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

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')
p 1
y 2
t 3
h 4
o 5
n 6

OK, but what if we do this:

KeyError                                  Traceback (most recent call last)
Cell In[2], line 1
----> 1 df.loc['p':'o']

File ~/.local/lib/python3.10/site-packages/pandas/core/indexing.py:1103, in _LocationIndexer.__getitem__(self, key)
   1100 axis = self.axis or 0
   1102 maybe_callable = com.apply_if_callable(key, self.obj)
-> 1103 return self._getitem_axis(maybe_callable, axis=axis)

File ~/.local/lib/python3.10/site-packages/pandas/core/indexing.py:1323, in _LocIndexer._getitem_axis(self, key, axis)
   1321 if isinstance(key, slice):
   1322     self._validate_key(key, axis)
-> 1323     return self._get_slice_axis(key, axis=axis)
   1324 elif com.is_bool_indexer(key):
   1325     return self._getbool_axis(key, axis=axis)

File ~/.local/lib/python3.10/site-packages/pandas/core/indexing.py:1355, in _LocIndexer._get_slice_axis(self, slice_obj, axis)
   1352     return obj.copy(deep=False)
   1354 labels = obj._get_axis(axis)
-> 1355 indexer = labels.slice_indexer(slice_obj.start, slice_obj.stop, slice_obj.step)
   1357 if isinstance(indexer, slice):
   1358     return self.obj._slice(indexer, axis=axis)

File ~/.local/lib/python3.10/site-packages/pandas/core/indexes/base.py:6344, in Index.slice_indexer(self, start, end, step)
   6300 def slice_indexer(
   6301     self,
   6302     start: Hashable | None = None,
   6303     end: Hashable | None = None,
   6304     step: int | None = None,
   6305 ) -> slice:
   6306     """
   6307     Compute the slice indexer for input labels and step.
   6342     slice(1, 3, None)
   6343     """
-> 6344     start_slice, end_slice = self.slice_locs(start, end, step=step)
   6346     # return a slice
   6347     if not is_scalar(start_slice):

File ~/.local/lib/python3.10/site-packages/pandas/core/indexes/base.py:6543, in Index.slice_locs(self, start, end, step)
   6541 end_slice = None
   6542 if end is not None:
-> 6543     end_slice = self.get_slice_bound(end, "right")
   6544 if end_slice is None:
   6545     end_slice = len(self)

File ~/.local/lib/python3.10/site-packages/pandas/core/indexes/base.py:6470, in Index.get_slice_bound(self, label, side)
   6468     slc = lib.maybe_booleans_to_slice(slc.view("u1"))
   6469     if isinstance(slc, np.ndarray):
-> 6470         raise KeyError(
   6471             f"Cannot get {side} slice bound for non-unique "
   6472             f"label: {repr(original_label)}"
   6473         )
   6475 if isinstance(slc, slice):
   6476     if side == "left":

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()
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
 df = pd.read_csv("https://runestone.academy/ns/books/published/httlads/_static/movies_metadata.csv").dropna(axis=1, how='all')
time_scheduler = df.set_index('runtime')
time_scheduler = time_scheduler[['title', 'release_date']]
title release_date
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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