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_episodes/04-more-dataframes.md

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@@ -278,7 +278,7 @@ surveys_df[0:3]
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>
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> _Remember: R begins indexing lists at 1._
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>
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> > ## Solution
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<!-- > > ## Solution
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> >
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> > With the same list:
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> > `['A','B','C','D','E','F','G','H','I','J']`,
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> > This is ignored when you use the syntax
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> > `x[0:5]`, and the elements in `1` through
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> > `5` are printed.
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> {: .solution}
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> {: .solution} -->
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{: .challenge}
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> Combine selecting a subset with column headings and slice syntax for rows. Get
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> every 5th row for rows 20-60, from the columns `plot_id`, `species_id`, and `sex`.
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>
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> > ## Solution
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<!-- > > ## Solution
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> >
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> > ~~~
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> > surveys_df[['plot_id', 'species_id','sex']][20:61:5]
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> > 60 23 DM M
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> > ~~~
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> > {: .output}
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> {: .solution}
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> {: .solution} -->
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{: .challenge}
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## Changing Values in a DataFrame
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>
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> 2. Use `.iloc[]` to view the same thing.
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>
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> > ## Solution
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<!-- > > ## Solution
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> >
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> > ~~~
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> > # 1
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> > surveys_df.iloc[1:6:2, [5,6]]
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> > ~~~
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> > {: .python}
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> {: .solution}
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> {: .solution} -->
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{: .challenge}
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> the year 1999 and that contain weight values less than or equal to 8. How
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> many rows did you end up with?
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>
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> > ## Solution
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<!-- > > ## Solution
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> >
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> > ~~~
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> > surveys_df[(surveys_df.weight <= 8.0) & (surveys_df.year == 1999)]
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> > 29905 4.0
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> > ~~~
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> > {: .output}
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> {: .solution}
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> {: .solution} -->
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{: .challenge}
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## Iterating Over a DataFrame

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