调用怎么解决这两个问题,本篇文章进行了详细的解答,同时本文还将给你拓展Python Flask,TypeError:“ dict”对象不可调用、Python TypeError:“ str”对象不可用于类、python – TypeError:’KFold’对象不可迭代、Python-TypeError:“ int”对象不可调用等相关知识,希望可以帮助到你。

本文目录一览:

如何修复TypeError:“ int”对象不可迭代?(int对象不可调用怎么解决)

如何修复TypeError:“ int”对象不可迭代?(int对象不可调用怎么解决)

我正在尝试编写一个程序,该程序允许您输入班级中的学生人数,然后为每个学生输入3个测试成绩以计算平均值。我是编程新手,但会不断收到错误消息,我不了解它的含义或解决方法。这是我到目前为止所拥有的:

students=int(input(''Please enter the number of students in the class: ''))for number in students:        first_grade=(input("Enter student''s first grade: "))        second_grade=(input("Enter student''s second grade: "))        third_grade=(input("Enter student''s third grade: "))

答案1

小编典典

当你写

for number in students:

您的意图是“运行这段代码students时间,students我刚才输入的值在哪里”。但是在Python,你传递给事情一个for声明需求是某种迭代对象的。在这种情况下,您想要的只是一个rangestatement。这将生成一个数字列表,并且对这些数字进行迭代将使您的for循环能够执行正确的次数:

for number in range(students):    # do stuff

在幕后,rangeJust会生成一个序列号列表:

>>> range(5)[0, 1, 2, 3, 4]

在您的情况下,数字到底是什么并不重要;以下两个for语句将执行相同的操作:

for number in range(5):for number in [1, 3, 97, 4, -32768]:

但是,range如果您需要在循环中更改某种列表(这可能是以后需要做的事情),则使用该版本被认为更惯用且更方便。

Python Flask,TypeError:“ dict”对象不可调用

Python Flask,TypeError:“ dict”对象不可调用

遇到一个似乎很普遍的问题,但我已经完成了研究,并且看不到它在任何地方都被完全重新创建了。当我打印时json.loads(rety.text),我看到了所需的输出。但是,当我打电话给return时,它显示了这个错误。有任何想法吗?非常感谢您的帮助,谢谢。我正在使用Flask
MethodHandler

class MHandler(MethodView):    def get(self):        handle = ''''        tweetnum = 100        consumer_token = ''''         consumer_secret = ''''        access_token = ''-''        access_secret = ''''        auth = tweepy.OAuthHandler(consumer_token,consumer_secret)        auth.set_access_token(access_token,access_secret)        api  = tweepy.API(auth)        statuses = api.user_timeline(screen_name=handle,                          count= tweetnum,                          include_rts=False)        pi_content_items_array = map(convert_status_to_pi_content_item, statuses)        pi_content_items = { ''contentItems'' : pi_content_items_array }        saveFile = open("static/public/text/en.txt",''a'')         for s in pi_content_items_array:             stat = s[''content''].encode(''utf-8'')            print stat            trat = ''''.join(i for i in stat if ord(i)<128)            print trat            saveFile.write(trat.encode(''utf-8'')+''\n''+''\n'')        try:            contentFile = open("static/public/text/en.txt", "r")            fr = contentFile.read()        except Exception as e:            print "ERROR: couldn''t read text file: %s" % e        finally:            contentFile.close()        return lookup.get_template("newin.html").render(content=fr)    def post(self):        try:            contentFile = open("static/public/text/en.txt", "r")            fd = contentFile.read()        except Exception as e:            print "ERROR: couldn''t read text file: %s" % e        finally:                contentFile.close()        rety = requests.post(''https://gateway.watsonplatform.net/personality-insights/api/v2/profile'',                 auth=(''---'', ''''),                headers = {"content-type": "text/plain"},                data=fd            )        print json.loads(rety.text)        return json.loads(rety.text)    user_view = MHandler.as_view(''user_api'')    app.add_url_rule(''/results2'', view_func=user_view, methods=[''GET'',])    app.add_url_rule(''/results2'', view_func=user_view, methods=[''POST'',])

这是Traceback(请记住上面的结果在打印):

Traceback (most recent call last):  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1836, in __call__    return self.wsgi_app(environ, start_response)  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1820, in wsgi_app    response = self.make_response(self.handle_exception(e))  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1403, in handle_exception    reraise(exc_type, exc_value, tb)  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1817, in wsgi_app    response = self.full_dispatch_request()  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1478, in full_dispatch_request    response = self.make_response(rv)  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1577, in make_response    rv = self.response_class.force_type(rv, request.environ)  File "/Users/RZB/anaconda/lib/python2.7/site-packages/werkzeug/wrappers.py", line 841, in force_type    response = BaseResponse(*_run_wsgi_app(response, environ))  File "/Users/RZB/anaconda/lib/python2.7/site-packages/werkzeug/test.py", line 867, in run_wsgi_app    app_rv = app(environ, start_response)

答案1

小编典典

Flask只希望视图返回类似响应的对象。
这表示Response描述主体,代码和标题的,字符串或元组。您正在返回一个dict,这不是其中之一。由于您要返回JSON,因此请返回响应,该响应的正文中包含JSON字符串,内容类型为application/json

return app.response_class(rety.content, content_type=''application/json'')

在您的示例中,您已经有一个JSON字符串,即您发出的请求返回的内容。但是,如果要将Python结构转换为JSON响应,请使用jsonify

data = {''name'': ''davidism''}return jsonify(data)

在幕后,Flask是一个WSGI应用程序,它希望传递可调用对象,这就是为什么您会收到该特定错误的原因:dict是不可调用的,并且Flask不知道如何将其转换为可调用对象。

Python TypeError:“ str”对象不可用于类

Python TypeError:“ str”对象不可用于类

请帮助我理解这一点。我创建了一个非常简单的程序来尝试理解类。

class One(object):
    def __init__(self,class2):
        self.name = 'Amy'
        self.age = 21
        self.class2 = class2

    def greeting(self):
        self.name = raw_input("What is your name?: ")
        print 'hi %s' % self.name

    def birthday(self):
        self.age = int(raw_input("What is your age?: "))
        print self.age

    def buy(self):
        print 'You buy ',self.class2.name

class Two(object):
    def __init__(self): 
        self.name = 'Polly'
        self.gender = 'female'

    def name(self):
        self.gender = raw_input("Is she male or female? ")
        if self.gender == 'male'.lower():
            self.gender = 'male'
        else:
            self.gender = 'female'

        self.name = raw_input("What do you want to name her? ")

        print "Her gender is %s and her name is %s" % (self.gender,self.name)

Polly = Two()
Amy = One(Polly) 
# I want it to print


Amy.greeting()
Amy.buy() 
Amy.birthday()

问题代码

Polly.name() # TypeError: 'str' object is not callable
Two.name(Polly)# Works. Why?

为什么在类实例Polly上调用方法无效?我很迷路。我已经看过http://mail.python.org/pipermail/tutor/2003-May/022128.html和其他与此类似的Stackoverflow问题,但我不明白。非常感谢。

python – TypeError:’KFold’对象不可迭代

python – TypeError:’KFold’对象不可迭代

我正在关注 Kaggle中的一个内核,主要是,我正在关注 A kernel for Credit Card Fraud Detection.

我到达了需要执行KFold的步骤,以便找到Logistic回归的最佳参数.

以下代码显示在内核本身,但由于某种原因(可能是旧版本的scikit-learn,给我一些错误).

def printing_Kfold_scores(x_train_data,y_train_data):
    fold = KFold(len(y_train_data),5,shuffle=False) 

    # Different C parameters
    c_param_range = [0.01,0.1,1,10,100]

    results_table = pd.DataFrame(index = range(len(c_param_range),2),columns = ['C_parameter','Mean recall score'])
    results_table['C_parameter'] = c_param_range

    # the k-fold will give 2 lists: train_indices = indices[0],test_indices = indices[1]
    j = 0
    for c_param in c_param_range:
        print('-------------------------------------------')
        print('C parameter: ',c_param)
        print('-------------------------------------------')
        print('')

        recall_accs = []
        for iteration,indices in enumerate(fold,start=1):

            # Call the logistic regression model with a certain C parameter
            lr = LogisticRegression(C = c_param,penalty = 'l1')

            # Use the training data to fit the model. In this case,we use the portion of the fold to train the model
            # with indices[0]. We then predict on the portion assigned as the 'test cross validation' with indices[1]
            lr.fit(x_train_data.iloc[indices[0],:],y_train_data.iloc[indices[0],:].values.ravel())

            # Predict values using the test indices in the training data
            y_pred_undersample = lr.predict(x_train_data.iloc[indices[1],:].values)

            # Calculate the recall score and append it to a list for recall scores representing the current c_parameter
            recall_acc = recall_score(y_train_data.iloc[indices[1],:].values,y_pred_undersample)
            recall_accs.append(recall_acc)
            print('Iteration ',iteration,': recall score = ',recall_acc)

            # The mean value of those recall scores is the metric we want to save and get hold of.
        results_table.ix[j,'Mean recall score'] = np.mean(recall_accs)
        j += 1
        print('')
        print('Mean recall score ',np.mean(recall_accs))
        print('')

    best_c = results_table.loc[results_table['Mean recall score'].idxmax()]['C_parameter']

    # Finally,we can check which C parameter is the best amongst the chosen.
    print('*********************************************************************************')
    print('Best model to choose from cross validation is with C parameter = ',best_c)
    print('*********************************************************************************')

    return best_c

我得到的错误如下:
对于这一行:fold = KFold(len(y_train_data),shuffle = False)
错误:

TypeError: init() got multiple values for argument ‘shuffle’

如果我从这一行中删除shuffle = False,我收到以下错误:

TypeError: shuffle must be True or False; got 5

如果我删除5并保持shuffle = False,我收到以下错误;

TypeError: ‘KFold’ object is not iterable
which is from this line: for iteration,start=1):

如果有人可以帮助我解决这个问题,并建议如何使用最新版本的scikit-learn来完成,我们将非常感激.

谢谢.

解决方法

KFold是一个分裂器,所以你必须给分裂.

示例代码:

X = np.array([1,1],[2,2,2],[3,3,3],[4,4,4]])
y = np.array([1,4])
# Now you create your Kfolds by the way you just have to pass number of splits and if you want to shuffle.
fold = KFold(2,shuffle=False)
# For iterate over the folds just use split
for train_index,test_index in fold.split(X):
    X_train,X_test = X[train_index],X[test_index]
    y_train,y_test = y[train_index],y[test_index]
    # Follow fitting the classifier

如果你想获得train / test循环的索引,只需添加enumerate

for i,train_index,test_index in enumerate(fold.split(X)):
    print('Iteration:',i)
    X_train,y[test_index]

我希望这有效

Python-TypeError:“ int”对象不可调用

Python-TypeError:“ int”对象不可调用

给定以下整数和计算

from __future__ import divisiona = 23b = 45c = 16round((a/b)*0.9*c)

结果是:

TypeError: ''int'' object is not callable.

如何将输出舍入为整数?

答案1

小编典典

给定以下整数和计算

from __future__ import divisiona = 23b = 45c = 16round((a/b)*0.9*c)

结果是:

TypeError: ''int'' object is not callable.

如何将输出舍入为整数?

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