Pandas与SQL对比


与SQL比较

大多数示例将使用tipspandas测试中找到的数据集。我们将数据读入名为tips的DataFrame中,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,使用您要选择的以逗号分隔的列列表(或* 选择所有列)来完成选择:

SELECT total_bill, tip, smoker, time
FROM tips
LIMIT 5;

使用pandas,通过将列名列表传递给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

在没有列名列表的情况下调用DataFrame将显示所有列(类似于SQL*)。

WHERE

SQL中的过滤是通过WHERE子句完成的。

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

DataFrame可以通过多种方式进行过滤; 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,返回所有带有True的行。

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|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notna()isna() 方法完成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

假设我们有一个与上面的DataFrame结构相同的表。我们只能col2通过以下查询看到IS NULL 的记录:

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

获取col1IS NOT NULL的项目可以完成notna()

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

在pandas中,SQL的GROUP BY操作使用类似命名的 groupby()方法执行。groupby()通常是指我们想要将数据集拆分成组,应用某些功能(通常是聚合),然后将这些组合在一起的过程。

常见的SQL操作是获取整个数据集中每个组中的记录数。例如,有一个需要向我们提供提示中的性别的数量的查询语句:

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

在 pandas 中可以这样:

In [17]: tips.groupby('sex').size()
Out[17]: 
sex
Female     87
Male      157
dtype: int64

请注意,在我们使用的pandas代码中size(),没有 count()。这是因为 count()将函数应用于每个列,返回每个列中的记录数。not 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,我们可以将该count()方法应用于单个列:

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的不同之处 - agg()允许您将字典传递给分组的DataFrame,指示要应用于特定列的函数。

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

通过将列列表传递给groupby()方法来完成多个列的分组 。

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()或执行JOIN merge()。默认情况下, join()将在其索引上加入DataFrame。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(列名称或索引)。

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DataFrames名称和结构相同的数据库表。

现在让我们来看看各种类型的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 -0.282863  1.212112
1   D -1.135632 -0.173215
2   D -1.135632  0.119209

merge() 当您想要将一个DataFrame列与另一个DataFrame索引连接时,还会为这些情况提供参数。

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 -0.282863  1.212112
3   D -1.135632 -0.173215
3   D -1.135632  0.119209

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  0.469112       NaN
1   B -0.282863  1.212112
2   C -1.509059       NaN
3   D -1.135632 -0.173215
4   D -1.135632  0.119209

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]: pd.merge(df1, df2, on='key', how='right')
Out[28]: 
  key   value_x   value_y
0   B -0.282863  1.212112
1   D -1.135632 -0.173215
2   D -1.135632  0.119209
3   E       NaN -1.044236

FULL JOIN

pandas还允许显示数据集两侧的FULL JOIN,无论连接列是否找到匹配项。在编写时,所有RDBMS(MySQL)都不支持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  0.469112       NaN
1   B -0.282863  1.212112
2   C -1.509059       NaN
3   D -1.135632 -0.173215
4   D -1.135632  0.119209
5   E       NaN -1.044236

UNION

UNION ALL可以使用concat()

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,但是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concat()结合使用 drop_duplicates()

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等同于某些SQL分析和聚合函数

带有偏移量的前N行

-- 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;

让我们找到每个性别组(等级<3)的提示(提示<2)。请注意,使用rank(method='min')函数时 rnk_min对于相同的提示保持不变 (如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

更新(UPDATE)

UPDATE tips
SET tip = tip*2
WHERE tip < 2;
In [38]: tips.loc[tips['tip'] < 2, 'tip'] *= 2

删除(DELETE)

DELETE FROM tips
WHERE tip > 9;

在pandas中,我们选择应保留的行,而不是删除它们

In [39]: tips = tips.loc[tips['tip'] <= 9]