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/main/pandas/tests/io/data/csv/tips.csv"
...: )
...:
In [4]: tips = pd.read_csv(url)
In [5]: tips
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
.. ... ... ... ... ... ... ...
239 29.03 5.92 Male No Sat Dinner 3
240 27.18 2.00 Female Yes Sat Dinner 2
241 22.67 2.00 Male Yes Sat Dinner 2
242 17.82 1.75 Male No Sat Dinner 2
243 18.78 3.00 Female No Thur Dinner 2
[244 rows x 7 columns]
Copies vs. in place operations¶
Most pandas operations return copies of the Series/DataFrame. To make the changes “stick”,
you’ll need to either assign to a new variable:
sorted_df = df.sort_values("col1")
or overwrite the original one:
df = df.sort_values("col1")
Note
You will see an inplace=True keyword argument available for some methods:
df.sort_values("col1", inplace=True)
Its use is discouraged. More information.
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;
With pandas, column selection is done by passing a list of column names to your DataFrame:
In [6]: tips[["total_bill", "tip", "smoker", "time"]]
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
.. ... ... ... ...
239 29.03 5.92 No Dinner
240 27.18 2.00 Yes Dinner
241 22.67 2.00 Yes Dinner
242 17.82 1.75 No Dinner
243 18.78 3.00 No Dinner
[244 rows x 4 columns]
Calling the DataFrame without the list of column names would display all columns (akin to SQL’s
*).
In SQL, you can add a calculated column:
SELECT *, tip/total_bill as tip_rate
FROM tips;
With pandas, you can use the DataFrame.assign() method of a DataFrame to append a new column:
In [7]: tips.assign(tip_rate=tips["tip"] / tips["total_bill"])
Out[7]:
total_bill tip sex smoker day time size tip_rate
0 16.99 1.01 Female No Sun Dinner 2 0.059447
1 10.34 1.66 Male No Sun Dinner 3 0.160542
2 21.01 3.50 Male No Sun Dinner 3 0.166587
3 23.68 3.31 Male No Sun Dinner 2 0.139780
4 24.59 3.61 Female No Sun Dinner 4 0.146808
.. ... ... ... ... ... ... ... ...
239 29.03 5.92 Male No Sat Dinner 3 0.203927
240 27.18 2.00 Female Yes Sat Dinner 2 0.073584
241 22.67 2.00 Male Yes Sat Dinner 2 0.088222
242 17.82 1.75 Male No Sat Dinner 2 0.098204
243 18.78 3.00 Female No Thur Dinner 2 0.159744
[244 rows x 8 columns]
WHERE¶
Filtering in SQL is done via a WHERE clause.
SELECT *
FROM tips
WHERE time = 'Dinner';
DataFrames can be filtered in multiple ways; the most intuitive of which is using boolean indexing.
In [8]: tips[tips["total_bill"] > 10]
Out[8]:
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
.. ... ... ... ... ... ... ...
239 29.03 5.92 Male No Sat Dinner 3
240 27.18 2.00 Female Yes Sat Dinner 2
241 22.67 2.00 Male Yes Sat Dinner 2
242 17.82 1.75 Male No Sat Dinner 2
243 18.78 3.00 Female No Thur Dinner 2
[227 rows x 7 columns]
The above statement is simply passing a Series of True/False objects to the DataFrame,
returning all rows with True.
In [9]: is_dinner = tips["time"] == "Dinner"
In [10]: is_dinner
Out[10]:
0 True
1 True
2 True
3 True
4 True
...
239 True
240 True
241 True
242 True
243 True
Name: time, Length: 244, dtype: bool
In [11]: is_dinner.value_counts()
Out[11]:
True 176
False 68
Name: time, dtype: int64
In [12]: tips[is_dinner]
Out[12]:
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
.. ... ... ... ... ... ... ...
239 29.03 5.92 Male No Sat Dinner 3
240 27.18 2.00 Female Yes Sat Dinner 2
241 22.67 2.00 Male Yes Sat Dinner 2
242 17.82 1.75 Male No Sat Dinner 2
243 18.78 3.00 Female No Thur Dinner 2
[176 rows x 7 columns]
Just like SQL’s OR and AND, multiple conditions can be passed to a DataFrame using |
(OR) and & (AND).
Tips of more than $5 at Dinner meals:
SELECT *
FROM tips
WHERE time = 'Dinner' AND tip > 5.00;
In [13]: tips[(tips["time"] == "Dinner") & (tips["tip"] > 5.00)]
Out[13]:
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;
In [14]: tips[(tips["size"] >= 5) | (tips["total_bill"] > 45)]
Out[14]:
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 [15]: frame = pd.DataFrame(
....: {"col1": ["A", "B", np.NaN, "C", "D"], "col2": ["F", np.NaN, "G", "H", "I"]}
....: )
....:
In [16]: frame
Out[16]:
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 [17]: frame[frame["col2"].isna()]
Out[17]:
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 [18]: frame[frame["col1"].notna()]
Out[18]:
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 [19]: tips.groupby("sex").size()
Out[19]:
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 [20]: tips.groupby("sex").count()
Out[20]:
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 [21]: tips.groupby("sex")["total_bill"].count()
Out[21]:
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
Thu 2.771452 62
*/
In [22]: tips.groupby("day").agg({"tip": np.mean, "day": np.size})
Out[22]:
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
Thu 45 2.673778
Yes Fri 15 2.714000
Sat 42 2.875476
Sun 19 3.516842
Thu 17 3.030000
*/
In [23]: tips.groupby(["smoker", "day"]).agg({"tip": [np.size, np.mean]})
Out[23]:
tip
size mean
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
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).
Warning
If both key columns contain rows where the key is a null value, those rows will be matched against each other. This is different from usual SQL join behaviour and can lead to unexpected results.
In [24]: df1 = pd.DataFrame({"key": ["A", "B", "C", "D"], "value": np.random.randn(4)})
In [25]: 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 [26]: pd.merge(df1, df2, on="key")
Out[26]:
key value_x value_y
0 B -0.282863 1.212112
1 D -1.135632 -0.173215
2 D -1.135632 0.119209
merge() also offers parameters for cases when you’d like to join one DataFrame’s
column with another DataFrame’s index.
In [27]: indexed_df2 = df2.set_index("key")
In [28]: pd.merge(df1, indexed_df2, left_on="key", right_index=True)
Out[28]:
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;
In [29]: pd.merge(df1, df2, on="key", how="left")
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
RIGHT JOIN¶
Show all records from df2.
SELECT *
FROM df1
RIGHT OUTER JOIN df2
ON df1.key = df2.key;
In [30]: pd.merge(df1, df2, on="key", how="right")
Out[30]:
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 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;
In [31]: pd.merge(df1, df2, on="key", how="outer")
Out[31]:
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 can be performed using concat().
In [32]: df1 = pd.DataFrame(
....: {"city": ["Chicago", "San Francisco", "New York City"], "rank": range(1, 4)}
....: )
....:
In [33]: 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 [34]: pd.concat([df1, df2])
Out[34]:
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 [35]: pd.concat([df1, df2]).drop_duplicates()
Out[35]:
city rank
0 Chicago 1
1 San Francisco 2
2 New York City 3
1 Boston 4
2 Los Angeles 5
LIMIT¶
SELECT * FROM tips
LIMIT 10;
In [36]: tips.head(10)
Out[36]:
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
5 25.29 4.71 Male No Sun Dinner 4
6 8.77 2.00 Male No Sun Dinner 2
7 26.88 3.12 Male No Sun Dinner 4
8 15.04 1.96 Male No Sun Dinner 2
9 14.78 3.23 Male No Sun Dinner 2
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 [37]: tips.nlargest(10 + 5, columns="tip").tail(10)
Out[37]:
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 [38]: (
....: tips.assign(
....: rn=tips.sort_values(["total_bill"], ascending=False)
....: .groupby(["day"])
....: .cumcount()
....: + 1
....: )
....: .query("rn < 3")
....: .sort_values(["day", "rn"])
....: )
....:
Out[38]:
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 [39]: (
....: tips.assign(
....: rnk=tips.groupby(["day"])["total_bill"].rank(
....: method="first", ascending=False
....: )
....: )
....: .query("rnk < 3")
....: .sort_values(["day", "rnk"])
....: )
....:
Out[39]:
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') function
rnk_min remains the same for the same tip
(as Oracle’s RANK() function)
In [40]: (
....: tips[tips["tip"] < 2]
....: .assign(rnk_min=tips.groupby(["sex"])["tip"].rank(method="min"))
....: .query("rnk_min < 3")
....: .sort_values(["sex", "rnk_min"])
....: )
....:
Out[40]:
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 [41]: 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 [42]: tips = tips.loc[tips["tip"] <= 9]