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Id
13389808
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2012-11-15T00:30:57.783
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2016-01-28T00:12:42.207
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2017-05-23T11:59:59.107
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13389203
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2
Score
24
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text
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As <a href="https://stackoverflow.com/a/31490891/1240268">Robbie-Clarken answers</a>, since 0.14 you can pass a <a href="http://pandas.pydata.org/pandas-docs/stable/advanced.html#using-slicers" rel="noreferrer">slice in the tuple you pass to loc</a>: <pre><code>In [11]: s.loc[('b', slice(2, 10))] Out[11]: b 2 -0.65394 4 0.08227 dtype: float64 </code></pre> Indeed, you can pass a slice for each level: <pre><code>In [12]: s.loc[(slice('a', 'b'), slice(2, 10))] Out[12]: a 5 0.27919 b 2 -0.65394 4 0.08227 dtype: float64 </code></pre> Note: the slice is inclusive. <hr> <h3>Old answer:</h3> You can also do this using: <pre><code>s.ix[1:10, "b"] </code></pre> (It's good practice to do in a single ix/loc/iloc since this version allows assignment.) This answer was written prior to the <a href="https://github.com/pydata/pandas/pull/2922" rel="noreferrer">introduction of iloc</a> in early 2013, i.e. position/integer location - which may be preferred in this case. The reason it was created was to remove the ambiguity from integer-indexed pandas objects, and be more descriptive: "I'm slicing on position". <pre><code>s["b"].iloc[1:10] </code></pre> That said, I kinda disagree with the docs that ix is: <blockquote> most robust and consistent way </blockquote> it's not, the most consistent way is to describe what you're doing: <ul> <li>use loc for labels</li> <li>use iloc for position</li> <li>use ix for both (if you really have to)</li> </ul> Remember the <a href="https://www.python.org/dev/peps/pep-0020/" rel="noreferrer">zen of python</a>: <blockquote> explicit is better than implicit </blockquote>
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