Pandas 与 SQL 对比
Pandas 与 SQL 对比
原文：http://pandas.pydata.org/pandas-docs/stable/comparison_with_sql.html
由于许多pandas用户使用SQL，因此本页面将提供一些使用pandas执行各种SQL操作的示例。如果你未接触过Pandas，你可能需要通过阅读10 Minutes to pandas，以熟悉一下这个库。
按照惯例，我们导入pandas和numpy如下：
In [1]: import pandas as pd
In [2]: import numpy as np
下面的大多数例子将使用在pandas测试集中的tips数据集。我们将数据读入一个名为tips的DataFrame，并假设我们也有一个具有相同名称和结构的数据库表tips。
In [3]: url = 'https://raw.github.com/pandas-dev/pandas/master/pandas/tests/data/tips.csv'
In [4]: tips = pd.read_csv(url)
In [5]: tips.head()
Out[5]:
total_bill tip sex smoker day time size
0 16.99 1.01 Female No Sun Dinner 2
1 10.34 1.66 Male No Sun Dinner 3
2 21.01 3.50 Male No Sun Dinner 3
3 23.68 3.31 Male No Sun Dinner 2
4 24.59 3.61 Female No Sun Dinner 4
SELECT
在SQL中，使用逗号分隔的列选择列（或 * 选择所有列）进行选择：
SELECT total_bill, tip, smoker, time
FROM tips
LIMIT 5;
而pandas，通过将列名列表传递到DataFrame来完成列的选择：
In [6]: tips[['total_bill', 'tip', 'smoker', 'time']].head(5)
Out[6]:
total_bill tip smoker time
0 16.99 1.01 No Dinner
1 10.34 1.66 No Dinner
2 21.01 3.50 No Dinner
3 23.68 3.31 No Dinner
4 24.59 3.61 No Dinner
调用没有列名称列表的DataFrame将显示所有列（类似于SQL的 *）。
WHERE
在SQL中的筛选是通过WHERE子句完成的。
SELECT *
FROM tips
WHERE time = 'Dinner'
LIMIT 5;
DataFrames可以以多种方式进行筛选；其中最直观的是使用关系表达式。
In [7]: tips[tips['time'] == 'Dinner'].head(5)
Out[7]:
total_bill tip sex smoker day time size
0 16.99 1.01 Female No Sun Dinner 2
1 10.34 1.66 Male No Sun Dinner 3
2 21.01 3.50 Male No Sun Dinner 3
3 23.68 3.31 Male No Sun Dinner 2
4 24.59 3.61 Female No Sun Dinner 4
上面的语句只是将一个Series的True 或 False对象传递给DataFrame，返回DataFrame中所有为True的行。
也可以这样来写：
In [8]: is_dinner = tips['time'] == 'Dinner'
In [9]: is_dinner.value_counts()
Out[9]:
True 176
False 68
Name: time, dtype: int64
In [10]: tips[is_dinner].head(5)
Out[10]:
total_bill tip sex smoker day time size
0 16.99 1.01 Female No Sun Dinner 2
1 10.34 1.66 Male No Sun Dinner 3
2 21.01 3.50 Male No Sun Dinner 3
3 23.68 3.31 Male No Sun Dinner 2
4 24.59 3.61 Female No Sun Dinner 4
就像SQL的OR和AND一样，可以使用 | （OR）和＆（AND）将多个条件传递给DataFrame。
注意条件要用圆括号。
-- tips of more than $5.00 at Dinner meals
SELECT *
FROM tips
WHERE time = 'Dinner' AND tip > 5.00;
# tips of more than $5.00 at Dinner meals
In [11]: tips[(tips['time'] == 'Dinner') & (tips['tip'] > 5.00)]
Out[11]:
total_bill tip sex smoker day time size
23 39.42 7.58 Male No Sat Dinner 4
44 30.40 5.60 Male No Sun Dinner 4
47 32.40 6.00 Male No Sun Dinner 4
52 34.81 5.20 Female No Sun Dinner 4
59 48.27 6.73 Male No Sat Dinner 4
116 29.93 5.07 Male No Sun Dinner 4
155 29.85 5.14 Female No Sun Dinner 5
170 50.81 10.00 Male Yes Sat Dinner 3
172 7.25 5.15 Male Yes Sun Dinner 2
181 23.33 5.65 Male Yes Sun Dinner 2
183 23.17 6.50 Male Yes Sun Dinner 4
211 25.89 5.16 Male Yes Sat Dinner 4
212 48.33 9.00 Male No Sat Dinner 4
214 28.17 6.50 Female Yes Sat Dinner 3
239 29.03 5.92 Male No Sat Dinner 3
-- tips by parties of at least 5 diners OR bill total was more than $45
SELECT *
FROM tips
WHERE size >= 5 OR total_bill > 45;
# tips by parties of at least 5 diners OR bill total was more than $45
In [12]: tips[(tips['size'] >= 5) | (tips['total_bill'] > 45)]
Out[12]:
total_bill tip sex smoker day time size
59 48.27 6.73 Male No Sat Dinner 4
125 29.80 4.20 Female No Thur Lunch 6
141 34.30 6.70 Male No Thur Lunch 6
142 41.19 5.00 Male No Thur Lunch 5
143 27.05 5.00 Female No Thur Lunch 6
155 29.85 5.14 Female No Sun Dinner 5
156 48.17 5.00 Male No Sun Dinner 6
170 50.81 10.00 Male Yes Sat Dinner 3
182 45.35 3.50 Male Yes Sun Dinner 3
185 20.69 5.00 Male No Sun Dinner 5
187 30.46 2.00 Male Yes Sun Dinner 5
212 48.33 9.00 Male No Sat Dinner 4
216 28.15 3.00 Male Yes Sat Dinner 5
In [13]: frame = pd.DataFrame({'col1': ['A', 'B', np.NaN, 'C', 'D'],
....: 'col2': ['F', np.NaN, 'G', 'H', 'I']})
....:
In [14]: frame
Out[14]:
col1 col2
0 A F
1 B NaN
2 NaN G
3 C H
4 D I
假设我们有一个与上面的DataFrame结构相同的表。通过以下查询，我们可以筛选出col2 为 NULL的记录：
SELECT *
FROM frame
WHERE col2 IS NULL;
In [15]: frame[frame['col2'].isna()]
Out[15]:
col1 col2
1 B NaN
使用notna()可以处理col1 为 NOT NULL的项目。
SELECT *
FROM frame
WHERE col1 IS NOT NULL;
In [16]: frame[frame['col1'].notna()]
Out[16]:
col1 col2
0 A F
1 B NaN
3 C H
4 D I
GROUP BY
SQL的GROUP BY操作，在pandas中使用groupby()方法执行。groupby()将数据集分组，应用一些函数（通常是聚类），然后组合在一起。
常见的SQL操作是获取数据集中每个组中的记录数。例如，查询tips表中不同性别的数量：
SELECT sex, count(*)
FROM tips
GROUP BY sex;
/*
Female 87
Male 157
*/
相当于pandas的如下操作：
In [17]: tips.groupby('sex').size()
Out[17]:
sex
Female 87
Male 157
dtype: int64
注意，在pandas代码中，我们使用size()而不是count()。这是因为count()将函数应用于每个列，返回每个列中不是Null的个数记录 。
In [18]: tips.groupby('sex').count()
Out[18]:
total_bill tip smoker day time size
sex
Female 87 87 87 87 87 87
Male 157 157 157 157 157 157
或者，我们可以将count()方法应用于单独的列：
In [19]: tips.groupby('sex')['total_bill'].count()
Out[19]:
sex
Female 87
Male 157
Name: total_bill, dtype: int64
也可以一次应用多个功能。例如，假设我们希望查看每天tip的总量 - agg()方法允许您将一个字典传递到已分组的DataFrame中，指明哪些函数应用于特定列。
SELECT day, AVG(tip), COUNT(*)
FROM tips
GROUP BY day;
/*
Fri 2.734737 19
Sat 2.993103 87
Sun 3.255132 76
Thur 2.771452 62
*/
In [20]: tips.groupby('day').agg({'tip': np.mean, 'day': np.size})
Out[20]:
tip day
day
Fri 2.734737 19
Sat 2.993103 87
Sun 3.255132 76
Thur 2.771452 62
通过将列列表传递到groupby()方法来对多个列进行分组。
SELECT smoker, day, COUNT(*), AVG(tip)
FROM tips
GROUP BY smoker, day;
/*
smoker day
No Fri 4 2.812500
Sat 45 3.102889
Sun 57 3.167895
Thur 45 2.673778
Yes Fri 15 2.714000
Sat 42 2.875476
Sun 19 3.516842
Thur 17 3.030000
*/
In [21]: tips.groupby(['smoker', 'day']).agg({'tip': [np.size, np.mean]})
Out[21]:
tip
size mean
smoker day
No Fri 4.0 2.812500
Sat 45.0 3.102889
Sun 57.0 3.167895
Thur 45.0 2.673778
Yes Fri 15.0 2.714000
Sat 42.0 2.875476
Sun 19.0 3.516842
Thur 17.0 3.030000
JOIN
可以使用join()或merge()执行JOIN。默认情况下，DataFramess使用索引JOIN。每个方法都有参数，允许您指定要执行的连接类型（LEFT，RIGHT，INNER，FULL）或要连接的列（列名或索引）。
In [22]: df1 = pd.DataFrame({'key': ['A', 'B', 'C', 'D'],
....: 'value': np.random.randn(4)})
....:
In [23]: df2 = pd.DataFrame({'key': ['B', 'D', 'D', 'E'],
....: 'value': np.random.randn(4)})
....:
假设我们有两个与我们的DataFrames具有相同名称和结构的数据库表。
现在让我们来讨论各种类型的JOIN。
INNER JOIN
SELECT *
FROM df1
INNER JOIN df2
ON df1.key = df2.key;
# 一般merge 相当于一个 INNER JOIN
In [24]: pd.merge(df1, df2, on='key')
Out[24]:
key value_x value_y
0 B -0.318214 0.543581
1 D 2.169960 -0.426067
2 D 2.169960 1.138079
merge()还提供了将DataFrame的列与另一个DataFrame的索引相连接的情形。
In [25]: indexed_df2 = df2.set_index('key')
In [26]: pd.merge(df1, indexed_df2, left_on='key', right_index=True)
Out[26]:
key value_x value_y
1 B -0.318214 0.543581
3 D 2.169960 -0.426067
3 D 2.169960 1.138079
LEFT OUTER JOIN
-- show all records from df1
SELECT *
FROM df1
LEFT OUTER JOIN df2
ON df1.key = df2.key;
# 从表 df1 选取所有记录
In [27]: pd.merge(df1, df2, on='key', how='left')
Out[27]:
key value_x value_y
0 A 0.116174 NaN
1 B -0.318214 0.543581
2 C 0.285261 NaN
3 D 2.169960 -0.426067
4 D 2.169960 1.138079
RIGHT JOIN
-- show all records from df2
SELECT *
FROM df1
RIGHT OUTER JOIN df2
ON df1.key = df2.key;
# 从表 df2 选取所有记录
In [28]: pd.merge(df1, df2, on='key', how='right')
Out[28]:
key value_x value_y
0 B -0.318214 0.543581
1 D 2.169960 -0.426067
2 D 2.169960 1.138079
3 E NaN 0.086073
FULL JOIN
pandas还允许FULL JOIN，它选取两侧数据集，无论连接的列是否找到匹配。记住：所有RDBMS（MySQL）不支持FULL JOIN。
-- show all records from both tables
SELECT *
FROM df1
FULL OUTER JOIN df2
ON df1.key = df2.key;
# show all records from both frames
In [29]: pd.merge(df1, df2, on='key', how='outer')
Out[29]:
key value_x value_y
0 A 0.116174 NaN
1 B -0.318214 0.543581
2 C 0.285261 NaN
3 D 2.169960 -0.426067
4 D 2.169960 1.138079
5 E NaN 0.086073
UNION
可以使用concat()执行UNION ALL。
In [30]: df1 = pd.DataFrame({'city': ['Chicago', 'San Francisco', 'New York City'],
....: 'rank': range(1, 4)})
....:
In [31]: df2 = pd.DataFrame({'city': ['Chicago', 'Boston', 'Los Angeles'],
....: 'rank': [1, 4, 5]})
....:
SELECT city, rank
FROM df1
UNION ALL
SELECT city, rank
FROM df2;
/*
city rank
Chicago 1
San Francisco 2
New York City 3
Chicago 1
Boston 4
Los Angeles 5
*/
In [32]: pd.concat([df1, df2])
Out[32]:
city rank
0 Chicago 1
1 San Francisco 2
2 New York City 3
0 Chicago 1
1 Boston 4
2 Los Angeles 5
SQL的UNION类似于UNION ALL，但UNION将删除重复的行。
SELECT city, rank
FROM df1
UNION
SELECT city, rank
FROM df2;
-- notice that there is only one Chicago record this time
/*
city rank
Chicago 1
San Francisco 2
New York City 3
Boston 4
Los Angeles 5
*/
在pandas中，您可以使用concat()与drop_duplicates()结合使用。
In [33]: pd.concat([df1, df2]).drop_duplicates()
Out[33]:
city rank
0 Chicago 1
1 San Francisco 2
2 New York City 3
1 Boston 4
2 Los Angeles 5
Pandas还有一些方法可对应于SQL分析和聚合函数
Top N rows with offset
-- MySQL
SELECT * FROM tips
ORDER BY tip DESC
LIMIT 10 OFFSET 5;
In [34]: tips.nlargest(10+5, columns='tip').tail(10)
Out[34]:
total_bill tip sex smoker day time size
183 23.17 6.50 Male Yes Sun Dinner 4
214 28.17 6.50 Female Yes Sat Dinner 3
47 32.40 6.00 Male No Sun Dinner 4
239 29.03 5.92 Male No Sat Dinner 3
88 24.71 5.85 Male No Thur Lunch 2
181 23.33 5.65 Male Yes Sun Dinner 2
44 30.40 5.60 Male No Sun Dinner 4
52 34.81 5.20 Female No Sun Dinner 4
85 34.83 5.17 Female No Thur Lunch 4
211 25.89 5.16 Male Yes Sat Dinner 4
每组前N行
-- Oracle's ROW_NUMBER() analytic function
SELECT * FROM (
SELECT
t.*,
ROW_NUMBER() OVER(PARTITION BY day ORDER BY total_bill DESC) AS rn
FROM tips t
)
WHERE rn < 3
ORDER BY day, rn;
In [35]: (tips.assign(rn=tips.sort_values(['total_bill'], ascending=False)
....: .groupby(['day'])
....: .cumcount() + 1)
....: .query('rn < 3')
....: .sort_values(['day','rn'])
....: )
....:
Out[35]:
total_bill tip sex smoker day time size rn
95 40.17 4.73 Male Yes Fri Dinner 4 1
90 28.97 3.00 Male Yes Fri Dinner 2 2
170 50.81 10.00 Male Yes Sat Dinner 3 1
212 48.33 9.00 Male No Sat Dinner 4 2
156 48.17 5.00 Male No Sun Dinner 6 1
182 45.35 3.50 Male Yes Sun Dinner 3 2
197 43.11 5.00 Female Yes Thur Lunch 4 1
142 41.19 5.00 Male No Thur Lunch 5 2
同样使用rank（method ='first'）函数
In [36]: (tips.assign(rnk=tips.groupby(['day'])['total_bill']
....: .rank(method='first', ascending=False))
....: .query('rnk < 3')
....: .sort_values(['day','rnk'])
....: )
....:
Out[36]:
total_bill tip sex smoker day time size rnk
95 40.17 4.73 Male Yes Fri Dinner 4 1.0
90 28.97 3.00 Male Yes Fri Dinner 2 2.0
170 50.81 10.00 Male Yes Sat Dinner 3 1.0
212 48.33 9.00 Male No Sat Dinner 4 2.0
156 48.17 5.00 Male No Sun Dinner 6 1.0
182 45.35 3.50 Male Yes Sun Dinner 3 2.0
197 43.11 5.00 Female Yes Thur Lunch 4 1.0
142 41.19 5.00 Male No Thur Lunch 5 2.0
-- Oracle's RANK() analytic function
SELECT * FROM (
SELECT
t.*,
RANK() OVER(PARTITION BY sex ORDER BY tip) AS rnk
FROM tips t
WHERE tip < 2
)
WHERE rnk < 3
ORDER BY sex, rnk;
我们来找tips中(rnk<3)(tips<2)（请注意，当使用rank(method='min')相当于Oracle的RANK()函数）
In [37]: (tips[tips['tip'] < 2]
....: .assign(rnk_min=tips.groupby(['sex'])['tip']
....: .rank(method='min'))
....: .query('rnk_min < 3')
....: .sort_values(['sex','rnk_min'])
....: )
....:
Out[37]:
total_bill tip sex smoker day time size rnk_min
67 3.07 1.00 Female Yes Sat Dinner 1 1.0
92 5.75 1.00 Female Yes Fri Dinner 2 1.0
111 7.25 1.00 Female No Sat Dinner 1 1.0
236 12.60 1.00 Male Yes Sat Dinner 2 1.0
237 32.83 1.17 Male Yes Sat Dinner 2 2.0
UPDATE
UPDATE tips
SET tip = tip*2
WHERE tip < 2;
In [38]: tips.loc[tips['tip'] < 2, 'tip'] *= 2
现在.loc已不推荐使用了，有警告信息。
DELETE
DELETE FROM tips
WHERE tip > 9;
在pandas中，我们选择应该保留的行，而不是删除它们
In [39]: tips = tips.loc[tips['tip'] <= 9]
