Comparison with SQL

由于许多潜在的pandas用户对SQL有一些熟悉,因此本页面将提供一些使用pandas执行各种SQL操作的示例。

如果你是新来的熊猫,你可能需要先阅读10 Minutes to pandas,以熟悉自己的图书馆。

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

大多数示例将使用在pandas测试中发现的tips数据集。我们将数据读入一个名为提示的DataFrame,并假设我们有一个具有相同名称和结构的数据库表。

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;

使用pandas,通过将列名列表传递到DataFrame来完成列选择:

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中的过滤是通过WHERE子句完成的。

SELECT *
FROM tips
WHERE time = 'Dinner'
LIMIT 5;

DataFrames可以以多种方式进行过滤;其中最直观的是使用布尔索引

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 / False对象传递给DataFrame,返回所有行为True。

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(5Out[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|将多个条件传递给DataFrame (OR)和&(AND)。

-- 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

使用notnull()isnull()方法进行NULL检查。

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

假设我们有一个与我们的DataFrame结构相同的表。通过以下查询,我们只能看到col2 IS NULL的记录:

SELECT *
FROM frame
WHERE col2 IS NULL;
In [15]: frame[frame['col2'].isnull()]
Out[15]: 
  col1 col2
1    B  NaN

使用notnull()可以处理col1 IS NOT NULL的项目。

SELECT *
FROM frame
WHERE col1 IS NOT NULL;
In [16]: frame[frame['col1'].notnull()]
Out[16]: 
  col1 col2
0    A    F
1    B  NaN
3    C    H
4    D    I

GROUP BY

在pandas中,SQL的GROUP BY操作使用类似命名的groupby()方法执行。groupby()通常指的是一个过程,其中我们要将数据集拆分成组,应用一些函数(通常是聚合),然后将组合在一起。

常见的SQL操作是获取数据集中每个组中的记录数。例如,一个查询获得我们按性别留下的提示数量:

SELECT sex, count(*)
FROM tips
GROUP BY sex;
/*
Female     87
Male      157
*/

熊猫相当于:

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,返回每个列中的而不是 null

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,假设我们希望查看提示量与星期几不同 - agg()允许您将字典传递到已分组的DataFrame,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()merge()执行JOIN。默认情况下,join()将在其索引上加入DataFrames。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;
# merge performs an INNER JOIN by default
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()还提供了您想要将DataFrame的列与另一个DataFrame的索引相连接的情况下的参数。

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;
# show all records from df1
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;
# show all records from df2
In [28]: p$L̫WxnME$L̫Wxn/span>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,它显示数据集的两侧,无论连接的列是否找到匹配。从写作,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()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 equivalents for some SQL analytic and aggregate functions

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

相同使用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;

让我们找到提示(排名请注意,对于相同的提示(作为Oracle的RANK()函数)使用rank(method='min')函数rnk_min

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

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]
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