标签: aggregate

require(raster)

## Function to aggregate
fun.patch <- function(x) {
  if (max(table(x)) >= 0.9 * length(x)) {
    return(as.vector(which.max(table(x))))
  }
  else
    return(NA)
}

r.lc <- raster(nrows = 100, ncols = 100)
r.lc[] <- 1:6
aggregate(r.lc, fact = c(5,5), fun.patch)

FUN(newX[, i], ...) 中的错误：未使用的参数（na.rm = TRUE）

我只是想摆脱典型的存储库/服务/演示的 N 层架构的舒适区，并开始研究 DDD 和聚合，我不得不承认我有点困惑，希望有人能澄清以下内容例子：

如果我有一个名为 News、NewsImage 和 Customer 的实体，它们都是 EF 可持久对象，如下所示：

public class Customer
{
    public virtual int Id { get; set; }
    public virtual string Name { get; set; }
}

public class NewsImage
{
    public virtual int Id { get; set; }
    public virtual byte[] Data { get; set; }
    public virtual News News { get; set; }
}

public class News
{ 
    public virtual int Id { get; set; }
    public virtual string Name { get; …

我想检查一组记录的所有（浮动）值是否相等。就像是

SELECT ..., equal(my_field) FROM my_table WHERE ... GROUP BY ...

哪里equal(my_field)的回报true，如果所有的值my_field相等。

问题在一个简单的数据帧（可下载的 csv）上尝试 groupby ，然后 agg 返回列的聚合值（大小、总和、平均值、标准偏差）。看似简单的问题却出现了出乎意料的具有挑战性的错误。

Top15.groupby('Continent')['Pop Est'].agg(np.mean, np.std...etc)
# returns 
ValueError: No axis named <function std at 0x7f16841512f0> for object type <class 'pandas.core.series.Series'>

我想要得到的是一个索引设置为大陆和列的 df ['size', 'sum', 'mean', 'std']

示例代码

import pandas as pd
import numpy as np

# Create df
df = pd.DataFrame({'Country':['Australia','China','America','Germany'],'Pop Est':['123','234','345','456'],'Continent':['Asia','Asia','North America','Europe']})

# group and agg
df = df.groupby('Continent')['Pop Est'].agg('size','sum','np.mean','np.std')

对于 T-SQL (SQL Server 2016)bit 类型，是否有某种方法可以实现逻辑 AND 和逻辑 OR 的聚合等效？例如，使用此表：

CREATE TABLE #Example       (id int, category int, isRed bit, isBlue bit)
INSERT INTO #Example VALUES (     1,            1,         1,          1)
INSERT INTO #Example VALUES (     2,            1,         0,          0)
INSERT INTO #Example VALUES (     3,            1,         1,          0)
INSERT INTO #Example VALUES (     4,            2,         0,          1)
INSERT INTO #Example VALUES (     5,            2,         0,          1)
INSERT INTO #Example VALUES (     6,            2,         0,          1)

我想创建一个查询，列表，每个类别，如果任何的isRed …

我正在尝试在 PostgreSql 中标准化日终股票价格。

假设我有一个这样定义的库存表：

create table eod (
  date date not null,
  stock_id int not null,
  split decimal(16,8) not null,
  close decimal(12,6) not null,
  constraint pk_eod primary key (date, stock_id)
);

此表中的数据可能如下所示：

"date","stock_id","eod_split","close"
"2014-06-13",14010920,"1.00000000","182.560000"
"2014-06-13",14010911,"1.00000000","91.280000"
"2014-06-13",14010923,"1.00000000","41.230000"
"2014-06-12",14010911,"1.00000000","92.290000"
"2014-06-12",14010920,"1.00000000","181.220000"
"2014-06-12",14010923,"1.00000000","40.580000"
"2014-06-11",14010920,"1.00000000","182.250000"
"2014-06-11",14010911,"1.00000000","93.860000"
"2014-06-11",14010923,"1.00000000","40.860000"
"2014-06-10",14010911,"1.00000000","94.250000"
"2014-06-10",14010923,"1.00000000","41.110000"
"2014-06-10",14010920,"1.00000000","184.290000"
"2014-06-09",14010920,"1.00000000","186.220000"
"2014-06-09",14010911,"7.00000000","93.700000"
"2014-06-09",14010923,"1.00000000","41.270000"
"2014-06-06",14010923,"1.00000000","41.480000"
"2014-06-06",14010911,"1.00000000","645.570000"
"2014-06-06",14010920,"1.00000000","186.370000"
"2014-06-05",14010920,"1.00000000","185.980000"
"2014-06-05",14010911,"1.00000000","647.350000"
"2014-06-05",14010923,"1.00000000","41.210000"
... 
"2005-03-04",14010920,"1.00000000","92.370000"
"2005-03-04",14010911,"1.00000000","42.810000"
"2005-03-04",14010923,"1.00000000","25.170000"
"2005-03-03",14010923,"1.00000000","25.170000"
"2005-03-03",14010911,"1.00000000","41.790000"
"2005-03-03",14010920,"1.00000000","92.410000"
"2005-03-02",14010920,"1.00000000","92.920000"
"2005-03-02",14010923,"1.00000000","25.260000"
"2005-03-02",14010911,"1.00000000","44.121000"
"2005-03-01",14010920,"1.00000000","93.300000"
"2005-03-01",14010923,"1.00000000","25.280000"
"2005-03-01",14010911,"1.00000000","44.500000"
"2005-02-28",14010923,"1.00000000","25.160000"
"2005-02-28",14010911,"2.00000000","44.860000"
"2005-02-28",14010920,"1.00000000","92.580000"
"2005-02-25",14010923,"1.00000000","25.250000"
"2005-02-25",14010920,"1.00000000","92.800000"
"2005-02-25",14010911,"1.00000000","88.990000"
"2005-02-24",14010923,"1.00000000","25.370000"
"2005-02-24",14010920,"1.00000000","92.640000"
"2005-02-24",14010911,"1.00000000","88.930000"
"2005-02-23",14010923,"1.00000000","25.200000"
"2005-02-23",14010911,"1.00000000","88.230000"
"2005-02-23",14010920,"1.00000000","92.100000"
... …

  aggregrated_table = df_input.groupBy('city', 'income_bracket') \
        .agg(
       count('suburb').alias('suburb'),
       sum('population').alias('population'),
       sum('gross_income').alias('gross_income'),
       sum('no_households').alias('no_households'))

想按城市和收入等级分组，但在每个城市内，某些郊区有不同的收入等级。我如何按每个城市最常出现的收入等级分组？

例如：

  aggregrated_table = df_input.groupBy('city', 'income_bracket') \
        .agg(
       count('suburb').alias('suburb'),
       sum('population').alias('population'),
       sum('gross_income').alias('gross_income'),
       sum('no_households').alias('no_households'))

将按income_bracket_10 分组

我有一个大的data.frame.data.frame包含很多值.

例如:

df <- data.frame(Company = c('A', 'A', 'B', 'C', 'A', 'B', 'B', 'C', 'C'), 
                 Name = c("Wayne", "Duane", "William", "Rafael", "John", "Eric", "James", "Pablo", "Tammy"), 
                 Age = c(26, 27, 28, 32, 28, 24, 34, 30, 25), 
                 Wages = c(50000, 70000, 70000, 60000, 50000, 70000, 65000, 50000, 50000), 
                 Education.University = c(1, 1, 1, 0, 0, 1, 1, 0, 1), 
                 Productivity = c(100, 120, 120, 95, 88, 115, 100, 90, 120))

我如何汇总我的data.frame？我想分析每家公司的价值观.它必须看起来像:

年龄 - >公司所有员工的平均年龄

工资 - >公司所有员工的平均工资

Education.University …

为了简单起见，我有四个表（A、B、Category 和 Relation），Relation 表将IntensityA 的 A存储在 B 中，Category 存储 B 的类型。

A <--- 关系 ---> B ---> 类别

（所以A和B的关系是n比n，当B和Category的关系是n比1时）

我的类别和需要一个ORM到组关系的记录，然后计算出Sum的Intensity每个（A类，A）（似乎简单到这里），然后我要诠释的最大计算Sum每个类别。

我的代码是这样的：

 A.objects.values('B_id').annotate(AcSum=Sum(Intensity)).annotate(Max(AcSum))

哪个抛出错误：

django.core.exceptions.FieldError: Cannot compute Max('AcSum'): 'AcSum' is an aggregate

具有相同错误的Django-group-by包。

有关更多信息，请参阅此 stackoverflow 问题。

我正在使用 Django 2 和 PostgreSQL。

有没有办法使用 ORM 来实现这一点，如果没有，使用原始 SQL 表达式的解决方案是什么？

更新

经过一番挣扎，我发现我写的确实是一个聚合，但是我想要的是找出每个类别中每个A的AcSum的最大值。所以我想我必须在 AcSum Calculation 之后再次对结果进行分组。基于这种见解，我发现了一个堆栈溢出问题，它提出了相同的概念（这个问题是在 1 年，2 个月前提出的，没有任何公认的答案）。将另一个值（'id'）链接到集合既不能作为 group_by 也不能作为输出属性的过滤器，它会从集合中删除 AcSum。由于按结果集分组的变化，将 AcSum 添加到 values() 也不是一个选项。我想我想做的是根据列内的字段（即id）重新分组查询分组。有什么想法吗？

我想对三列进行分组，然后找到在前三列中重复的所有行的第四个数字列的平均值。我可以通过以下功能实现这一点：

df2 = df.groupby(['col1', 'col2', 'col3'], as_index=False)['col4'].mean()

问题是我还想要第五列，它将聚合由 groupby 函数分组的所有行，我不知道如何在前一个函数之上执行此操作。例如：

df 
index    col1        col2       col3       col4       col5
0        Week_1      James      John       1          when and why?
1        Week_1      James      John       3          How?
2        Week_2      James      John       2          Do you know when?
3        Week_2      Mark       Jim        3          What time?
4        Week_2      Andrew     Simon      1          How far is it?
5        Week_2      Andrew     Simon      2          Are you going?


CURRENT(with above function):
index    col1        col2       col3       col4
0        Week_1      James      John       2
1        Week_2      James …