# 5.5. Numbers as Indices¶

Enough about movie budgets, it’s time to budget my time instead. Because I schedule my day to the minute, I like to be able to look up movies by their runtime, so that when I have a spare two hours and 34 minutes, I can find all the movies that would fit precisely in that time slot. (Popcorn-making time is budgeted separately).

Before you start, here is a refresher on the index operator in Pandas.

Selecting Columns of a DataFrame

• df[<string>] gets me a column and returns the Series corresponding to that column.

• df[<list of strings>] gets me a bunch of columns and returns a DataFrame.

Selecting Rows of a DataFrame

• df[<series/list of Boolean>] gets me the rows for each element in the list like thing you passed me that is True. However, I think this is confusing and whenever you want to select some rows of a DataFrame you should use df.loc[].

• df.loc[<series/list of Boolean>] behaves just like df[<series/list of Boolean>].

• df.loc[<string>] uses the non-numeric row index and returns the row(s) for that index value.

• df.loc[<string1>:<string2>] uses the non-numeric index and returns a data frame composed of the rows starting with string1 and ending with each string2.

• df.loc[<list/Series of strings>] returns a data frame composed of each row from df with an index value that matches a string in the list.

If you use an integer in any of the last four examples, it works just like the string, but the index values are numeric instead. What is important (and confusing) about this is that they use the index, not the position. So, if you create a data frame with 4 rows of some data, it will have an index that is created by default where the first row starts with 0, the next row is 1 and so on. However, if you sort the data frame such that the last row becomes the first and the first row becomes the last, using df.loc[0] on the sorted data frame will return the last row.

If you want to be strictly positional, you should use df.iloc[0], which will return the first row regardless of the index value. df.iloc[0:5] is the same as doing df.head(), and df.iloc[[1, 3, 5, 7]] will return four rows: the 2nd, 4th, 6th and 8th.

import pandas as pd
df = pd.DataFrame({'a':list("pythonrocks"), 'b':[1,2,3,4,5,6,7,8,9,10,11]})
df = df.set_index('a')
df.loc['p':'n']

b
a
p 1
y 2
t 3
h 4
o 5
n 6

OK, but what if we do this:

df.loc['p':'o']

---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)
<ipython-input-2-2a2fe84d1c0c> in <module>
----> 1 df.loc['p':'o']

~/.local/lib/python3.7/site-packages/pandas/core/indexing.py in __getitem__(self, key)
1766
1767             maybe_callable = com.apply_if_callable(key, self.obj)
-> 1768             return self._getitem_axis(maybe_callable, axis=axis)
1769
1770     def _is_scalar_access(self, key: Tuple):

~/.local/lib/python3.7/site-packages/pandas/core/indexing.py in _getitem_axis(self, key, axis)
1910         if isinstance(key, slice):
1911             self._validate_key(key, axis)
-> 1912             return self._get_slice_axis(key, axis=axis)
1913         elif com.is_bool_indexer(key):
1914             return self._getbool_axis(key, axis=axis)

~/.local/lib/python3.7/site-packages/pandas/core/indexing.py in _get_slice_axis(self, slice_obj, axis)
1795         labels = obj._get_axis(axis)
1796         indexer = labels.slice_indexer(
-> 1797             slice_obj.start, slice_obj.stop, slice_obj.step, kind=self.name
1798         )
1799

~/.local/lib/python3.7/site-packages/pandas/core/indexes/base.py in slice_indexer(self, start, end, step, kind)
4710         slice(1, 3)
4711         """
-> 4712         start_slice, end_slice = self.slice_locs(start, end, step=step, kind=kind)
4713
4714         # return a slice

~/.local/lib/python3.7/site-packages/pandas/core/indexes/base.py in slice_locs(self, start, end, step, kind)
4929         end_slice = None
4930         if end is not None:
-> 4931             end_slice = self.get_slice_bound(end, "right", kind)
4932         if end_slice is None:
4933             end_slice = len(self)

~/.local/lib/python3.7/site-packages/pandas/core/indexes/base.py in get_slice_bound(self, label, side, kind)
4856             if isinstance(slc, np.ndarray):
4857                 raise KeyError(
-> 4858                     f"Cannot get {side} slice bound for non-unique "
4859                     f"label: {repr(original_label)}"
4860                 )

KeyError: "Cannot get right slice bound for non-unique label: 'o'"


Pandas raises an error because there are two ‘o’s in the index. It doesn’t know which one you mean, first? last? If you argue it should use the last then consider the performance implications if this was a really large index? In that case it would be very time consuming to search the index for the last occurance.

On the other hand, if we sort the index then the last instance can be found quite quickly, and with a sorted index loc will work for this example.

df = df.sort_index()
df.loc['c':'o']

b
a
c 9
h 4
k 10
n 6
o 5
o 8

## 5.5.1. Practice Questions¶

Create a Series called time_scheduler that is indexed by runtime and has the movie’s title as its values. Note that you will need to use sort_index() in order to be able to look up movies by their duration. Base yourself on df rather than budget_df.

While you’re at it, remove any movie that is less than 10 minutes (you can’t get into it if it’s too short) or longer than 3 hours (who’s got time for that?).

Hint: You may have to use pd.to_numeric to force the runtimes to be numbers (instead of numbers in a string).

Here is a simpler example that shows the movies that are 7 minutes long

 import pandas as pd
time_scheduler = df.set_index('runtime')
time_scheduler = time_scheduler[['title', 'release_date']]

title release_date
runtime
7.0 Balance 1989-01-01
7.0 Killer Bean 2: The Party 2000-08-08
7.0 The Employment 2008-01-01
7.0 Moscow Clad in Snow 1909-04-09
7.0 Paperman 2012-11-02

Now let’s find all those two-hour-and-34-minute movies.

But what is the 155th shortest movie in this collection?

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