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# 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.githubusercontent.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` or `copy=False` keyword argument available for some methods:

    df.replace(5, inplace=True)

There is an active discussion about deprecating and removing `inplace` and `copy` for most methods (e.g. `dropna`) except for a very small subset of methods (including `replace`). Both keywords won’t be necessary anymore in the context of Copy-on-Write. The proposal can be found here.

## 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]:
    time
    True     176
    False     68
    Name: count, 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 `DataFrameGroupBy.size()` and not `DataFrameGroupBy.count()`. This is because `DataFrameGroupBy.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 `DataFrameGroupBy.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 - `DataFrameGroupBy.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": "mean", "day": "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": ["size", "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

`JOIN`s 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 `JOIN`s.

### 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

`merge()` also supports joining on multiple columns by passing a list of column names.

    SELECT *
    FROM df1_multi
    INNER JOIN df2_multi
      ON df1_multi.key1 = df2_multi.key1
        AND df1_multi.key2 = df2_multi.key2;

    In [29]: df1_multi = pd.DataFrame({
       ....:     "key1": ["A", "B", "C", "D"],
       ....:     "key2": [1, 2, 3, 4],
       ....:     "value": np.random.randn(4)
       ....: })
       ....:

    In [30]: df2_multi = pd.DataFrame({
       ....:     "key1": ["B", "D", "D", "E"],
       ....:     "key2": [2, 4, 4, 5],
       ....:     "value": np.random.randn(4)
       ....: })
       ....:

    In [31]: pd.merge(df1_multi, df2_multi, on=["key1", "key2"])
    Out[31]:
      key1  key2   value_x   value_y
    0    B     2 -2.104569  0.721555
    1    D     4  1.071804 -0.706771
    2    D     4  1.071804 -1.039575

If the columns have different names between DataFrames, on can be replaced with left_on and right_on.

    In [32]: df2_multi = pd.DataFrame({
       ....:     "key_1": ["B", "D", "D", "E"],
       ....:     "key_2": [2, 4, 4, 5],
       ....:     "value": np.random.randn(4)
       ....: })
       ....:

    In [33]: pd.merge(df1_multi, df2_multi, left_on=["key1", "key2"], right_on=["key_1", "key_2"])
    Out[33]:
      key1  key2   value_x key_1  key_2   value_y
    0    B     2 -2.104569     B      2 -0.424972
    1    D     4  1.071804     D      4  0.567020
    2    D     4  1.071804     D      4  0.276232

### LEFT OUTER JOIN

Show all records from `df1`.

    SELECT *
    FROM df1
    LEFT OUTER JOIN df2
      ON df1.key = df2.key;

    In [34]: pd.merge(df1, df2, on="key", how="left")
    Out[34]:
      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 [35]: pd.merge(df1, df2, on="key", how="right")
    Out[35]:
      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 JOIN`s, which display both sides of the dataset, whether or not the joined columns find a match. As of writing, `FULL JOIN`s 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 [36]: pd.merge(df1, df2, on="key", how="outer")
    Out[36]:
      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 [37]: df1 = pd.DataFrame(
       ....:     {"city": ["Chicago", "San Francisco", "New York City"], "rank": range(1, 4)}
       ....: )
       ....:

    In [38]: 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 [39]: pd.concat([df1, df2])
    Out[39]:
                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 [40]: pd.concat([df1, df2]).drop_duplicates()
    Out[40]:
                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 [41]: tips.head(10)
    Out[41]:
       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 [42]: tips.nlargest(10 + 5, columns="tip").tail(10)
    Out[42]:
         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 [43]: (
       ....:     tips.assign(
       ....:         rn=tips.sort_values(["total_bill"], ascending=False)
       ....:         .groupby(["day"])
       ....:         .cumcount()
       ....:         + 1
       ....:     )
       ....:     .query("rn < 3")
       ....:     .sort_values(["day", "rn"])
       ....: )
       ....:
    Out[43]:
         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 [44]: (
       ....:     tips.assign(
       ....:         rnk=tips.groupby(["day"])["total_bill"].rank(
       ....:             method="first", ascending=False
       ....:         )
       ....:     )
       ....:     .query("rnk < 3")
       ....:     .sort_values(["day", "rnk"])
       ....: )
       ....:
    Out[44]:
         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 [45]: (
       ....:     tips[tips["tip"] < 2]
       ....:     .assign(rnk_min=tips.groupby(["sex"])["tip"].rank(method="min"))
       ....:     .query("rnk_min < 3")
       ....:     .sort_values(["sex", "rnk_min"])
       ....: )
       ....:
    Out[45]:
         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 [46]: 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 the rows that should be removed:

    In [47]: tips = tips.loc[tips["tip"] <= 9]