如何转储数据,只转储数据库的某些SQLite3表(不是所有表)的数据,而不是模式?转储应该是SQL格式的,因为它应该稍后可以很容易地重新输入到数据库中,并且应该从命令行完成.就像是
sqlite3 db .dump
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但是没有转储模式并选择要转储的表.
我是Apache Spark的新手,我刚刚了解到Spark支持三种类型的集群:
由于我是Spark的新手,我想我应该首先尝试Standalone.但我想知道哪一个是推荐的.说,将来我需要构建一个大型集群(数百个实例),我应该去哪个集群类型?
我想更好地了解Gridview工作方式,特别是auto_fit.这是XML布局:
<?xml version="1.0" encoding="utf-8"?>
<GridView xmlns:android="http://schemas.android.com/apk/res/android"
android:id="@+id/gridview"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:columnWidth="60dp"
android:numColumns="auto_fit"
/>
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它可以通过一系列六个缩略图(48*48像素)正常工作.在纵向模式下,它显示一行,六列.

我不明白为什么这条线android:columnWidth="60dp"是必要的,因为auto_fit应该找到正确数量的列.
如果没有该行android:columnWidth="60dp",它将显示一个3行和2列的网格.

这是ImageAdapter班级:
package com.examples.HelloGridView;
import android.content.Context;
import android.view.View;
import android.view.ViewGroup;
import android.widget.BaseAdapter;
import android.widget.ImageView;
public class ImageAdapter extends BaseAdapter {
private Context mContext;
public ImageAdapter(Context c) {
mContext = c;
}
public int getCount() {
return mThumbIds.length;
}
public Object getItem(int position) {
return null;
}
public long getItemId(int position) {
return 0;
} …Run Code Online (Sandbox Code Playgroud) 我想知道什么是JAVA_HOME.我在哪里设置javac.exe和java.exe的路径.它在环境变量中.当我从命令提示符编译Java程序时,JVM如何找到javac.exe?
我正试图用FileProvider私人路径播放视频.面对
java.lang.IllegalArgumentException: Failed to find configured root that contains /data/data/XXXXX(Package)/files/Videos/final.mp4
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码:
<paths>
<files-path path="my_docs" name="Videos/" />
</paths>
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Java代码:
File imagePath = new File(getFilesDir(), "Videos");
File newFile = new File(imagePath, "final.mp4");
Log.d(TAG, "-------------newFile:"+newFile.exists());//True here
//Exception in below line
Uri contentUri = FileProvider.getUriForFile(this,"com.wow.fileprovider", newFile);
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的Manifest.xml
<provider
android:name="android.support.v4.content.FileProvider"
android:authorities="com.wow.fileprovider"
android:exported="false"
android:grantUriPermissions="true">
<meta-data
android:name="android.support.FILE_PROVIDER_PATHS"
android:resource="@xml/file_paths" />
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这有什么线索吗?
谢谢Nitz
有人可以向我解释name_scopeTensorFlow中的工作原理吗?
假设我有以下代码:
import tensorflow as tf
g1 = tf.Graph()
with g1.as_default() as g:
with g.name_scope( "g1" ) as scope:
matrix1 = tf.constant([[3., 3.]])
matrix2 = tf.constant([[2.],[2.]])
product = tf.matmul(matrix1, matrix2)
tf.reset_default_graph()
g2 = tf.Graph()
with g2.as_default() as g:
with g.name_scope( "g2" ) as scope:
matrix1 = tf.constant([[4., 4.]])
matrix2 = tf.constant([[5.],[5.]])
product = tf.matmul(matrix1, matrix2)
tf.reset_default_graph()
with tf.Session( graph = g1 ) as sess:
result = sess.run( product )
print( result )
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当我运行此代码时,我收到以下错误消息:
Tensor Tensor("g2/MatMul:0", shape=(1, 1), dtype=float32) …Run Code Online (Sandbox Code Playgroud) 我安装pandas和matplotlib使用pip3 install.然后我运行了这个脚本:
import pandas as pd
import matplotlib.pyplot as plt
data = pd.ExcelFile("Obes-phys-acti-diet-eng-2014-tab.xls")
print (data.sheet_names)
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并收到此错误:
dhcp-169-233-172-97:Obesity juliushamilton$ python3 ob.py
Traceback (most recent call last):
File "ob.py", line 4, in <module>
data = pd.ExcelFile("Obes-phys-acti-diet-eng-2014-tab.xls")
File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/site-packages/pandas/io/excel.py", line 169, in __init__
import xlrd # throw an ImportError if we need to
ImportError: No module named 'xlrd'
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为什么xlrd缺少必要的?
我有一棵树作为广度优先搜索的输入,我想知道算法在哪个级别进展?
# Breadth First Search Implementation
graph = {
'A':['B','C','D'],
'B':['A'],
'C':['A','E','F'],
'D':['A','G','H'],
'E':['C'],
'F':['C'],
'G':['D'],
'H':['D']
}
def breadth_first_search(graph,source):
"""
This function is the Implementation of the breadth_first_search program
"""
# Mark each node as not visited
mark = {}
for item in graph.keys():
mark[item] = 0
queue, output = [],[]
# Initialize an empty queue with the source node and mark it as explored
queue.append(source)
mark[source] = 1
output.append(source)
# while queue is not empty
while queue:
# …Run Code Online (Sandbox Code Playgroud) 假设我struct在ANSI C中有这个:
typedef struct _point
{
float x;
float y;
} Point;
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这个函数来创建这个struct:
Point createpoint(float x, float y)
{
Point p;
p.x = x;
p.y = y;
return p;
}
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这允许我创建一个struct具有此功能,即:
int main()
{
Point pointOne = createpoint(5, 6);
Point pointTwo = createpoint(10, 4);
float distance = calculatedistancefunc(pointOne, pointTwo);
/* ...other stuff */
return 0;
}
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有人告诉我这段代码无效,因为在返回之前它struct没有malloc在createpoint(float x, float y)函数中获取,并且struct将被删除.但是,当我使用我struct这样的时候,它似乎没有被删除.
所以我的问题是:我必须malloc这样做struct …
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