这是“如何像计算机科学家一样思考”的练习。我正在学习 Python/编程，但不确定如何完成此任务。
这是书中的一个例子，向前显示字母，我不知道如何获得相反的效果。必须使用while循环。
fruit = 'banana'
index = 0
while index > len(fruit):
letter = fruit[index]
print letter
index = index + 1
说我有以下数组:
Array
(
[0] => Array
(
[category_id] => 1
[name] => foo
[parent_id] => 0
)
[1] => Array
(
[category_id] => 2
[name] => bar
[parent_id] => 1
)
[2] => Array
(
[category_id] => 3
[name] => baz
[parent_id] => 0
)
[3] => Array
(
[category_id] => 4
[name] => test
[parent_id] => 2
)
[4] => Array
(
[category_id] => 5
[name] => test2
[parent_id] => 4
)
)
我正在尝试获得以下输出:
foo
foo > bar
baz …我正在尝试使用以下json数据获取JSON:
{
"count": 96,
"value": {
"title": "cotripWeatherStations",
"description": "Pipes Output",
"link": "http:\/\/pipes.yahoo.com\/pipes\/pipe.info?_id=f43ccc14ab9d1c88d537236e93c1525e",
"pubDate": "Mon, 03 Sep 2012 02:17:48 +0000",
"generator": "http:\/\/pipes.yahoo.com\/pipes\/",
"callback": "",
"items": [
{
"ws:Location": {
"global:Latitude": "37.086807",
"global:Longitude": "-104.521294"
},
"ws:Device": {
"global:DeviceId": "6796",
"global:CommonName": "025N008 RATON PASS (RWIS)",
"global:Status": "enabled",
"global:EntityId": "1",
"global:RoadId": "31",
"global:RoadName": "I-25",
"global:Direction": "North",
"global:MileMarker": "8",
"global:LastUpdateDate": "2012-08-26T13:52:00.000-06:00",
"global:Icon": "theme\/cotrip.org\/images\/devices\/icon_device_weather_station_with_cam_16x23.gif"
}
}
]
}
以下代码为我提供了输出警报,但没有数据:
var weather = $.getJSON("http://pipes.yahoo.com/pipes/pipe.run?_id=f43ccc14ab9d1c88d537236e93c1525e&_render=json&_callback=?", function(json) {
alert("JSON Data: " + json.value["items"]);
});
我如何进一步遍历这个JSON到'ws:Devise - "global:CommonName"
我有一个嵌套的地图,看起来像这样:
{"a" {:points 2} "b" {:points 7} "c" {:points 1} "d" {:points 3}}
我想把它变成一个地图序列,以便排序(排序)并在之后打印.
({:name "a" :points 2}
{:name "b" :points 7}
{:name "c" :points 1}
{:name "d" :points 3})
从文档中我发现我需要像邮路一样的东西,但我无法绕过它.
我有一个关于URL Dispatch或Traversal的初学者问题.我想要做的是使用Mako渲染器渲染我的观点.所以我有这个文件夹/ includes /所有的html和css和javascript.这是我的金字塔主要代码:
from pyramid.config import Configurator
def main(global_config, **settings):
config = Configurator(settings=settings)
config.add_static_view('includes', 'includes', cache_max_age=3600)
config.add_renderer(".html", "pyramid.mako_templating.renderer_factory")
config.add_route('home', '/')
config.scan()
return config.make_wsgi_app()
这些是我的看法:
@view_config(route_name='home', renderer='index.html')
def my_view(request):
return {'name':'Netherdrake'}
@view_config(name = 'login', renderer='login.html')
def login(request):
return {'name':'Netherdrake'}
问题是,当我访问ip:port/login网站工作正常,但当我尝试ip:port/login/css,javascript和图像不起作用.原因是,我的绝对路径无效.
这是我的login.html片段(真的是mako模板)而没有/登录时:
<link href="includes/css/twitter/bootstrap.css" rel="stylesheet">
<link href="includes/css/base.css" rel="stylesheet">
<link href="includes/css/twitter/responsive.css" rel="stylesheet">
<link href="includes/css/jquery-ui-1.8.23.custom.css" rel="stylesheet">
<script src="includes/js/plugins/modernizr.custom.32549.js"></script>
在这里,我正在登录/路径(在这种情况下网站被破坏,没有css,图像,js ......):
<link href="includes/css/twitter/bootstrap.css" rel="stylesheet">
<link href="includes/css/base.css" rel="stylesheet">
<link href="includes/css/twitter/responsive.css" rel="stylesheet">
<link href="includes/css/jquery-ui-1.8.23.custom.css" rel="stylesheet">
<script src="includes/js/plugins/modernizr.custom.32549.js"></script>
login.html也在/包含.
我如何解决这个问题,并使其在两条路径下工作,有无斜线？我尝试了几个和url调度,两个问题都是一样的.
如何从下往上逐行遍历文件？例如,这是我从上到下遍历它的代码:
void *readFileTB(char *str1)
{
int size = 1024;
char *buffer = malloc(size);
FILE *fp;
fp = fopen("testfile.txt", "r");
while(fgets(buffer, 1024, fp)) //get a line from a file
{
printf(buffer);
}
return 0;
}
如果文件包含:
line1onetest
line2twotest
line3threetest
如果执行此功能将打印以下内容:
line1onetest
line2twotest
line3threetest
我怎样才能编写一个能够完成上述操作但功能相反的函数,以便输出以下内容？
line3threetest
line2twotest
line1onetest
有什么想法吗？
我做了这个小班:
class Analyzer {
public static function analyze($phrases) {
$sortedPhrases = array();
array_walk($phrases, array('self', 'splitByLength'), $sortedPhrases);
var_dump($sortedPhrases);
}
private static function splitByLength($item, $key, &$resArr) {
// line stolen from here: http://stackoverflow.com/a/4786840/603003
// thanks to arnaud576875 <http://stackoverflow.com/users/576875/arnaud576875>
$len = count( preg_split('#\PL+#u', $item, -1, PREG_SPLIT_NO_EMPTY) );
if (!isset($resArr[$len])) {
$resArr[$len] = array();
}
$resArr[$len][] = $item;
var_dump($resArr);
}
}
$phrases = array(
"I can't believe the great content",
"I can't understand the superior information",
"I can't comprehend the amazing data",
"I cannot …我不确定我将如何攻击我的霍夫曼树的穿越.树是正确的,我只是很难确定如何以一种好的方式遍历它.出于某种原因,我的遍历方法没有结果......
更新:清理代码,使其更加面向对象
节点类:
public class Node
{
public int frekvens; //Frequency
public char tegn; //Symbol
public Node venstre; //Left child
public Node høyre; //Right child
public string s; //result string
public string resultat;
public Node (char c) // Node constructor containing symbol.
{
frekvens = 1;
tegn = c;
}
public Node (int f, Node venstre, Node høyre) // Node Constructor containing frequency and children
{
frekvens = f;
this.venstre = venstre;
this.høyre = høyre;
}
public Node (Node node) …我有一个庞大的推荐系统（超过 500k 个条目），其工作原理如下
[{id: 1, name: "John", ref: 0},
{id: 2, name: "Jack", ref: 1},
{id: 3, name: "Bill", ref: 1},
{id: 5, name: "Jason", ref: 2},
{id: 6, name: "James", ref: 3},
{id: 7, name: "Tom", ref: 0}]
每当用户使用其他用户的推荐代码加入时，推荐人都会获得一些积分，并且它适用于所有级别，因此在此示例中，John该 ID 会获得积分[2, 3, 5, 6]
我使用此方法根据推荐 ID 来统计和组织所有条目
const countRefs = (list) => {
return list.reduce((acc, cur) => {
if(!acc[cur.ref]) acc[cur.ref] = [];
acc[cur.ref].push(cur.id);
return acc;
},{});
}
然后使用此递归函数通过 ID 获取所有引用的用户。
let depth = 0;
// Keep …假设我想从字符串中删除重复项.我决定使用长度为256的布尔数组来存储特定字符是否已经出现.我可以遍历字符串,并可以借助此辅助布尔数组删除所有重复项.
我的问题是"这个算法是否就位？"
我认为它使用的是恒定的空间量,它不会随着它应该就地输入的大小而改变.如果我错了,请纠正.