我一直在研究一种算法()来计算包含属性P的序列的子序列数。例如,有多少个子序列认为“左边的正数比右边的多”。
但这只是为了提供一些背景。
重要的功能是下面这个CountAux函数。给出了一个命题P (例如,x is positive),一个序列的sequ序列,一个在序列中移动的索引i,以及一个上界j。
function CountAux<T>(P: T -> bool, sequ: seq<T>, i: int, j:int): int
requires 0 <= i <= j <= |sequ|
decreases j - i //
这个信息会有点长,但这是因为我想尽可能最好地解释它。
在Dafny中,我遇到了下一个问题:给定一个数组,计算发生这种情况的长度为k的段数;左半段中的正数更大,或者等于右半部分。
举个例子(想象段只能是偶数,这样就不会讨论什么是一半):
k=2 ---> count(array[-4,-2,2,1],k) ---> 2, as [-4,-2] fulfills and also [2,1]
k=4 ---> count(array[-4,-2,2,1],k) ---> 0, as [-4,-2,2,1] does not fulfil.
k=6 ---> count
with
t1 as (
select 1 as poss, 1 as sequ, 'jon' as name union all
select 1 as poss, 2 as sequ, 'nick' as name union all
select 1 as poss, 3 as sequ, null as name union all
select 1 as poss, 4 as sequ, null as name union all
select 1 as poss, 5 as sequ, 'tom&
我想根据存储在元素sequi和sequi中的两个值对数组进行排序。
如果sequi具有相似的值,则按sequi值对其进行排序。
vector< vector<int> > sequ;
int m = 1024, n = 32;
sequ.resize(m);
for(int i = 0 ; i < m ; ++i){
sequ[i].resize(n);
}
sort(sequ[0].begin(),sequ[0].end());
不幸的是,我只知道如何将数组作为一个整体进行排序,而不知道如何对特定元素进行排序。我该怎么做?
我正在使用下面的查询
select id,
number_sequ,
startvalue
lead(startvalue,1,0) over (partition by id order by number_sequ) AS End_value
from mytable
填充以下输出
id number_sequ startvalue End_value
---- ----- ---------- -----------
AAA
我有一个大约需要2秒加载的查询:
SELECT OUTPUT_VAL.NEXTVAL VAR1_R_ID,A.R_ID,A.VAR1,A.SEQU,A.OUTPUT,B.VAR1 DATATYPE_VAR1
FROM
(
SELECT A.R_ID,A.VAR1,A.SEQU,A.OUTPUT,B.D_TYPE
FROM
(
select A.R_ID, 2484 VAR1,1 SEQU, A.USER OUTPUT
from R_TB_1 A
WHERE A.R_ID BETWEEN 2457854437 AND
在Dafny中,我试图创建一个Max多态和高阶函数,该函数在给定序列和谓词的情况下,返回包含它的最长的子序列。例如,最长的递增子序列,或其中所有元素都为零的最长子序列。
为此,我设计了一个较慢的算法(给定P谓词和S序列):
1. Start an i pivot in the left and a j pivot in the same place.
2. Start the max_sequence = [] and the max_sequence_length = 0.
3. While i<S.length:
counter = 0
我有以下代码
let rec consume() : Async<unit> = async {
.....
listA
|> Seq.iter(fun i ->
.....
let listB : seq<...> option =
let c = getListB a b
match c with
| Some d -> Seq.filter(....) |> Some
| None -
我的目标是在两个文本段落中提取对齐匹配序列。下面是我的短信:
txt1='the heavy lorry crashed into the building at midnight'
txt2='what a heavy lorry it is that crashed into the building'
预期产出:
'heavy lorry'
'crashed into the building'
我的尝试:
def sequ(s1,s2):
_split1=s1.split()
_split2=s2.spli
我有关于在交谈中提问时瞳孔反应的数据。瞳孔数据在A*、B*和C*列中串在一起,以*intpl_new结尾的列表示插值瞳孔面积值,以*dur结尾的列给出每个瞳孔观察的持续时间。Dataframe df包含两个说明性问题Sequ的数据:
df <- structure(list(Speaker = c("ID08.A", "ID08.A"), Utterance = c("what so you're going to a £!party! in Italy£=",
我可以看到DBA团队建议在性能优化时将序列缓存设置为更高的值。若要将值从20增加到1000或5000,oracle文档表示缓存值,
Specify how many values of the sequence the database preallocates and keeps in memory for faster access.
在我能看到的报告的某个地方,
select SEQ_MY_SEQU_EMP_ID.nextval from dual
如果我增加SEQ_MY_SEQU_EMP_ID的缓存值,可以看到性能的改善吗?
我的问题是:
序列缓存是否在性能中发挥任何重要作用?如果是这
我是否可以声明"IF“语句或类似MVC模型的内容,例如,下面是我的代码:
namespace SolrWeb.Models
{
public class SolrVariables
{
[SolrField("sequentialid")]
public long Sequ { get; set; }
[SolrField("exchangestatus")]
public int ExtId { get; set; }
[SolrField("msgtype")]
pu