Comparison with SQL¶
Since many potential pandas users have some familiarity with SQL, this page is meant to provide some examples of how various SQL operations would be performed using pandas.
If you’re new to pandas, you might want to first read through 10 Minutes to pandas to familiarize yourself with the library.
As is customary, we import pandas and numpy as follows:
In [1]: import pandas as pd
In [2]: import numpy as np
Most of the examples will utilize the tips dataset found within pandas tests. We’ll read the data into a DataFrame called tips and assume we have a database table of the same name and structure.
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
In SQL, selection is done using a comma-separated list of columns you’d like to select (or a * to select all columns):
SELECT total_bill, tip, smoker, time FROM tips LIMIT 5;
With pandas, column selection is done by passing a list of column names to your DataFrame:
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
Calling the DataFrame without the list of column names would display all columns (akin to SQL’s *).
WHERE
Filtering in SQL is done via a WHERE clause.
SELECT *
FROM tips
WHERE time = 'Dinner' LIMIT 5;
DataFrames can be filtered in multiple ways; the most intuitive of which is using boolean indexing.
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
The above statement is simply passing a Series of True/False objects to the DataFrame, returning all rows with 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(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
Just like SQL’s OR and AND, multiple conditions can be passed to a DataFrame using | (OR) and & (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
NULL checking is done using the notna() and isna() methods.
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
Assume we have a table of the same structure as our DataFrame above. We can see only the records where col2 IS NULL with the following query:
SELECT *
FROM frame
WHERE col2 IS NULL;
In [15]: frame[frame['col2'].isna()] Out[15]: col1 col2 1 B NaN
Getting items where col1 IS NOT NULL can be done with 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
In pandas, SQL’s GROUP BY operations are performed using the similarly named groupby() method. groupby()typically refers to a process where we’d like to split a dataset into groups, apply some function (typically aggregation) , and then combine the groups together.
A common SQL operation would be getting the count of records in each group throughout a dataset. For instance, a query getting us the number of tips left by sex:
SELECT sex, count(*) FROM tips GROUP BY sex; /* Female 87 Male 157 */
The pandas equivalent would be:
In [17]: tips.groupby('sex').size() Out[17]: sex Female 87 Male 157 dtype: int64
Notice that in the pandas code we used size() and not count(). This is because count() applies the function to each column, returning the number of not null records within each.
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
Alternatively, we could have applied the count() method to an individual column:
In [19]: tips.groupby('sex')['total_bill'].count() Out[19]: sex Female 87 Male 157 Name: total_bill, dtype: int64
Multiple functions can also be applied at once. For instance, say we’d like to see how tip amount differs by day of the week - agg() allows you to pass a dictionary to your grouped DataFrame, indicating which functions to apply to specific columns.
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
Grouping by more than one column is done by passing a list of columns to the groupby() method.
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
JOINs can be performed with join() or merge(). By default, join() will join the DataFrames on their indices. Each method has parameters allowing you to specify the type of join to perform (LEFT, RIGHT, INNER, FULL) or the columns to join on (column names or indices).
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)}) ....:
Assume we have two database tables of the same name and structure as our DataFrames.
Now let’s go over the various types of JOINs.
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() also offers parameters for cases when you’d like to join one DataFrame’s column with another DataFrame’s index.
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]: 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 also allows for FULL JOINs, which display both sides of the dataset, whether or not the joined columns find a match. As of writing, FULL JOINs are not supported in all RDBMS (MySQL).
-- 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
UNION ALL can be performed using 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’s UNION is similar to UNION ALL, however UNION will remove duplicate rows.
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 */
In pandas, you can use concat() in conjunction with 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 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
Top N rows per group
-- 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
the same using rank(method=’first’) function
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;
Let’s find tips with (rank < 3) per gender group for (tips < 2). Notice that when using rank(method='min') functionrnk_min remains the same for the same tip (as Oracle’s RANK() function)
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;
In pandas we select the rows that should remain, instead of deleting them
In [39]: tips = tips.loc[tips['tip'] <= 9]