类型错误：“模块”对象在 MLP 回归器中不可调用

Question

尝试运行此命令时，我收到以下错误。我真的不知道如何解决这个问题。由于我对此很陌生，所以我非常感谢任何帮助。在这里，我使用不同的模型来检查哪一个在这里最好。检查此导入代码：import numpy as np import matplotlib.pyplot as plt import matplotlib.colors as mcolorsimport pandas as pd import plotly.express as pxfrom datetime import date, timedeltaimport randomimport mathimport timeimport operator import foliumimport plotly.offline as pyfrom sklearn.linear_model import LinearRegression, BayesianRidgefrom sklearn.model_selection import RandomizedSearchCV, train_test_splitfrom sklearn.preprocessing import PolynomialFeaturesfrom fbprophet import Prophetfrom fbprophet.plot import plot_plotly, add_changepoints_to_plotfrom sklearn import preprocessing, cross_validationfrom sklearn.tree import DecisionTreeRegressorfrom sklearn.neural_network import MLPRegressorfrom sklearn.svm import SVRfrom sklearn.metrics import mean_squared_error, mean_absolute_errorfrom statsmodels.tsa.arima_model import ARIMAfrom statsmodels.graphics.tsaplots import plot_acf, plot_pacfimport statsmodels.api as smfrom keras.models import Sequentialfrom keras.layers import LSTM,Densefrom keras.layers import Dropoutfrom sklearn.preprocessing import MinMaxScalerfrom tensorflow.keras.preprocessing.sequence import TimeseriesGeneratorplt.style.use('fivethirtyeight')%matplotlib inline 多项式回归的模型开发data1 = confirmed_df.melt(value_vars=dates1, var_name='Date', value_name='Confirmed')data1 = data1.groupby('Date')['Confirmed'].sum().reset_index()data1.head()X = pd.DataFrame(data=data1, columns=data1.Date)y = data1.Confirmedy -= y.mean()#cross_validation.train_test_split(x, y, test_size=0.1,random_state=0)X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.1,                                                    random_state=0)

Answer 1

我想你想跟踪你的脚本的性能，但是你需要指定要从中调用方法time（）的模块，所以要解决这个问题，只需调用，请参阅下面证明的示例tic=time()tic = time.time()

#!/usr/bin/python

import time;  # This is required to include time module.

ticks = time.time()

print "Number of ticks since 12:00am, January 1, 1970:", ticks

Answer 2

错误告诉您这是一个模块，而不是一个函数。你在剧本的顶部有这样的台词吗？time

import time

如果是这样，那么要获取要在模块中调用函数的当前时间：timetime

tic = time.time()

如果使用，则可以在运行代码之前捕获此类错误。在调用模块的代码上运行，就好像它是一个函数一样，会产生 .使用宏可以显示差异：mypymypytimeerror: Module not callablereveal_type

import time

reveal_type(time)       # note: Revealed type is '_importlib_modulespec.ModuleType'

reveal_type(time.time)  # note: Revealed type is 'def () -> builtins.float'

如果您只想从模块中获取函数，则可以专门导入该函数，以便在代码范围内引用该函数而不是整个模块：timetimetime

from time import time

reveal_type(time)       # note: Revealed type is 'def () -> builtins.float'

reveal_type(time.time)  # error: "Callable[[], float]" has no attribute "time"

因此，现在我们已经将函数直接导入到我们的作用域中，是正确的（因为 是一个函数），而不是（因为该函数没有名为 的成员）。timetime()timetime.time()timetime