Question

我正在寻找一种方法来编写一个SQL查询，该查询将第一个离散差异应用于原始系列。使用Pandas的.diff()方法在Python中这非常容易：

import pandas as pd
import numpy as np

df = pd.DataFrame(np.random.randint(0,100,size=(10, 2)), columns=list('AB'))

df["diff_A"]=df["A"].diff()
df["diff_B"]=df["B"].diff()

print(df)

我希望的输出在"diff_A"和"diff_B"列中展示：

    A   B  diff_A  diff_B
0  36  14     NaN     NaN
1  32  13    -4.0    -1.0
2  31  87    -1.0    74.0
3  58  88    27.0     1.0
4  44  34   -14.0   -54.0
5   2  43   -42.0     9.0
6  15  94    13.0    51.0
7  46  74    31.0   -20.0
8  60   9    14.0   -65.0
9  43  57   -17.0    48.0

我使用的是Oracle，但我绝对更喜欢干净的ANSI解决方案。

Answer 1

IIUC您可以使用分析LAG功能：

with v as (
  select rowid as rn, a, b from tab
)
select
  a, b,
  a - lag(a, 1) over(order by rn) as diff_a,
  b - lag(b, 1) over(order by rn) as diff_b
from v
order by rn;

PS使用真实列（如日期）进行排序会更好，因为rowid can be changed。

例如：

select
  a, b,
  a - lag(a, 1) over(order by inserted) as diff_a,
  b - lag(b, 1) over(order by inserted) as diff_b
from tab;

@MatBailie has posted a very good explanation：

SQL中的数据集是无序的。对于LAG中的确定性结果（）始终使用足够的ORDER BY子句。（如果不存在这样的字段，则一个应该在插入到SQL数据中的数据之前/之前创建组。 SQL数据集的无序特性允许大量的数据集可扩展性选项和优化选项。）

SQL Fiddle test

PS Windowing functions were added to the ANSI/ISO Standard SQL:2003 and then extended in ANSI/ISO Standard SQL:2008. Microsoft was late to this game. DB2, Oracle, Sybase, PostgreSQL and other products have had full implementations for years. SQL Server did not catch up until SQL 2012.

Answer 2

我发布这个答案只是因为我能够在接受答案中的评论之后在SQLFiddle中复制结果。除了事后rowid改变之外，还有一个有效的论据，为什么这个简单的答案不起作用。

select
  a, b,
  a - lag(a, 1) over(order by rowid) as diff_a,
  b - lag(b, 1) over(order by rowid) as diff_b
from tab;

Pandas diff（）函数的SQL模拟（第一个离散差异）[LAG函数]

2 个答案: