小编ubu*_*oob的帖子

这是完整的错误 -

Error:Execution failed for task ':app:preDebugAndroidTestBuild'.
> Conflict with dependency 'com.android.support:support-annotations' in project ':app'. Resolved versions for app (26.1.0) and test app (27.1.1) differ. See https://d.android.com/r/tools/test-apk-dependency-conflicts.html for details.

我知道这类答案有很多解决方案,但我是android studio的绝对初学者,我无法理解那些解决方案,比如命令行界面与gradle交互等等......

如果有的话,我一直在寻找这个问题的简单解决方案.非常感谢!

给定具有节点对(p,q)的有向图G,我们得到了

我想计算这个递归函数的值,其中L(p)表示节点p的无向链路邻居集.此函数用于第(k + 1)个值.

我知道如何计算L(p)和L(q).

这是我的尝试 -

from __future__ import division
import copy

import numpy as np
import scipy
import scipy.sparse as sp
import time
import warnings




class Algo():
    def __init__(self, c=0.8, max_iter=5, is_verbose=False, eps=1e-3):
        self.c = c
        self.max_iter = max_iter
        self.is_verbose = is_verbose
        self.eps = eps

    def compute_similarity(self, a):
        '''

        Note: The adjacency matrix is a (0,1)-matrix with zeros on its 
        diagonal.
        '''                                 
        a_transpose=np.transpose(a)
        a_undirected=np.add(a,a_transpose)

        self.num_nodes = np.shape(a)[0]
        current_sim = np.eye(self.num_nodes)
        prev_sim = np.zeros(shape=(self.num_nodes, self.num_nodes))

        #Determine the set of …

import pandas as pd
import numpy as np
from datetime import datetime

data = {'date': ['1998-03-01 00:00:01', '2001-04-01 00:00:01','1998-06-01 00:00:01','2001-08-01 00:00:01','2001-05-03 00:00:01','1994-03-01 00:00:01'], 
        'node1': [1, 1, 2,2,3,2],
     'node2': [8,316,26,35,44,56],
     'weight': [1,1,1,1,1,1], }

df2 = pd.DataFrame(data, columns = ['date', 'node1','node2','weight'])


df2['date'] = pd.to_datetime(df2['date'])

l1 = [1990,1991,1992,1993,1994,1995,1996,1997,1998]
l2 = [1999,2000,2001]
ndf = df2[df2['date'].dt.year.isin(l1+l2)]

mask = ndf.groupby('node1','node2').apply(lambda x : (x['date'].dt.year.isin(l1)).any())
mask2 = ndf.groupby('node1','node2').apply(lambda x : (x['date'].dt.year.isin(l2)).any())

我得到的错误 -

Traceback (most recent call last):
  File "datanew.py", line 32, in <module>
    mask = ndf.groupby('node1','node2').apply(lambda x …

data1 = { 'node1': [2,2,3,6],
     'node2': [6,7,7,28],
     'weight': [1,2,1,1], }
df1 = pd.DataFrame(data1, columns = ['node1','node2','weight'])

我想按递增顺序重命名data1中的node1和node2.节点是2 3 6 7 28,因此它们分别变为1 2 3 4 5.

所以数据框变成了 -

data1 = { 'node1': [1,1,2,3],
     'node2': [3,4,4,5],
     'weight': [1,2,1,1], }
df1 = pd.DataFrame(data1, columns = ['node1','node2','weight'])

之前的数据看起来像这样

但现在看起来像这样

我有一个名为的 pandas 列'date' ，其值和类型类似于2014-07-30 00:00:00 <class 'datetime.datetime'>. 我想从日期中删除时间。最终结果是 datetime.datetime 格式的“2014-07-30”。

我尝试了很多解决方案，例如 -

df['PSG Date '] = df['PSG Date '].dt.date

但它给了我错误-

AttributeError: Can only use .dt accessor with datetimelike values

我有-

data=pd.read_csv('data.csv')
if data.empty==False:
    do something

在我的代码中，根据某些条件，生成的数据框数据有时为空。现在，当 csv 文件为空时，它会抛出错误 -

EmptyDataError: No columns to parse from file

我应该怎么做才能避免这个错误？

我的模特-

from tensorflow.keras.layers import ReLU
from keras.layers import Dropout
from tensorflow.keras.utils import plot_model
from matplotlib import pyplot

# define encoder
visible = Input(shape=(n_inputs,))



# encoder level 1
e = Dense(300)(visible)
e = ReLU()(e)
e = Dropout(0.05)(e)

# encoder level 1
e = Dense(200)(visible)
e = ReLU()(e)
e = Dropout(0.05)(e)

# encoder level 1
e = Dense(100)(visible)
e = ReLU()(e)
e = Dropout(0.05)(e)

# encoder level 1
e = Dense(50)(visible)
e = ReLU()(e)
e = Dropout(0.05)(e)



# bottleneck
n_bottleneck = round(float(n_inputs)) …

我创建了一个列表数据类型，其中包含三个文件夹的路径，其中每个文件夹都有很多 .txt 文件。我试图通过将文件夹中的每个文件设置为熊猫数据框来处理文件夹中的每个文件，但我收到了列出的错误。

代码-

for l in list: 
    for root, dirs, files in os.walk(l, topdown=False):
        for name in files:
            #print(os.path.join(root, name))

            df = pd.read_csv(os.path.join(root, name))   

错误-

Traceback (most recent call last):
      File "feature_drebin.py", line 18, in <module>
        df = pd.read_csv(os.path.join(root, name))
      File "E:\anaconda\lib\site-packages\pandas\io\parsers.py", line 709, in parser_f
        return _read(filepath_or_buffer, kwds)
      File "E:\anaconda\lib\site-packages\pandas\io\parsers.py", line 449, in _read
        parser = TextFileReader(filepath_or_buffer, **kwds)
      File "E:\anaconda\lib\site-packages\pandas\io\parsers.py", line 818, in __init__
        self._make_engine(self.engine)
      File "E:\anaconda\lib\site-packages\pandas\io\parsers.py", line 1049, in _make_engine
        self._engine = CParserWrapper(self.f, **self.options)
      File "E:\anaconda\lib\site-packages\pandas\io\parsers.py", line …

 if  not (sp.csc_matrix.transpose(a) == a).all():        
            a_transpose=sp.csc_matrix.transpose(a)
            a=np.add(a,a_transpose)

我正在检查稀疏矩阵是否对称，但出现以下错误 - AttributeError: all not found

我读的数据库系统,并在(组比较)它是书面的话题嵌套子查询的书基本面some和in相同,而<>some并not in没有.根据我的<>some意思,"不是至少一个",not in意思是"不在集合中......"所以我认为它们应该是一样的.

我有一个这样的清单 -

list=[137,136,135,134,119,118,-14,-89,-208,-291,-491,-513,-596,-699]

现在我想计算列表中的最小数量而不考虑符号,但在最终答案中应该保留符号.

例如,答案是-14.

现在我首先将列表分为正数和负数,分别计算最小值和最大值,然后比较绝对值和返回答案值.