我开始学习Elixir,这也是我的第一个动态语言,所以我真的失去了没有类型声明的函数.
我想做什么:def create_training_data(file_path, indices_path, result_path) do
file_path
|> File.stream!
|> Stream.with_index
|> filter_data_with_indices(indices_path)
|> create_output_file(result_path)
end
def filter_data_with_indices(raw_data, indices_path) do
Stream.filter raw_data, fn {_elem, index} ->
index_match?(index, indices_path)
end
end
defp index_match?(index, indices_path) do
indices_path
|> File.stream!
|> Enum.any? fn elem ->
(elem
|> String.replace(~r/\n/, "")
|> String.to_integer
|> (&(&1 == index)).())
end
end
defp create_output_file(data, path) do
File.write(path, data)
end
当我调用该函数时:
create_training_data("./resources/data/usps.csv","./resources/indices/17.csv","./output.txt")
它返回{:error,:badarg}.我已经检查过,错误发生在create_output_file函数上.
如果我注释掉函数create_output_file,我得到的是一个流(有点意义).问题可能是我不能给Stream to File.write吗？如果是问题,我该怎么办？我在文档上没有找到任何相关内容.
所以,问题是File.write的路径应该没问题,我修改了这个函数:
defp create_output_file(data, path) do
IO.puts("You are trying to write to: …我正在使用a webpack-dev-server,我正在尝试包含bootstrap.
我有这个项目结构:
?? css
? ??? bootstrap.min.css
??? js
| ??? bootstrap.min.js
??? dist
??? index.html
??? package.json
??? server.js
??? src
? ??? actions.js
? ??? App.js
? ??? components
? ??? constants
? ??? index.js
? ??? reducers.js
??? webpack.config.js
这是index.html:
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" href="css/bootstrap.min.css">
</head>
<body>
<div id='root'></div>
<script src="/static/bundle.js"></script>
</body>
</html>
每当我运行服务器时,我都会收到类型错误:
Cannot GET /css/bootstrap.min.css
这是webpack.config.js:
var path = require('path');
var webpack = require('webpack');
module.exports = {
devtool: 'eval', …javascript twitter-bootstrap webpack webpack-dev-server webpack-style-loader
我编译了库GLEW.它似乎工作正常,这是make install的输出:
install -d -m 0755 "/usr/include/GL"
install -m 0644 include/GL/wglew.h "/usr/include/GL/"
install -m 0644 include/GL/glew.h "/usr/include/GL/"
install -m 0644 include/GL/glxew.h "/usr/include/GL/"
sed \
-e "s|@prefix@|/usr|g" \
-e "s|@libdir@|/usr/lib64|g" \
-e "s|@exec_prefix@|/usr/bin|g" \
-e "s|@includedir@|/usr/include/GL|g" \
-e "s|@version@|1.11.0|g" \
-e "s|@cflags@||g" \
-e "s|@libname@|GLEW|g" \
-e "s|@requireslib@|glu|g" \
< glew.pc.in > glew.pc
install -d -m 0755 "/usr/lib64"
install -m 0644 lib/libGLEW.so.1.11.0 "/usr/lib64/"
ln -sf libGLEW.so.1.11.0 "/usr/lib64/libGLEW.so.1.11"
ln -sf libGLEW.so.1.11.0 "/usr/lib64/libGLEW.so"
install -m 0644 lib/libGLEW.a "/usr/lib64/"
install -d -m 0755 "/usr/lib64" …我只是在学习Rust.我正在尝试为我的Game结构创建一个构建器结构.这是代码:
struct Input {
keys_pressed: HashMap<VirtualKeyCode, bool>,
}
pub struct GameBuilder {
settings: GameSettings,
input: Input,
}
impl GameBuilder {
pub fn new() -> GameBuilder {
GameBuilder {
settings: GameSettings {
window_dimensions: (800, 600),
title: "".to_string(),
},
input: Input {
keys_pressed: HashMap::new(),
}
}
}
pub fn with_dimensions(&mut self, width: u32, height: u32) -> &mut GameBuilder {
self.settings.window_dimensions = (width, height);
self
}
pub fn with_title(&mut self, title: &str) -> &mut GameBuilder {
self.settings.title = title.to_string();
self
}
pub fn …我尝试在java中创建一个二维泛型数组.我没有编译错误,但运行代码时出现异常:
Exception in thread "main" java.lang.ClassCastException: Cannot cast [[[Ljava.lang.String; to [[Ljava.lang.String;
at java.lang.Class.cast(Unknown Source)
at Tabela.<init>(Tabela.java:8)
at TabelaTest.main(TabelaTest.java:4)
这是代码:
import java.lang.reflect.Array;
public class Tabela<T> {
private T[][] data;
public Tabela(Class<T[][]> c,int sizeX, int sizeY) {
this.data = c.cast(Array.newInstance(c.getComponentType(), sizeX, sizeY));
}
public void setInfoAt(T info, int x, int y) {
this.data[x][y] = info;
}
public T getObjectAt(int x, int y) {
return this.data[x][y];
}
}
public class TabelaTest {
public static void main(String args[]) {
Tabela<String> tabela = new Tabela<String>(String[][].class, …我正在通过 Learn You A Haskell 一书了解 Writer Monad。
这是一段代码：
import Control.Monad.Writer
logNumber :: Int -> Writer [String] Int
logNumber num = writer (num, ["Got number: " ++ show num])
multWithLog :: Writer [String] Int
multWithLog = do
a <- logNumber 3
b <- logNumber 5
return (a * b)
运行时multWithLog，结果如下：
*Main> runWriter multWithLog
(15,["Got number: 3","Got number: 5"])
在这一行：
a <- logNumber 3
b <- logNumber 5
很容易看出a = 3and b = 5，因为它们都在return函数上相乘。 …
我正在尝试使用QtConcurrent::mapped到QVector<QString>. 我已经尝试了很多方法，但似乎总是存在重载的问题。
QVector<QString> words = {"one", "two", "three", "four"};\n\nusing StrDouble = std::pair<QString, double>;\n\nQFuture<StrDouble> result = QtConcurrent::mapped<StrDouble>(words, [](const QString& word) -> StrDouble {\n return std::make_pair(word + word, 10);\n});\n该片段返回以下错误：
\n\n/home/lhahn/dev/cpp/TestLambdaConcurrent/mainwindow.cpp:23: error: no matching function for call to \xe2\x80\x98mapped(QVector<QString>&, MainWindow::MainWindow(QWidget*)::<lambda(const QString&)>)\xe2\x80\x99\n });\n ^\n我看到这篇文章，它说 Qt 找不到 lambda 的返回值，所以你必须使用std::bind它。如果我尝试这个：
using StrDouble = std::pair<QString, double>;\nusing std::placeholders::_1;\n\nauto map_fn = [](const QString& word) -> StrDouble {\n return std::make_pair(word + word, 10.0);\n};\n\nauto wrapper_map_fn = …我有以下两个类:
struct Entity
{
unsigned id;
Entity(unsigned id);
Entity();
};
class EntityManager
{
public:
Entity create();
...
private:
Entity make_entity(unsigned index, unsigned generation);
};
这个目前工作正常.问题是封装.我不想允许直接创建类Entity.
因此,我的目标是使构造者成为Entity私有的.然后,我可以(从我的理解)十个分量的功能在EntityManager通过使Entity一个friend的EntityManager.
因此,进行更改将是结果:
struct Entity
{
unsigned id;
private:
Entity(unsigned id);
Entity();
};
class EntityManager
{
friend struct Entity;
public:
Entity create();
private:
Entity make_entity(unsigned index, unsigned generation);
};
这打破了代码.我得到的错误就是这个:
entity_manager.cpp: In member function ‘Entity EntityManager::make_entity(unsigned int, unsigned int)’:
entity_manager.cpp:12:1: error: ‘Entity::Entity(unsigned int)’ is …我在Java中有这个代码:
List<String> unSorted = new ArrayList<String>();
List<String> beforeHash = new ArrayList<String>();
String[] unSortedAux, beforeHashAux;
String line = null;
BufferedReader reader = new BufferedReader(new FileReader("C:\\CPD\\temp0.txt"));
while ((line = reader.readLine()) != null){
unSorted.add(line);
beforeHash.add(line.split("#")[0]);
}
reader.close();
Collections.sort(beforeHash);
beforeHashAux = beforeHash.toArray(new String[beforeHash.size()]);
unSortedAux = unSorted.toArray(new String[unSorted.size()]);
System.out.println(Arrays.toString(beforeHashAux));
System.out.println(Arrays.toString(unSortedAux));
它读取名为temp0.txt的文件,其中包含:
Carlos Magno#261
Mateus Carl#12
Analise Soares#151
Giancarlo Tobias#150
我的目标是在字符串中对名称进行排序,而不是"#"之后的字符串.我正在使用beforeHash.add(line.split("#")[0]); 去做这个.问题是它正确读取文件,但它按错误的顺序排序.相应的产出是:
[Analise Soares, Giancarlo Tobias, Mateus Carl, Carlos Magno]
[Carlos Magno#261, Mateus Carl#12, Analise Soares#151, Giancarlo Tobias#150]
第一个结果是"排序的",注意"Carlos Magno"来自"Mateus Carl".我在代码中找不到问题.
我有两个数组:
import numpy as np
import pylab as pl
x = np.array([-36., -34.95522388, -33.91044776, -32.86567164,
-31.82089552, -30.7761194 , -29.73134328, -28.68656716,
-27.64179104, -26.59701493, -25.55223881, -24.50746269,
-23.46268657, -22.41791045, -21.37313433, -20.32835821,
-19.28358209, -18.23880597, -17.19402985, -16.14925373,
-15.10447761, -14.05970149, -13.01492537, -11.97014925,
-10.92537313, -9.88059701, -8.8358209 , -7.79104478,
-6.74626866, -5.70149254, -4.65671642, -3.6119403 ,
-2.56716418, -1.52238806, -0.47761194, 0.56716418,
1.6119403 , 2.65671642, 3.70149254, 4.74626866,
5.79104478, 6.8358209 , 7.88059701, 8.92537313,
9.97014925, 11.01492537, 12.05970149, 13.10447761,
14.14925373, 15.19402985, 16.23880597, 17.28358209,
18.32835821, 19.37313433, 20.41791045, 21.46268657,
22.50746269, 23.55223881, 24.59701493, 25.64179104,
26.68656716, 27.73134328, 28.7761194 , …