5.4. 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

Selecting Rows of a DataFrame

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 thsi 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

5.4.1. 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
 df = pd.read_csv("https://media.githubusercontent.com/media/bnmnetp/httlads/master/Data/movies_metadata.csv").dropna(axis=1, how='all')
time_scheduler = df.set_index('runtime')
time_scheduler = time_scheduler[['title', 'release_date']]
time_scheduler.loc[7].head()
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.

Q-1: How many movies lasting 154 minutes are there?

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

Q-2: Copy and paste the name of the 155th shortest movie in this collection, without quotes.

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