Customer Year Customer Lost/Retained
A 2009 Retained
A 2010 Retained
A 2011 Lost
B 2008 Lost
C 2008 Retained
C 2009 lost 我已经使用itterrows()根据上面的逻辑创建了Customer Lost/Retained列。 如果客户连续一年重复,则保留该客户,否
我在dataframe中有一个列,如下所示...
retention_completion_variable_name <- data.frame(
retention_completion_variable_name = c(
"Completed Degree in 1 Year",
"Retained to Midyear Year 1",
"Completed Degree in 2 Years",
"Retained to Midyear Year 2",
"Re
目前,我正试图以百分比的形式返回一系列聚合函数。有人对怎么做有什么建议吗?
下面是SQL。
SELECT
et.Term
,count(et.employeeid) as 'Total Enrolled'
,sum(et.retained) as 'Retained'
,sum(et.EnrollButSwitchedDept) as 'EnrollButSwitched'
,sum(et.NotEnrolled) as 'Not Enrolled'
,sum(et.Graduated) as 'Graduated
我正在努力将以下PostgreSQL查询转换为MariaDB。转换后的查询不正确地返回“新建用户”字段。我做错了什么?
原始,PostgreSQL工作查询:
WITH
users AS (
SELECT user_id,
MIN(occurred_at) AS activated_at
FROM modeanalytics.retention_events
WHERE occurred_at <= NOW()
GROUP BY 1
),
events AS (
SELECT user_id,
event_name,
我正在编写一个MariaDB查询,并遇到了以下错误: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'GROUP BY 1,2
) x
) z
WHERE user_period != 0
ORDER BY 1,2
-- {"us' at line 42 下面是查询: WITH
users AS (
SEL
能否将ICollection映射到单个实体BonusDto?
我使用Customer类中的ICollection获取所有行,并从行中计算要映射到PreviouslyPaid的部分加成类中保留的&PreviouslyPaid字段。
public class BonusDto
{
public virtual long CustomerId { get; set; }
public virtual decimal Retained { get; set; }
public virtual decimal PreviouslyPai
我需要在每个用户内部使用一个case语句进行搜索,该语句在结束为true之前具有多个条件。该语句需要做的是搜索每个用户,看看他们是否连续出现,排除了其他日子。
select
user_id,
case
when dates = 'Monday' and dates = 'Wednesday' then 'not_retained'
when dates = 'Monday' and dates = 'Tuesday' and dates = 'Wednesda
我有这个问题
--Retention by DOC,Users created >= Jan 1,2012--
Select
One.Date_Of_Concern,
Two.Users,
One.Retained,
Round(One.Retained/Two.Users,4) as Perc_Retained
From
(
Select
To_Date('2012-sep-09','yyyy-mon-dd')As Date_Of_Concern,
Count(P.Player_Id) As Retained
From Player P
Wher
intercatedData n
1 Completed Degree in 2 Years.0 2
2 Retained to Midyear Year 2.0 1
3 Retained to Start of Year 2.0 1
4 Retained to Midyear Year 2.1 1
5 Retained to Start of Year 2.1 1
这是我正在使用的数据frame(mydf)。
现在,我希望通过名称(Completed Degree in 2 Years.0)访问第一行值(即2)。我可
我有一个PostgreSQL表,其中包含: person_identifier、period_identifier、status
person | period | status
-------+--------+--------
Bob | Jan | new
Bob | Feb | retained
Bob | Mar | retained
Bob | Apr | dormant
Bob | May | dormant
Bob | Jun | resurected
Bob
我正在尝试删除c++字符串数组中的所有重复项。我有代码,但它导致我的程序什么都不做。
int removeDups(string a[], int n)
{
if (n < 0)
{
return -1;
}
int k = 0;
int retained = 0;
while (k<n)
{
if (a[k] == a[k + 1])
{
for (int j = k+1; j < (n-k); j++)
{
我定义了以下UDTF,其中我已经确认查询工作独立于函数。
CREATE OR REPLACE FUNCTION UDF_RETAINED_CUST(period_end date)
RETURNS TABLE(RETAINED_CUSTOMERS number)
LANGUAGE SQL
AS
$$
SELECT count(DISTINCT po.id_customer)
FROM STR_PS_ORDERS po
INNER JOIN (
SELECT DISTINCT ID_CUSTOMER
F
我有一个简单的活动和保留的片段,就像一样。
我的活动在onCreate()里面是这样的
FragmentManager fm = getSupportFragmentManager();
retainedFragment = (GridFragment) fm.findFragmentByTag(RETAINED_FRAGMENT_TAG);
// If the Fragment is non-null, then it is currently being
// retained across a configuration change.
if (reta
我正在使用Komodo,需要查找/替换大量代码。
我需要替换所有出现的以下内容,但保留其中的一部分,如下所示:
<th title="THIS IS THE STRING THAT NEEDS TO BE RETAINED">StartDate</th>
替换为::
<th><?=$this->Html->image('bm.acp.help.png', array('alt' => 'Help', 'title'=>'THIS IS THE
我正在努力将PostgreSQL查询转换为MariaDB,需要帮助理解如何在PostgreSQL中重写这一行: retained_users / MAX(CASE WHEN user_period = 0
THEN retained_users
ELSE NULL END)
OVER (PARTITION BY "Signup Date")::FLOAT AS retention_rate, MariaDB不理解::FLOAT -如何为MariaDB重写这一点?
我想以这样一种方式对状态标志的“失效”值进行排名,当它遇到任何其他值时,它会重新设置自己,当它再次遇到“失效”时,它应该再次从1开始排名。
我搞不懂我们怎么用sql做这样的事情。我试过分析函数,但没有成功
数据如下所示。
id yr_mon status
-------------------------
1002 201703 New
1002 201704 Retained
1002 201705 Unretained
1002 201706 Lapsed
1002 201707 Lapsed
1002 201708 Reacti
请回顾一下这个。
我想要做的是在结果中添加一个领先的"$“。标签"in Retained “运行良好,但是当我向其添加"$”时,它只会将数字添加到输出中,而不是显示正在运行的总数。
因此,添加2000 + 4000 + 5000,显示为"$200040005000 in Retained Aided“(带有美元符号,但实际上并没有添加数字)。
var updateTotal = function () {
var input1 = parseInt($('#earnedAid1').val() || 0);
var input2 = pa
下面的查询给出两列的输出:
Day Num_Users_Retained
0 209312
1 22139
2 11457
以此类推,一直到259 (2012年的每一天),我希望第0天包含num_users_retained中从0到259...and的所有值的和,然后我希望第1天包含从1到259的所有值的和,依此类推,直到我到达最后一天。以下是原始查询:
--Retention since January 1,2012--
select retention as Day,count(retention) as Num_Users_Retained
from (select play